Rd Sharma 2021 Solutions for Class 10 Maths Chapter 4 Quadratic Equations are provided here with simple step-by-step explanations. These solutions for Quadratic Equations are extremely popular among Class 10 students for Maths Quadratic Equations Solutions come handy for quickly completing your homework and preparing for exams. All questions and answers from the Rd Sharma 2021 Book of Class 10 Maths Chapter 4 are provided here for you for free. You will also love the ad-free experience on Meritnationâ€™s Rd Sharma 2021 Solutions. All Rd Sharma 2021 Solutions for class Class 10 Maths are prepared by experts and are 100% accurate.

#### Page No 4.19:

#### Question 1:

Solve the following quadratic equations by factorization:

(*x* − 4) (*x *+ 2) = 0

#### Answer:

We have been given,

Therefore,

or

Therefore, or .

#### Page No 4.19:

#### Question 2:

Solve the following quadratic equations by factorization:

(2*x* + 3)(3*x* − 7) = 0

#### Answer:

We have been given,

Therefore,

or

Therefore, or .

#### Page No 4.19:

#### Question 3:

Solve the following quadratic equations by factorization:

3*x*^{2} − 14*x* − 5 = 0

#### Answer:

We have been given

Therefore,

or,

Hence, or .

#### Page No 4.19:

#### Question 4:

Solve the following quadratic equations by factorization:

9*x*^{2} − 3*x* − 2 = 0

#### Answer:

We have been given,

or,

Hence, or .

#### Page No 4.19:

#### Question 5:

Solve the following quadratic equations by factorization:

$\frac{1}{x-1}-\frac{1}{x+5}=\frac{6}{7},x\ne 1,-5$

#### Answer:

We have been given

Therefore,

or,

Hence, *x* = 2 or *x* = −6.

#### Page No 4.19:

#### Question 6:

Solve the following quadratic equations by factorization:

6*x*^{2} + 11*x* + 3 = 0

#### Answer:

We have been given

or,

Hence .

#### Page No 4.19:

#### Question 7:

Solve the following quadratic equations by factorization:

5*x*^{2} − 3*x* − 2 = 0

#### Answer:

We have been given

Therefore,

or,

Hence, or .

#### Page No 4.19:

#### Question 8:

Solve the following quadratic equations by factorization:

48*x*^{2} − 13*x* − 1 = 0

#### Answer:

We have been given

Therefore,

or,

Hence, or .

#### Page No 4.19:

#### Question 9:

Solve the following quadratic equations by factorization:

3*x*^{2} = −11*x* − 10

#### Answer:

We have been given

Therefore,or,

Hence, or .

#### Page No 4.19:

#### Question 10:

Solve the following quadratic equations by factorization:

25*x* (*x* + 1) = −4

#### Answer:

We have been given

Therefore,

or,

Hence, or .

#### Page No 4.19:

#### Question 11:

Solve the following quadratic equations by factorization:

$16x-\frac{10}{x}=27$

#### Answer:

$16x-\frac{10}{x}=27\phantom{\rule{0ex}{0ex}}\Rightarrow 16{x}^{2}-10=27x\phantom{\rule{0ex}{0ex}}\Rightarrow 16{x}^{2}-27x-10=0\phantom{\rule{0ex}{0ex}}\Rightarrow 16{x}^{2}-32x+5x-10=0\phantom{\rule{0ex}{0ex}}\Rightarrow 16x\left(x-2\right)+5\left(x-2\right)=0\phantom{\rule{0ex}{0ex}}\Rightarrow \left(16x+5\right)\left(x-2\right)=0\phantom{\rule{0ex}{0ex}}\Rightarrow 16x+5=0\mathrm{or}x-2=0\phantom{\rule{0ex}{0ex}}\Rightarrow x=-\frac{5}{16}\mathrm{or}x=2$

Hence, the factors are $2$ and $-\frac{5}{16}$.

#### Page No 4.19:

#### Question 12:

Solve the following quadratic equations by factorization:

$\frac{1}{x}-\frac{1}{x-2}=3,x\ne 0,2$

#### Answer:

We have been given

Therefore,

or,

Hence, or .

#### Page No 4.19:

#### Question 13:

Solve the following quadratic equations by factorization:

$x-\frac{1}{x}=3,x\ne 0$

#### Answer:

We have been given

Therefore,

or,

Hence, or .

#### Page No 4.19:

#### Question 14:

Solve the following quadratic equations by factorization:

$\frac{1}{x+4}-\frac{1}{x-7}=\frac{11}{30},x\ne 4,7$

#### Answer:

We have been given

Therefore,

or,

Hence, or .

#### Page No 4.19:

#### Question 15:

Solve the following quadratic equations by factorization:

$\frac{1}{x-3}+\frac{2}{x-2}=\frac{8}{x};x\ne 0,2,3$

#### Answer:

$\frac{1}{x-3}+\frac{2}{x-2}=\frac{8}{x}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{\left(x-2\right)+2\left(x-3\right)}{\left(x-3\right)\left(x-2\right)}=\frac{8}{x}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{x-2+2x-6}{{x}^{2}-2x-3x+6}=\frac{8}{x}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{3x-8}{{x}^{2}-5x+6}=\frac{8}{x}\phantom{\rule{0ex}{0ex}}\Rightarrow x\left(3x-8\right)=8\left({x}^{2}-5x+6\right)\phantom{\rule{0ex}{0ex}}\Rightarrow 3{x}^{2}-8x=8{x}^{2}-40x+48\phantom{\rule{0ex}{0ex}}\Rightarrow 5{x}^{2}-32x+48=0\phantom{\rule{0ex}{0ex}}\Rightarrow 5{x}^{2}-20x-12x+48=0\phantom{\rule{0ex}{0ex}}\Rightarrow 5x\left(x-4\right)-12\left(x-4\right)=0\phantom{\rule{0ex}{0ex}}\Rightarrow \left(5x-12\right)\left(x-4\right)=0\phantom{\rule{0ex}{0ex}}\Rightarrow 5x-12=0\mathrm{or}x-4=0\phantom{\rule{0ex}{0ex}}\Rightarrow x=\frac{12}{5}\mathrm{or}x=4$

Hence, the factors are 4 and $\frac{12}{5}$.

#### Page No 4.19:

#### Question 16:

Solve the following quadratic equations by factorization:

${a}^{2}{x}^{2}-3abx+2{b}^{2}=0$

#### Answer:

We have been given

Therefore,

or,

Hence, or .

#### Page No 4.19:

#### Question 17:

Solve the following quadratic equations by factorization:

$9{x}^{2}-6{b}^{2}x-\left({a}^{4}-{b}^{4}\right)=0$

#### Answer:

$9{x}^{2}-6{b}^{2}x-\left({a}^{4}-{b}^{4}\right)=0\phantom{\rule{0ex}{0ex}}\Rightarrow 9{x}^{2}-6{b}^{2}x-\left({a}^{2}-{b}^{2}\right)\left({a}^{2}+{b}^{2}\right)=0\phantom{\rule{0ex}{0ex}}\Rightarrow 9{x}^{2}+3({a}^{2}-{b}^{2})x-3\left({a}^{2}+{b}^{2}\right)x-\left({a}^{2}-{b}^{2}\right)\left({a}^{2}+{b}^{2}\right)=0\phantom{\rule{0ex}{0ex}}\Rightarrow 3x\left[3x+\left({a}^{2}-{b}^{2}\right)\right]-\left({a}^{2}+{b}^{2}\right)\left[3x+\left({a}^{2}-{b}^{2}\right)\right]=0\phantom{\rule{0ex}{0ex}}\Rightarrow \left[3x-\left({a}^{2}+{b}^{2}\right)\right]\left[3x+\left({a}^{2}-{b}^{2}\right)\right]=0\phantom{\rule{0ex}{0ex}}\Rightarrow 3x-\left({a}^{2}+{b}^{2}\right)=0\mathrm{or}3x+\left({a}^{2}-{b}^{2}\right)=0\phantom{\rule{0ex}{0ex}}\Rightarrow x=\frac{{a}^{2}+{b}^{2}}{3}\mathrm{or}x=-\frac{{a}^{2}-{b}^{2}}{3}\phantom{\rule{0ex}{0ex}}\Rightarrow x=\frac{{a}^{2}+{b}^{2}}{3}\mathrm{or}x=\frac{{b}^{2}-{a}^{2}}{3}$

Hence, the factors are $\frac{{a}^{2}+{b}^{2}}{3}$ and $\frac{{b}^{2}-{a}^{2}}{3}$.

#### Page No 4.19:

#### Question 18:

Solve the following quadratic equations by factorization:

$4{x}^{2}+4bx-\left({a}^{2}-{b}^{2}\right)=0$

#### Answer:

We have been given

Therefore,

or,

Hence, or.

#### Page No 4.19:

#### Question 19:

Solve the following quadratic equations by factorization:

$a{x}^{2}+\left(4{a}^{2}-3b\right)x-12ab=0$

#### Answer:

We have been given

Therefore,

or

Hence, or .

#### Page No 4.19:

#### Question 20:

Solve the following quadratic equations by factorization:

$2{x}^{2}+ax-{a}^{2}=0$

#### Answer:

$2{x}^{2}+ax-{a}^{2}=0\phantom{\rule{0ex}{0ex}}\Rightarrow 2{x}^{2}+2ax-ax-{a}^{2}=0\phantom{\rule{0ex}{0ex}}\Rightarrow 2x\left(x+a\right)-a\left(x+a\right)=0\phantom{\rule{0ex}{0ex}}\Rightarrow \left(2x-a\right)\left(x+a\right)=0\phantom{\rule{0ex}{0ex}}\Rightarrow 2x-a=0\mathrm{or}x+a=0\phantom{\rule{0ex}{0ex}}\Rightarrow x=\frac{a}{2}\mathrm{or}x=-a$

Hence, the factors are $\frac{a}{2}$ and $-a$.

#### Page No 4.19:

#### Question 21:

Solve the following quadratic equations by factorization:

$\frac{16}{x}-1=\frac{15}{x+1};x\ne 0,-1$

#### Answer:

$\frac{16}{x}-1=\frac{15}{x+1}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{16-x}{x}=\frac{15}{x+1}\phantom{\rule{0ex}{0ex}}\Rightarrow \left(16-x\right)\left(x+1\right)=15x\phantom{\rule{0ex}{0ex}}\Rightarrow 16x+16-{x}^{2}-x=15x\phantom{\rule{0ex}{0ex}}\Rightarrow -{x}^{2}+16+15x=15x\phantom{\rule{0ex}{0ex}}\Rightarrow -{x}^{2}+16=0\phantom{\rule{0ex}{0ex}}\Rightarrow {x}^{2}-16=0\phantom{\rule{0ex}{0ex}}\Rightarrow \left(x-4\right)\left(x+4\right)=0\phantom{\rule{0ex}{0ex}}\Rightarrow x-4=0\mathrm{or}x+4=0\phantom{\rule{0ex}{0ex}}\Rightarrow x=4\mathrm{or}x=-4$

Hence, the factors are 4 and −4.

#### Page No 4.20:

#### Question 22:

Solve the following quadratic equations by factorization:

$\frac{x+3}{x+2}=\frac{3x-7}{2x-3}$

#### Answer:

We have been given

Therefore,

or,

Hence, or .

#### Page No 4.20:

#### Question 23:

Solve the following quadratic equations by factorization:

$\frac{2x}{x-4}+\frac{2x-5}{x-3}=\frac{25}{3}$

#### Answer:

We have been given

Therefore,or,

Hence, or .

#### Page No 4.20:

#### Question 24:

Solve the following quadratic equations by factorization:

$\frac{x+3}{x-2}-\frac{1-x}{x}=\frac{17}{4}$

#### Answer:

We have been given

Therefore,

or,

Hence, or .

#### Page No 4.20:

#### Question 25:

Solve the following quadratic equations by factorization:

$\frac{x-3}{x+3}-\frac{x+3}{x-3}=\frac{48}{7},x\ne 3,x\ne -3$

#### Answer:

We have been given

Therefore,

Therefore,

or,

Hence, or .

#### Page No 4.20:

#### Question 26:

Solve the following quadratic equations by factorization:

$\frac{1}{x-2}+\frac{2}{x-1}=\frac{6}{x},x\ne 0$

#### Answer:

We have been given

Therefore,

Therefore,

or,

Hence, or .

#### Page No 4.20:

#### Question 27:

Solve the following quadratic equations by factorization:

$\frac{x+1}{x-1}-\frac{x-1}{x+1}=\frac{5}{6},x\ne 1,-1$

#### Answer:

We have been given

Therefore,

or,

Hence, or .

#### Page No 4.20:

#### Question 28:

Solve the following quadratic equations by factorization:

$\frac{x-1}{2x+1}+\frac{2x+1}{x-1}=\frac{5}{2},x\ne -\frac{1}{2},1$

#### Answer:

We have been given

Therefore,

Hence, .

#### Page No 4.20:

#### Question 29:

Solve the following quadratic equations by factorization:

$\frac{4}{x}-3=\frac{5}{2x+3},x\ne 0,-\frac{3}{2}$

#### Answer:

$\frac{4}{x}-3=\frac{5}{2x+3}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{4-3x}{x}=\frac{5}{2x+3}\phantom{\rule{0ex}{0ex}}\Rightarrow \left(4-3x\right)\left(2x+3\right)=5x\phantom{\rule{0ex}{0ex}}\Rightarrow 8x+12-6{x}^{2}-9x=5x\phantom{\rule{0ex}{0ex}}\Rightarrow -6{x}^{2}-6x+12=0\phantom{\rule{0ex}{0ex}}\Rightarrow {x}^{2}+x-2=0\phantom{\rule{0ex}{0ex}}\Rightarrow {x}^{2}+2x-x-2=0\phantom{\rule{0ex}{0ex}}\Rightarrow x(x+2)-1(x+2)=0\phantom{\rule{0ex}{0ex}}\Rightarrow (x-1)(x+2)=0\phantom{\rule{0ex}{0ex}}\Rightarrow x-1=0\mathrm{or}x+2=0\phantom{\rule{0ex}{0ex}}\Rightarrow x=1\mathrm{or}x=-2$

Hence, the factors are 1 and −2.

#### Page No 4.20:

#### Question 30:

Solve the following quadratic equations by factorization:

$\frac{x-4}{x-5}+\frac{x-6}{x-7}=\frac{10}{3};x\ne 5,7$

#### Answer:

$\frac{x-4}{x-5}+\frac{x-6}{x-7}=\frac{10}{3}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{\left(x-4\right)\left(x-7\right)+\left(x-6\right)\left(x-5\right)}{\left(x-5\right)\left(x-7\right)}=\frac{10}{3}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{{x}^{2}-11x+28+{x}^{2}-11x+30}{{x}^{2}-12x+35}=\frac{10}{3}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{2{x}^{2}-22x+58}{{x}^{2}-12x+35}=\frac{10}{3}\phantom{\rule{0ex}{0ex}}\Rightarrow 3\left(2{x}^{2}-22x+58\right)=10\left({x}^{2}-12x+35\right)\phantom{\rule{0ex}{0ex}}\Rightarrow 6{x}^{2}-66x+174=10{x}^{2}-120x+350\phantom{\rule{0ex}{0ex}}\Rightarrow 4{x}^{2}-54x+176=0\phantom{\rule{0ex}{0ex}}\Rightarrow 2{x}^{2}-27x+88=0\phantom{\rule{0ex}{0ex}}\Rightarrow 2{x}^{2}-11x-16x+88=0\phantom{\rule{0ex}{0ex}}\Rightarrow x\left(2x-11\right)-8\left(2x-11\right)=0\phantom{\rule{0ex}{0ex}}\Rightarrow \left(x-8\right)\left(2x-11\right)=0\phantom{\rule{0ex}{0ex}}\Rightarrow x-8=0\mathrm{or}2x-11=0\phantom{\rule{0ex}{0ex}}\Rightarrow x=8\mathrm{or}x=\frac{11}{2}$

Hence, the factors are 8 and $\frac{11}{2}$.

#### Page No 4.20:

#### Question 31:

Solve the following quadratic equations by factorization:

$\frac{x-2}{x-3}+\frac{x-4}{x-5}=\frac{10}{3};x\ne 3,5$

#### Answer:

$\frac{x-2}{x-3}+\frac{x-4}{x-5}=\frac{10}{3}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{x-2}{x-3}-\frac{10}{3}=-\frac{x-4}{x-5}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{3\left(x-2\right)-10\left(x-3\right)}{3\left(x-3\right)}=-\frac{x-4}{x-5}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{3x-6-10x+30}{3x-9}=-\frac{x-4}{x-5}\phantom{\rule{0ex}{0ex}}\Rightarrow -\frac{7x-24}{3x-9}=-\frac{x-4}{x-5}\phantom{\rule{0ex}{0ex}}\Rightarrow \left(7x-24\right)\left(x-5\right)=\left(3x-9\right)\left(x-4\right)\phantom{\rule{0ex}{0ex}}\Rightarrow 7{x}^{2}-59x+120=3{x}^{2}-21x+36\phantom{\rule{0ex}{0ex}}\Rightarrow 4{x}^{2}-38x+84=0\phantom{\rule{0ex}{0ex}}\Rightarrow 2{x}^{2}-19x+42=0\phantom{\rule{0ex}{0ex}}\Rightarrow 2{x}^{2}-12x-7x+42=0\phantom{\rule{0ex}{0ex}}\Rightarrow 2x(x-6)-7(x-6)=0\phantom{\rule{0ex}{0ex}}\Rightarrow (2x-7)(x-6)=0\phantom{\rule{0ex}{0ex}}\Rightarrow 2x-7=0\mathrm{or}x-6=0\phantom{\rule{0ex}{0ex}}\Rightarrow x=\frac{7}{2}\mathrm{or}x=6$

Hence, the factors are 6 and $\frac{7}{2}$.

#### Page No 4.20:

#### Question 32:

Solve the following quadratic equations by factorization:

$\frac{5+x}{5-x}-\frac{5-x}{5+x}=3\frac{3}{4};x\ne 5,-5$

#### Answer:

$\frac{5+x}{5-x}-\frac{5-x}{5+x}=3\frac{3}{4}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{{\left(5+x\right)}^{2}-{\left(5-x\right)}^{2}}{\left(5+x\right)\left(5-x\right)}=\frac{15}{4}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{25+{x}^{2}+10x-25-{x}^{2}+10x}{25-{x}^{2}}=\frac{15}{4}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{20x}{25-{x}^{2}}=\frac{15}{4}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{4x}{25-{x}^{2}}=\frac{3}{4}\phantom{\rule{0ex}{0ex}}\Rightarrow 16x=75-3{x}^{2}\phantom{\rule{0ex}{0ex}}\Rightarrow 3{x}^{2}+16x-75=0\phantom{\rule{0ex}{0ex}}\Rightarrow 3{x}^{2}+25x-9x-75=0\phantom{\rule{0ex}{0ex}}\Rightarrow x(3x+25)-3(3x+25)=0\phantom{\rule{0ex}{0ex}}\Rightarrow (x-3)(3x+25)=0\phantom{\rule{0ex}{0ex}}\Rightarrow x-3=0\mathrm{or}3x+25=0\phantom{\rule{0ex}{0ex}}\Rightarrow x=3\mathrm{or}x=-\frac{25}{3}$

Hence, the factors are 3 and $-\frac{25}{3}$.

#### Page No 4.20:

#### Question 33:

Solve the following quadratic equations by factorization:

$\frac{3}{x+1}-\frac{1}{2}=\frac{2}{3x-1},x\ne -1,\frac{1}{3}$

#### Answer:

$\frac{3}{x+1}-\frac{1}{2}=\frac{2}{3x-1}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{6-\left(x+1\right)}{2(x+1)}=\frac{2}{3x-1}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{6-x-1}{2x+2}=\frac{2}{3x-1}\phantom{\rule{0ex}{0ex}}\Rightarrow \left(5-x\right)\left(3x-1\right)=2\left(2x+2\right)\phantom{\rule{0ex}{0ex}}\Rightarrow 15x-5-3{x}^{2}+x=4x+4\phantom{\rule{0ex}{0ex}}\Rightarrow -3{x}^{2}+16x-5-4x-4=0\phantom{\rule{0ex}{0ex}}\Rightarrow -3{x}^{2}+12x-9=0\phantom{\rule{0ex}{0ex}}\Rightarrow 3{x}^{2}-12x+9=0\phantom{\rule{0ex}{0ex}}\Rightarrow {x}^{2}-4x+3=0\phantom{\rule{0ex}{0ex}}\Rightarrow {x}^{2}-3x-x+3=0\phantom{\rule{0ex}{0ex}}\Rightarrow x(x-3)-1(x-3)=0\phantom{\rule{0ex}{0ex}}\Rightarrow (x-1)(x-3)=0\phantom{\rule{0ex}{0ex}}\Rightarrow x-1=0\mathrm{or}x-3=0\phantom{\rule{0ex}{0ex}}\Rightarrow x=1\mathrm{or}x=3$

Hence, the factors are 3 and 1.

#### Page No 4.20:

#### Question 34:

Solve the following quadratic equations by factorization:

$\frac{3}{x+1}+\frac{4}{x-1}=\frac{29}{4x-1};x\ne 1,-1,\frac{1}{4}$

#### Answer:

$\frac{3}{x+1}+\frac{4}{x-1}=\frac{29}{4x-1}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{3\left(x-1\right)+4\left(x+1\right)}{\left(x+1\right)\left(x-1\right)}=\frac{29}{4x-1}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{3x-3+4x+4}{{x}^{2}-1}=\frac{29}{4x-1}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{7x+1}{{x}^{2}-1}=\frac{29}{4x-1}\phantom{\rule{0ex}{0ex}}\Rightarrow \left(7x+1\right)\left(4x-1\right)=29\left({x}^{2}-1\right)\phantom{\rule{0ex}{0ex}}\Rightarrow 28{x}^{2}-7x+4x-1=29{x}^{2}-29\phantom{\rule{0ex}{0ex}}\Rightarrow 29{x}^{2}-28{x}^{2}+3x-28=0\phantom{\rule{0ex}{0ex}}\Rightarrow {x}^{2}+3x-28=0\phantom{\rule{0ex}{0ex}}\Rightarrow {x}^{2}+7x-4x-28=0\phantom{\rule{0ex}{0ex}}\Rightarrow x(x+7)-4(x+7)=0\phantom{\rule{0ex}{0ex}}\Rightarrow (x-4)(x+7)=0\phantom{\rule{0ex}{0ex}}\Rightarrow x-4=0\mathrm{or}x+7=0\phantom{\rule{0ex}{0ex}}\Rightarrow x=4\mathrm{or}x=-7$

Hence, the factors are 4 and −7.

#### Page No 4.20:

#### Question 35:

Solve the following quadratic equations by factorization:

$\frac{2}{x+1}+\frac{3}{2(x-2)}=\frac{23}{5x};x\ne 0,-1,2$

#### Answer:

$\frac{2}{x+1}+\frac{3}{2(x-2)}=\frac{23}{5x}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{4(x-2)+3(x+1)}{2(x-2)(x+1)}=\frac{23}{5x}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{4x-8+3x+3}{2({x}^{2}+x-2x-2)}=\frac{23}{5x}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{7x-5}{2{x}^{2}-2x-4}=\frac{23}{5x}\phantom{\rule{0ex}{0ex}}\Rightarrow 5x\left(7x-5\right)=23\left(2{x}^{2}-2x-4\right)\phantom{\rule{0ex}{0ex}}\Rightarrow 35{x}^{2}-25x=46{x}^{2}-46x-92\phantom{\rule{0ex}{0ex}}\Rightarrow 46{x}^{2}-35{x}^{2}-46x+25x-92=0\phantom{\rule{0ex}{0ex}}\Rightarrow 11{x}^{2}-21x-92=0\phantom{\rule{0ex}{0ex}}\Rightarrow 11{x}^{2}-44x+23x-92=0\phantom{\rule{0ex}{0ex}}\Rightarrow 11x(x-4)+23(x-4)=0\phantom{\rule{0ex}{0ex}}\Rightarrow (11x+23)(x-4)=0\phantom{\rule{0ex}{0ex}}\Rightarrow 11x+23=0\mathrm{or}x-4=0\phantom{\rule{0ex}{0ex}}\Rightarrow x=-\frac{23}{11}\mathrm{or}x=4$

Hence, the factors are 4 and $-\frac{23}{11}$.

#### Page No 4.20:

#### Question 36:

Solve the following quadratic equations by factorisation:

${x}^{2}-\left(\sqrt{3}+1\right)x+\sqrt{3}=0$

#### Answer:

Consider the equation ${x}^{2}-\left(\sqrt{3}+1\right)x+\sqrt{3}=0$

$\Rightarrow {x}^{2}-(\sqrt{3}+1)x+\sqrt{3}=0\phantom{\rule{0ex}{0ex}}\Rightarrow {x}^{2}-\sqrt{3}x-x+\sqrt{3}=0\phantom{\rule{0ex}{0ex}}\Rightarrow x\left(x-\sqrt{3}\right)-1\left(x-\sqrt{3}\right)=0\phantom{\rule{0ex}{0ex}}\Rightarrow \left(x-\sqrt{3}\right)\left(x-1\right)=0\phantom{\rule{0ex}{0ex}}\Rightarrow \left(x-\sqrt{3}\right)=0\mathrm{or}\left(x-1\right)=0\phantom{\rule{0ex}{0ex}}\Rightarrow x=\sqrt{3}\mathrm{or}x=1$

#### Page No 4.20:

#### Question 37:

Solve the following quadratic equations by factorization:

$3\sqrt{5}{x}^{2}+25x-10\sqrt{5}=0$

#### Answer:

Consider the equation $3\sqrt{5}{x}^{2}+25x-10\sqrt{5}=0$

$\Rightarrow 3\sqrt{5}{x}^{2}+30x-5x-10\sqrt{5}=0\phantom{\rule{0ex}{0ex}}\Rightarrow 3\sqrt{5}x\left(x+2\sqrt{5}\right)-5\left(x+2\sqrt{5}\right)=0\phantom{\rule{0ex}{0ex}}\Rightarrow \left(3\sqrt{5}x-5\right)\left(x+2\sqrt{5}\right)=0\phantom{\rule{0ex}{0ex}}\Rightarrow \left(3\sqrt{5}x-5\right)=0\mathrm{or}\left(x+2\sqrt{5}\right)=0\phantom{\rule{0ex}{0ex}}\Rightarrow x=\frac{\sqrt{5}}{3}\mathrm{or}x=-2\sqrt{5}$

#### Page No 4.20:

#### Question 38:

Solve the following quadratic equations by factorization:

$\sqrt{3}{x}^{2}-2\sqrt{2}x-2\sqrt{3}=0$

#### Answer:

$\sqrt{3}{x}^{2}-2\sqrt{2}x-2\sqrt{3}=0\phantom{\rule{0ex}{0ex}}\Rightarrow \sqrt{3}{x}^{2}-3\sqrt{2}x+\sqrt{2}x-2\sqrt{3}=0\phantom{\rule{0ex}{0ex}}\Rightarrow \sqrt{3}x\left(x-\sqrt{6}\right)+\sqrt{2}\left(x-\sqrt{6}\right)=0\phantom{\rule{0ex}{0ex}}\Rightarrow \left(\sqrt{3}x+\sqrt{2}\right)\left(x-\sqrt{6}\right)=0\phantom{\rule{0ex}{0ex}}\Rightarrow \sqrt{3}x+\sqrt{2}=0\mathrm{or}x-\sqrt{6}=0\phantom{\rule{0ex}{0ex}}\Rightarrow x=-\sqrt{\frac{2}{3}}\mathrm{or}x=\sqrt{6}$

Hence, the factors are $\sqrt{6}$ and $-\sqrt{\frac{2}{3}}$.

#### Page No 4.20:

#### Question 39:

Solve the following quadratic equations by factorization:

$4\sqrt{3}{x}^{2}+5x-2\sqrt{3}=0$

#### Answer:

We have been given

Therefore,

or,

Hence, or.

#### Page No 4.20:

#### Question 40:

Solve the following quadratic equations by factorization:

$\sqrt{2}{x}^{2}-3x-2\sqrt{2}=0$

#### Answer:

We have been given

Therefore,

or,

Hence, or .

#### Page No 4.20:

#### Question 41:

Solve the following quadratic equations by factorisation:

${x}^{2}-\left(\sqrt{2}+1\right)x+\sqrt{2}=0$

#### Answer:

Consider the equation ${x}^{2}-\left(\sqrt{2}+1\right)x+\sqrt{2}=0$

$\Rightarrow {x}^{2}-\sqrt{2}x-x+\sqrt{2}=0\phantom{\rule{0ex}{0ex}}\Rightarrow x\left(x-\sqrt{2}\right)-1\left(x-\sqrt{2}\right)=0\phantom{\rule{0ex}{0ex}}\Rightarrow \left(x-1\right)\left(x-\sqrt{2}\right)=0\phantom{\rule{0ex}{0ex}}\Rightarrow x-1=0\mathrm{or}x-\sqrt{2}=0\phantom{\rule{0ex}{0ex}}\Rightarrow x=1\mathrm{or}x=\sqrt{2}$

#### Page No 4.20:

#### Question 42:

Solve the following quadratic equations by factorization:

$3{x}^{2}-2\sqrt{6}x+2=0$

#### Answer:

We have been given

Therefore,

or,

Hence, or .

#### Page No 4.20:

#### Question 43:

Find the roots of the quadratic equation $\sqrt{2}{x}^{2}+7x+5\sqrt{2}=0$.

#### Answer:

We write, $7x=5x+2x\mathrm{as}$ $\sqrt{2}{x}^{2}\times 5\sqrt{2}=10{x}^{2}=5x\times 2x$

$\therefore \sqrt{2}{x}^{2}+7x+5\sqrt{2}=0\phantom{\rule{0ex}{0ex}}\Rightarrow \sqrt{2}{x}^{2}+5x+2x+5\sqrt{2}=0\phantom{\rule{0ex}{0ex}}\Rightarrow x\left(\sqrt{2}x+5\right)+\sqrt{2}\left(\sqrt{2}x+5\right)=0\phantom{\rule{0ex}{0ex}}\Rightarrow \left(\sqrt{2}x+5\right)\left(x+\sqrt{2}\right)=0$

$\Rightarrow x+\sqrt{2}=0\mathrm{or}\sqrt{2}x+5=0\phantom{\rule{0ex}{0ex}}\Rightarrow x=-\sqrt{2}\mathrm{or}x=-\frac{5}{\sqrt{2}}=-\frac{5\sqrt{2}}{2}$

Hence, the roots of the given equation are $-\sqrt{2}\mathrm{and}$ $-\frac{5\sqrt{2}}{2}$.

#### Page No 4.20:

#### Question 44:

Solve the following quadratic equations by factorization:

$\frac{m}{n}{x}^{2}+\frac{n}{m}=1-2x$

#### Answer:

We have been given

${m}^{2}{x}^{2}+mnx+mnx+\left[{n}^{2}-{\left(\sqrt{mn}\right)}^{2}\right]=0\phantom{\rule{0ex}{0ex}}{m}^{2}{x}^{2}+mnx+mnx+\left(n+\sqrt{mn}\right)\left(n-\sqrt{nm}\right)+\left(m\sqrt{mn}x-m\sqrt{mn}x\right)=0$

Therefore,

or,

Hence, or.

#### Page No 4.20:

#### Question 45:

Solve the following quadratic equations by factorization:

$\frac{x-a}{x-b}+\frac{x-b}{x-a}=\frac{a}{b}+\frac{b}{a}$

#### Answer:

We have been given

Therefore,

or,

Hence, or .

#### Page No 4.21:

#### Question 46:

Solve the following quadratic equations by factorization:

$\frac{1}{\left(x-1\right)\left(x-2\right)}+\frac{1}{\left(x-2\right)\left(x-3\right)}+\frac{1}{\left(x-3\right)\left(x-4\right)}=\frac{1}{6}$

#### Answer:

We have been given,

Therefore,

or,

Hence, or .

#### Page No 4.21:

#### Question 47:

Solve the following quadratic equation by factorization:

$\frac{a}{x-b}+\frac{b}{x-a}=2$

#### Answer:

$\frac{a}{x-b}+\frac{b}{x-a}=2$

$\Rightarrow \frac{ax-{a}^{2}+bx-{b}^{2}}{\left(x-a\right)\left(x-b\right)}=2\phantom{\rule{0ex}{0ex}}\Rightarrow ax-{a}^{2}+bx-{b}^{2}=2{x}^{2}-2bx-2ax+2ab\phantom{\rule{0ex}{0ex}}\Rightarrow 2{x}^{2}-2bx-2ax+2ab-ax+{a}^{2}-bx+{b}^{2}=0\phantom{\rule{0ex}{0ex}}\Rightarrow 2{x}^{2}+x\left[-2b-2a-a-b\right]+{a}^{2}+{b}^{2}+2ab=0\phantom{\rule{0ex}{0ex}}\Rightarrow 2{x}^{2}-3x\left[a+b\right]+{\left(a+b\right)}^{2}=0$

$\Rightarrow 2{x}^{2}-2\left(a+b\right)x-\left(a+b\right)x+{\left(a+b\right)}^{2}=0\phantom{\rule{0ex}{0ex}}\Rightarrow 2x\left[x-\left(a+b\right)\right]-\left(a+b\right)\left[x-\left(a+b\right)\right]=0\phantom{\rule{0ex}{0ex}}\Rightarrow \left[2x-\left(a+b\right)\right]\left[x-\left(a+b\right)\right]=0$

So, the value of *x* will be

$x=\frac{a+b}{2},a+b$

#### Page No 4.21:

#### Question 48:

Solve the following quadratic equations by factorization:

$\frac{x+1}{x-1}+\frac{x-2}{x+2}=4-\frac{2x+3}{x-2};x\ne 1,-2,2\phantom{\rule{0ex}{0ex}}$

#### Answer:

$\frac{x+1}{x-1}+\frac{x-2}{x+2}=4-\frac{2x+3}{x-2}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{(x+1)(x+2)+(x-1)(x-2)}{(x-1)(x+2)}=\frac{4(x-2)-(2x+3)}{x-2}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{({x}^{2}+2x+x+2)+({x}^{2}-2x-x+2)}{{x}^{2}+2x-x-2}=\frac{4x-8-2x-3}{x-2}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{{x}^{2}+3x+2+{x}^{2}-3x+2}{{x}^{2}+x-2}=\frac{2x-11}{x-2}$

$\Rightarrow \frac{2{x}^{2}+4}{{x}^{2}+x-2}=\frac{2x-11}{x-2}\phantom{\rule{0ex}{0ex}}\Rightarrow (2{x}^{2}+4)(x-2)=(2x-11)({x}^{2}+x-2)\phantom{\rule{0ex}{0ex}}\Rightarrow 2{x}^{3}-4{x}^{2}+4x-8=2{x}^{3}+2{x}^{2}-4x-11{x}^{2}-11x+22\phantom{\rule{0ex}{0ex}}\Rightarrow 2{x}^{3}-4{x}^{2}+4x-8=2{x}^{3}-9{x}^{2}-15x+22\phantom{\rule{0ex}{0ex}}\Rightarrow 2{x}^{3}-2{x}^{3}-4{x}^{2}+9{x}^{2}+4x+15x-8-22=0$

$\Rightarrow 5{x}^{2}+19x-30=0\phantom{\rule{0ex}{0ex}}\Rightarrow 5{x}^{2}+25x-6x-30=0\phantom{\rule{0ex}{0ex}}\Rightarrow 5x(x+5)-6(x+5)=0\phantom{\rule{0ex}{0ex}}\Rightarrow (5x-6)(x+5)=0\phantom{\rule{0ex}{0ex}}\Rightarrow 5x-6=0,x+5=0\phantom{\rule{0ex}{0ex}}\Rightarrow x=\frac{6}{5},x=-5$

#### Page No 4.21:

#### Question 49:

Solve the following quadratic equations by factorization:

$\frac{a}{x-a}+\frac{b}{x-b}=\frac{2c}{x-c}$

#### Answer:

We have been given,

Therefore,

or,

Hence, or .

#### Page No 4.21:

#### Question 50:

Solve the following quadratic equations by factorization:

*x*^{2} + 2*ab* = (2*a* + *b*)*x*

#### Answer:

We have been given

Therefore,

or,

Hence, or .

#### Page No 4.21:

#### Question 51:

Solve the following quadratic equations by factorization:

${\left(a+b\right)}^{2}{x}^{2}-4abx-{\left(a-b\right)}^{2}=0$

#### Answer:

We have been given

Therefore,

or,

Hence, or .

#### Page No 4.21:

#### Question 52:

Solve the following quadratic equations by factorization:

$a\left({x}^{2}+1\right)-x\left({a}^{2}+1\right)=0$

#### Answer:

We have been given

Therefore,

or,

Hence, or .

#### Page No 4.21:

#### Question 53:

Solve the following quadratic equations by factorization:

*x*^{2} − *x* − *a*(*a* + 1) = 0

#### Answer:

We have been given

Therefore,

or,

Hence, or.

#### Page No 4.21:

#### Question 54:

Solve the following quadratic equations by factorization:

${x}^{2}+\left(a+\frac{1}{a}\right)x+1=0$

#### Answer:

We have been given

Therefore,Therefore,

or,

Hence, or .

#### Page No 4.21:

#### Question 55:

Solve the following quadratic equations by factorization:

$ab{x}^{2}+\left({b}^{2}-ac\right)x-bc=0$

#### Answer:

We have been given

Therefore,or,

Hence, or .

#### Page No 4.21:

#### Question 56:

Solve the following quadratic equations by factorization:

${a}^{2}{b}^{2}{x}^{2}+{b}^{2}x-{a}^{2}x-1=0$

#### Answer:

We have been given

Therefore,

or,

Hence, or .

#### Page No 4.21:

#### Question 57:

Solve the following quadratic equations by factorization:

$\frac{x-1}{x-2}+\frac{x-3}{x-4}=3\frac{1}{3},x\ne 2,4$

#### Answer:

We have been given

Therefore,

or,

Hence, or .

#### Page No 4.21:

#### Question 58:

Solve the following quadratic equations by factorization:

$\frac{1}{2a+b+2x}=\frac{1}{2a}+\frac{1}{b}+\frac{1}{2x}$

#### Answer:

$\frac{1}{2a+b+2x}=\frac{1}{2a}+\frac{1}{b}+\frac{1}{2x}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{1}{2a+b+2x}-\frac{1}{2a}=\frac{1}{b}+\frac{1}{2x}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{2a-\left(2a+b+2x\right)}{\left(2a+b+2x\right)\left(2a\right)}=\frac{2x+b}{2bx}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{-b-2x}{4{a}^{2}+2ab+4ax}=\frac{2x+b}{2bx}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{-1\left(2x+b\right)}{4{a}^{2}+2ab+4ax}=\frac{2x+b}{2bx}\phantom{\rule{0ex}{0ex}}\Rightarrow -2bx\left(2x+b\right)=\left(4{a}^{2}+2ab+4ax\right)\left(2x+b\right)\phantom{\rule{0ex}{0ex}}\Rightarrow \left(4{a}^{2}+2ab+4ax\right)\left(2x+b\right)+2bx\left(2x+b\right)=0\phantom{\rule{0ex}{0ex}}\Rightarrow \left(2x+b\right)\left(4{a}^{2}+2ab+4ax+2bx\right)=0\phantom{\rule{0ex}{0ex}}\Rightarrow 2x+b=0\mathrm{or}4{a}^{2}+2ab+\left(4a+2b\right)x=0\phantom{\rule{0ex}{0ex}}\Rightarrow x=-\frac{b}{2}\mathrm{or}x=-\frac{4{a}^{2}+2ab}{4a+2b}\phantom{\rule{0ex}{0ex}}\Rightarrow x=-\frac{b}{2}\mathrm{or}x=-\frac{a\left(4a+2b\right)}{4a+2b}\phantom{\rule{0ex}{0ex}}\Rightarrow x=-\frac{b}{2}\mathrm{or}x=-a$

Hence, the factors are $-a$ and $-\frac{b}{2}$.

#### Page No 4.21:

#### Question 59:

Solve the following quadratic equations by factorization:

$3\left(\frac{3x-1}{2x+3}\right)-2\left(\frac{2x+3}{3x-1}\right)=5;x\ne \frac{1}{3},-\frac{3}{2}$

#### Answer:

$3\left(\frac{3x-1}{2x+3}\right)-2\left(\frac{2x+3}{3x-1}\right)=5\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{3(3x-1{)}^{2}-2{\left(2x+3\right)}^{2}}{\left(2x+3\right)\left(3x-1\right)}=5\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{3\left(9{x}^{2}+1-6x\right)-2\left(4{x}^{2}+9+12x\right)}{6{x}^{2}-2x+9x-3}=5\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{27{x}^{2}+3-18x-8{x}^{2}-18-24x}{6{x}^{2}+7x-3}=5\phantom{\rule{0ex}{0ex}}\Rightarrow 19{x}^{2}-42x-15=5\left(6{x}^{2}+7x-3\right)\phantom{\rule{0ex}{0ex}}\Rightarrow 19{x}^{2}-42x-15=30{x}^{2}+35x-15\phantom{\rule{0ex}{0ex}}\Rightarrow 30{x}^{2}-19{x}^{2}+35x+42x-15+15=0\phantom{\rule{0ex}{0ex}}\Rightarrow 11{x}^{2}+77x=0\phantom{\rule{0ex}{0ex}}\Rightarrow {x}^{2}+7x=0\phantom{\rule{0ex}{0ex}}\Rightarrow x\left(x+7\right)=0\phantom{\rule{0ex}{0ex}}\Rightarrow x=0\mathrm{or}x+7=0\phantom{\rule{0ex}{0ex}}\Rightarrow x=0\mathrm{or}x=-7$

Hence, the factors are 0 and −7.

#### Page No 4.21:

#### Question 60:

Solve the following quadratic equations by factorization:

$3\left(\frac{7x+1}{5x-3}\right)-4\left(\frac{5x-3}{7x+1}\right)=11;x\ne \frac{3}{5},-\frac{1}{7}$

#### Answer:

$3\left(\frac{7x+1}{5x-3}\right)-4\left(\frac{5x-3}{7x+1}\right)=11\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{3(7x+1{)}^{2}-4{\left(5x-3\right)}^{2}}{\left(5x-3\right)\left(7x+1\right)}=11\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{3\left(49{x}^{2}+1+14x\right)-4\left(25{x}^{2}+9-30x\right)}{35{x}^{2}+5x-21x-3}=11\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{147{x}^{2}+3+42x-100{x}^{2}-36+120x}{35{x}^{2}-16x-3}=11\phantom{\rule{0ex}{0ex}}\Rightarrow 47{x}^{2}+162x-33=11\left(35{x}^{2}-16x-3\right)\phantom{\rule{0ex}{0ex}}\Rightarrow 47{x}^{2}+162x-33=385{x}^{2}-176x-33\phantom{\rule{0ex}{0ex}}\Rightarrow 385{x}^{2}-47{x}^{2}-176x-162x-33+33=0\phantom{\rule{0ex}{0ex}}\Rightarrow 338{x}^{2}-338x=0\phantom{\rule{0ex}{0ex}}\Rightarrow {x}^{2}-x=0\phantom{\rule{0ex}{0ex}}\Rightarrow x\left(x-1\right)=0\phantom{\rule{0ex}{0ex}}\Rightarrow x=0\mathrm{or}x-1=0\phantom{\rule{0ex}{0ex}}\Rightarrow x=0\mathrm{or}x=1$

Hence, the factors are 0 and 1.

#### Page No 4.21:

#### Question 61:

Solve the following quadratic equations by factorization:

$\left(x-5\right)\left(x-6\right)=\frac{25}{{\left(24\right)}^{2}}$

#### Answer:

We have been given that,

Therefore,

or,

Hence, or .

#### Page No 4.21:

#### Question 62:

Solve the following quadratic equations by factorization:

$7x+\frac{3}{x}=35\frac{3}{5}$

#### Answer:

We have been given,

Therefore,Therefore,

or,

Hence, or .

#### Page No 4.26:

#### Question 1:

Find the roots of the following quadratic equations (if they exist) by the method of completing the square.

${x}^{2}-4\sqrt{2x}+6=0$

#### Answer:

We have been given that,

Now we take the constant term to the right hand side and we get

Now add square of half of co-efficient of ‘*x*’ on both the sides. We have,

Since right hand side is a positive number, the roots of the equation exist.

So, now take the square root on both the sides and we get

Now, we have the values of ‘*x*’ as

Also we have,

Therefore the roots of the equation are and.

#### Page No 4.26:

#### Question 2:

Find the roots of the following quadratic equations (if they exist) by the method of completing the square.

2*x*^{2} − 7*x* + 3 = 0

#### Answer:

We have to find the roots of given quadratic equation by the method of completing the square. We have,

We should make the coefficient ofunity. So,

Now shift the constant to the right hand side,

Now add square of half of coefficient ofon both the sides,

We can now write it in the form of perfect square as,

Taking square root on both sides,

So the required solution of,

#### Page No 4.26:

#### Question 3:

Find the roots of the following quadratic equations (if they exist) by the method of completing the square.

$3{x}^{2}+11x+10=0$

#### Answer:

We have been given that,

Now divide throughout by 3. We get,

Now take the constant term to the RHS and we get

Now add square of half of co-efficient of ‘*x*’ on both the sides. We have,

Since RHS is a positive number, therefore the roots of the equation exist.

So, now take the square root on both the sides and we get

Now, we have the values of ‘*x*’ as

Also we have,

Therefore the roots of the equation are and.

#### Page No 4.26:

#### Question 4:

Find the roots of the following quadratic equations (if they exist) by the method of completing the square.

2*x*^{2} + *x* − 4 = 0

#### Answer:

We have been given that,

Now divide throughout by 2. We get,

Now take the constant term to the RHS and we get

Now add square of half of co-efficient of ‘*x*’ on both the sides. We have,

Since RHS is a positive number, therefore the roots of the equation exist.

So, now take the square root on both the sides and we get

Now, we have the values of ‘*x*’ as

Also we have,

Therefore the roots of the equation are and.

#### Page No 4.26:

#### Question 5:

Find the roots of the following quadratic equations (if they exist) by the method of completing the square.

2*x*^{2} + *x* + 4 = 0

#### Answer:

We have been given that,

Now divide throughout by 2. We get,

Now take the constant term to the RHS and we get

Now add square of half of co-efficient of ‘*x*’ on both the sides. We have,

Since RHS is a negative number, therefore the roots of the equation do not exist as the square of a number cannot be negative.

Therefore the roots of the equation do not exist.

#### Page No 4.26:

#### Question 6:

Find the roots of the following quadratic equations (if they exist) by the method of completing the square.

*x*^{2} – 8*x* + 18 = 0

#### Answer:

Given: *x*^{2} – 8*x* + 18 = 0

${x}^{2}-8x+18=0\phantom{\rule{0ex}{0ex}}\mathrm{Adding}\mathrm{and}\mathrm{subtracting}{\left(\frac{1}{2}\times 8\right)}^{2},\mathrm{we}\mathrm{get}\phantom{\rule{0ex}{0ex}}\Rightarrow {x}^{2}-8x+18+{4}^{2}-{4}^{2}=0\phantom{\rule{0ex}{0ex}}\Rightarrow {x}^{2}-8x+16+18-16=0\phantom{\rule{0ex}{0ex}}\Rightarrow {\left(x-4\right)}^{2}+2=0\phantom{\rule{0ex}{0ex}}\Rightarrow {\left(x-4\right)}^{2}=-2\phantom{\rule{0ex}{0ex}}\Rightarrow \left(x-4\right)=\pm \sqrt{-2}\phantom{\rule{0ex}{0ex}}\mathrm{But},\sqrt{-2}\mathrm{is}\mathrm{not}\mathrm{a}\mathrm{real}\mathrm{number}.$

Hence, the roots of the quadratic equation doesn't exist.

#### Page No 4.26:

#### Question 7:

Find the roots of the following quadratic equations (if they exist) by the method of completing the square.

$4{x}^{2}+4\sqrt{3}x+3=0$

#### Answer:

We have been given that,

Now divide throughout by 4. We get,

Now take the constant term to the RHS and we get

Now add square of half of co-efficient of ‘*x*’ on both the sides. We have,

Since RHS is a positive number, therefore the roots of the equation exist.

So, now take the square root on both the sides and we get

Now, we have the values of ‘*x*’ as

Also we have,

Therefore the roots of the equation are and.

#### Page No 4.26:

#### Question 8:

Find the roots of the following quadratic equations (if they exist) by the method of completing the square.

$\sqrt{2}{x}^{2}-3x-2\sqrt{2}=0$

#### Answer:

We have been given that,

Now divide throughout by. We get,

Now take the constant term to the RHS and we get

Now add square of half of co-efficient of ‘*x*’ on both the sides. We have,

Since RHS is a positive number, therefore the roots of the equation exist.

So, now take the square root on both the sides and we get

Now, we have the values of ‘*x*’ as

Also we have,

Therefore the roots of the equation are and.

#### Page No 4.26:

#### Question 9:

Find the roots of the following quadratic equations (if they exist) by the method of completing the square.

$\sqrt{3}{x}^{2}+10x+7\sqrt{3}=0$

#### Answer:

We have been given that,

Now divide throughout by. We get,

Now take the constant term to the RHS and we get

Now add square of half of co-efficient of ‘*x*’ on both the sides. We have,

Since RHS is a positive number, therefore the roots of the equation exist.

So, now take the square root on both the sides and we get

Now, we have the values of ‘*x*’ as

Also we have,

Therefore the roots of the equation are and.

#### Page No 4.26:

#### Question 10:

Find the roots of the following quadratic equations (if they exist) by the method of completing the square.

${x}^{2}-\left(\sqrt{2}+1\right)x+\sqrt{2}=0$

#### Answer:

We have been given that,

Now take the constant term to the RHS and we get

Now add square of half of co-efficient of ‘*x*’ on both the sides. We have,

Since RHS is a positive number, therefore the roots of the equation exist.

So, now take the square root on both the sides and we get

Now, we have the values of ‘*x*’ as

Also we have,

Therefore the roots of the equation are and.

#### Page No 4.26:

#### Question 11:

Find the roots of the following quadratic equations (if they exist) by the method of completing the square.

${x}^{2}-4ax+4{a}^{2}-{b}^{2}=0$

#### Answer:

We have to find the roots of given quadratic equation by the method of completing the square. We have,

Now shift the constant to the right hand side,

Now add square of half of coefficient ofon both the sides,

We can now write it in the form of perfect square as,

Taking square root on both sides,

So the required solution of,

#### Page No 4.32:

#### Question 1:

Write the discriminant of the following quadratic equations:

(i) 2*x*^{2} − 5*x* + 3 = 0

(ii) *x*^{2} + 2*x* + 4 = 0

(iii) (*x* − 1) (2*x* − 1) = 0

(iv) *x*^{2} − 2*x* + *k* = 0, *k* ∈ R

(v) $\sqrt{3}{x}^{2}+2\sqrt{2}x-2\sqrt{3}=0$

(vi) *x*^{2} − *x* + 1 = 0

(vii) (*x* + 5)^{2} = 2(5*x *– 3)

#### Answer:

We have to find the discriminant of the following quadratic equations

(i) We have been given,

Now we also know that for an equation, the discriminant is given by the following equation:

Now, according to the equation given to us, we have,, and.

Therefore, the discriminant is given as,

Therefore, the discriminant of the equation is.

(ii) We have been given,

Now we also know that for an equation, the discriminant is given by the following equation:

Now, according to the equation given to us, we have,, and.

Therefore, the discriminant is given as,

Therefore, the discriminant of the equation is.

(iii) We have been given,

Now, simplify the equation to be represented in the quadratic form, so we have

Now we also know that for an equation, the discriminant is given by the following equation:

Now, according to the equation given to us, we have,, and.

Therefore, the discriminant is given as,

Therefore, the discriminant of the equation is.

(iv) We have been given,

Now we also know that for an equation, the discriminant is given by the following equation:

Now, according to the equation given to us, we have,, and.

Therefore, the discriminant is given as,

Therefore, the discriminant of the equation is.

(v) We have been given,

Now we also know that for an equation, the discriminant is given by the following equation:

Now, according to the equation given to us, we have,, and.

Therefore, the discriminant is given as,

Therefore, the discriminant of the equation is.

(vi) We have been given,

Now we also know that for an equation, the discriminant is given by the following equation:

Now, according to the equation given to us, we have,, and.

Therefore, the discriminant is given as,

Therefore, the discriminant of the equation is.

(vii) (*x*+ 5)

^{2}= 2(5

*x*– 3)

${\left(x+5\right)}^{2}=2\left(5x-3\right)\phantom{\rule{0ex}{0ex}}\Rightarrow {x}^{2}+{5}^{2}+2\left(x\right)\left(5\right)=10x-6\phantom{\rule{0ex}{0ex}}\Rightarrow {x}^{2}+25+10x=10x-6\phantom{\rule{0ex}{0ex}}\Rightarrow {x}^{2}+25+10x-10x+6=0\phantom{\rule{0ex}{0ex}}\Rightarrow {x}^{2}+31=0\phantom{\rule{0ex}{0ex}}\mathrm{Discriminant}={b}^{2}-4ac\phantom{\rule{0ex}{0ex}}={0}^{2}-4\left(1\right)\left(31\right)\phantom{\rule{0ex}{0ex}}=-124$

Hence, the discriminant is –124.

#### Page No 4.32:

#### Question 2:

In the following, determine whether the given quadratic equations have real roots and if so, find the roots:

(i) 16*x*^{2} = 24*x* + 1

(ii) *x*^{2} + *x* + 2 = 0

(iii) $\sqrt{3}{x}^{2}+10x-8\sqrt{3}=0$

(iv) 3*x*^{2} − 2*x* + 2 = 0

(v) $2{x}^{2}-2\sqrt{6}x+3=0$

(vi) $3{a}^{2}{x}^{2}+8abx+4{b}^{2}=0,a\ne 0$

(vii) $3{x}^{2}+2\sqrt{5}x-5=0$

(viii) *x*^{2} − 2*x* + 1 = 0

(ix) $2{x}^{2}+5\sqrt{3}x+6=0$

(x) $\sqrt{2}{x}^{2}+7x+5\sqrt{2}=0$

(xi) $2{x}^{2}-2\sqrt{2}x+1=0$

(xii) 3*x*^{2} − 5*x* + 2 = 0

#### Answer:

In the following parts we have to find the real roots of the equations

(i) We have been given,

Now we also know that for an equation, the discriminant is given by the following equation:

Now, according to the equation given to us, we have,, and.

Therefore, the discriminant is given as,

Since, in order for a quadratic equation to have real roots,.Here we find that the equation satisfies this condition, hence it has real roots.

Now, the roots of an equation is given by the following equation,

Therefore, the roots of the equation are given as follows,

Now we solve both cases for the two values of *x*. So, we have,

Also,

Therefore, the roots of the equation are and

(ii) We have been given,

Now we also know that for an equation, the discriminant is given by the following equation:

Now, according to the equation given to us, we have,, and.

Therefore, the discriminant is given as,

Since, in order for a quadratic equation to have real roots,.Here we find that the equation does not satisfies this condition, hence it does not have real roots.

(iii) We have been given,

Now we also know that for an equation, the discriminant is given by the following equation:

Now, according to the equation given to us, we have,, and.

Therefore, the discriminant is given as,

Since, in order for a quadratic equation to have real roots,.Here we find that the equation satisfies this condition, hence it has real roots.

Now, the roots of an equation is given by the following equation,

Therefore, the roots of the equation are given as follows,

Now we solve both cases for the two values of *x*. So, we have,

Also,

Therefore, the roots of the equation are and.

(iv) We have been given,

Now we also know that for an equation, the discriminant is given by the following equation:

Now, according to the equation given to us, we have,, and.

Therefore, the discriminant is given as,

Since, in order for a quadratic equation to have real roots,.Here we find that the equation does not satisfies this condition, hence it does not have real roots.

(v) We have been given,

Now we also know that for an equation, the discriminant is given by the following equation:

Now, according to the equation given to us, we have,, and.

Therefore, the discriminant is given as,

Since, in order for a quadratic equation to have real roots,.Here we find that the equation satisfies this condition, hence it has real and equal roots.

Now, the roots of an equation is given by the following equation,

Therefore, the roots of the equation are given as follows,

Therefore, the roots of the equation are real and equal and its value is.

(vi) We have been given,

Now we also know that for an equation, the discriminant is given by the following equation:

Now, according to the equation given to us, we have,, and.

Therefore, the discriminant is given as,

Since, in order for a quadratic equation to have real roots,.Here we find that the equation satisfies this condition, hence it has real roots.

Now, the roots of an equation is given by the following equation,

Therefore, the roots of the equation are given as follows,

Now we solve both cases for the two values of x. So, we have,

Also,

Therefore, the roots of the equation are and.

(vii) We have been given,

Now we also know that for an equation, the discriminant is given by the following equation:

Now, according to the equation given to us, we have,, and.

Therefore, the discriminant is given as,

Now, the roots of an equation is given by the following equation,

Therefore, the roots of the equation are given as follows,

Now we solve both cases for the two values of *x*. So, we have,

Also,

Therefore, the roots of the equation are and.

(viii) We have been given,

Now we also know that for an equation, the discriminant is given by the following equation:

Now, according to the equation given to us, we have,, and.

Therefore, the discriminant is given as,

Since, in order for a quadratic equation to have real roots,.Here we find that the equation satisfies this condition, hence it has real and equal roots.

Now, the roots of an equation is given by the following equation,

Therefore, the roots of the equation are given as follows,

Therefore, the roots of the equation are real and equal and its value is.

(ix) We have been given,

Now we also know that for an equation, the discriminant is given by the following equation:

Now, according to the equation given to us, we have,, and.

Therefore, the discriminant is given as,

Now, the roots of an equation is given by the following equation,

Therefore, the roots of the equation are given as follows,

Now we solve both cases for the two values of x. So, we have,

Also,

Therefore, the roots of the equation are and.

(x) We have been given,

Now we also know that for an equation, the discriminant is given by the following equation:

Now, according to the equation given to us, we have,, and.

Therefore, the discriminant is given as,

Now, the roots of an equation is given by the following equation,

Therefore, the roots of the equation are given as follows,

Now we solve both cases for the two values of *x*. So, we have,

Also,

Therefore, the roots of the equation are and

(xi) We have been given,

Now we also know that for an equation, the discriminant is given by the following equation:

Now, according to the equation given to us, we have,, and.

Therefore, the discriminant is given as,

Since, in order for a quadratic equation to have real roots,.Here we find that the equation satisfies this condition, hence it has real and equal roots.

Now, the roots of an equation is given by the following equation,

Therefore, the roots of the equation are given as follows,

Therefore, the roots of the equation are real and equal and its value is

(xii) We have been given,

Now we also know that for an equation, the discriminant is given by the following equation:

Now, according to the equation given to us, we have,, and.

Therefore, the discriminant is given as,

Now, the roots of an equation is given by the following equation,

Therefore, the roots of the equation are given as follows,

Now we solve both cases for the two values of *x*. So, we have,

Also,

Therefore, the roots of the equation are and.

#### Page No 4.32:

#### Question 3:

Solve for *x*:

(i) $\frac{x-1}{x-2}+\frac{x-3}{x-4}=3\frac{1}{3};x\ne 2,4$

(ii) $\frac{1}{x}-\frac{1}{x-2}=3,x\ne 0,2$

(iii) $x+\frac{1}{x}=3,x\ne 0$

(iv) $\frac{16}{x}-1=\frac{15}{x+1},x\ne 0,-1$

(v) $\frac{1}{x-3}-\frac{1}{x+5}=\frac{1}{6},x\ne 3,-5$

#### Answer:

(i) We have been given,

,

Now we solve the above equation as follows,

Now we also know that for an equation, the discriminant is given by the following equation:

Now, according to the equation given to us, we have,, and.

Therefore, the discriminant is given as,

Now, the roots of an equation is given by the following equation,

Therefore, the roots of the equation are given as follows,

Now we solve both cases for the two values of *x*. So, we have,

Also,

Therefore, the value of

(ii) We have been given,

,

Now we solve the above equation as follows,

Now we also know that for an equation, the discriminant is given by the following equation:

Now, according to the equation given to us, we have,, and.

Therefore, the discriminant is given as,

Now, the roots of an equation is given by the following equation,

Therefore, the roots of the equation are given as follows,

Now we solve both cases for the two values of *x*. So, we have,

Also,

Therefore, the value of

(iii) We have been given,

,

Now, we solve the equation as follows:

Now we also know that for an equation, the discriminant is given by the following equation:

Now, according to the equation given to us, we have,, and.

Therefore, the discriminant is given as,

Now, the roots of an equation is given by the following equation,

Therefore, the roots of the equation are given as follows,

Now we solve both cases for the two values of *x*. So, we have,

Also,

Therefore, the value of

(iv) We have been given,

$\frac{16}{x}-1=\frac{15}{x+1},x\ne 0,-1$

Now we solve the above equation as follows,

$\frac{16-x}{x}=\frac{15}{x+1}\phantom{\rule{0ex}{0ex}}\Rightarrow (16-x)(x+1)=15x\phantom{\rule{0ex}{0ex}}\Rightarrow 16x+16-{x}^{2}-x=15x\phantom{\rule{0ex}{0ex}}\Rightarrow 15x+16-{x}^{2}-15x=0\phantom{\rule{0ex}{0ex}}\Rightarrow 16-{x}^{2}=0\phantom{\rule{0ex}{0ex}}\Rightarrow {x}^{2}-16=0$

Now we also know that for an equation $a{x}^{2}+bx+c=0$, the discriminant is given by the following equation:

$D={b}^{2}-4ac$

Now, according to the equation given to us, we have, $a=1$, $b=0$ and $c=-16$.

Therefore, the discriminant is given as,

$D=(0{)}^{2}-4\left(1\right)\left(-16\right)\phantom{\rule{0ex}{0ex}}=64$

Now, the roots of an equation is given by the following equation,

$x=\frac{-b\pm \sqrt{D}}{2a}$

Therefore, the roots of the equation are given as follows,

$x=\frac{-0\pm \sqrt{64}}{2\left(1\right)}\phantom{\rule{0ex}{0ex}}=\frac{\pm 8}{2}\phantom{\rule{0ex}{0ex}}=\pm 4$

Therefore, the value of $x=\pm 4.$

(v) $\frac{1}{x-3}-\frac{1}{x+5}=\frac{1}{6},x\ne 3,-5$

$\frac{1}{x-3}-\frac{1}{x+5}=\frac{1}{6}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{x+5-x+3}{\left(x-3\right)\left(x+5\right)}=\frac{1}{6}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{8}{\left(x-3\right)\left(x+5\right)}=\frac{1}{6}$

$\Rightarrow 48={x}^{2}+2x-15\phantom{\rule{0ex}{0ex}}\Rightarrow {x}^{2}+2x-15-48=0\phantom{\rule{0ex}{0ex}}\Rightarrow {x}^{2}+2x-63=0\phantom{\rule{0ex}{0ex}}\Rightarrow {x}^{2}+9x-7x-63=0$

$\Rightarrow x\left(x+9\right)-7\left(x+9\right)=0\phantom{\rule{0ex}{0ex}}\Rightarrow \left(x-7\right)\left(x+9\right)=0\phantom{\rule{0ex}{0ex}}\Rightarrow x=7,-9$

#### Page No 4.4:

#### Question 1:

Which of the following are quadratic equations?

(i) *x*^{2} + 6*x* − 4 = 0

(ii) $\sqrt{3}{x}^{2}-2x+\frac{1}{2}=0$

(iii) ${x}^{2}+\frac{1}{{x}^{2}}=5$

(iv) $x-\frac{3}{x}={x}^{2}$

(v) $2{x}^{2}-\sqrt{3x}+9=0$

(vi) ${x}^{2}-2x-\sqrt{x}-5=0$

(vii) 3*x*^{2} − 5*x* + 9 = *x*^{2} − 7*x* + 3

(viii) $x+\frac{1}{x}=1$

(ix) *x*^{2} − 3*x* = 0

(x) ${\left(x+\frac{1}{x}\right)}^{2}=3\left(x+\frac{1}{x}\right)+4$

(xi) (2*x* + 1) (3*x* + 2) = 6(*x* − 1) (*x* − 2)

(xii) $x+\frac{1}{x}={x}^{2},x\ne 0$

(xiii) 16*x*^{2} − 3 = (2*x* + 5) (5*x* − 3)

(xiv) (*x* + 2)^{3} = *x*^{3} − 4

(xv) *x*(*x* + 1) + 8 = (*x* + 2) (*x* − 2)

#### Answer:

We are given the following algebraic expressions and are asked to find out which one is quadratic.

(i) Here it has been given that,

Now, the above equation clearly represents a quadratic equation of the form , where , and .

Hence, the above equation is a quadratic equation.

(ii) Here it has been given that,

Now, solving the above equation further we get,

Now, the above equation clearly represents a quadratic equation of the form , where , and .

Hence, the above equation is a quadratic equation.

(iii) Here it has been given that,

Now, solving the above equation further we get,

$\frac{{x}^{4}+1}{{x}^{2}}=5\phantom{\rule{0ex}{0ex}}\Rightarrow {x}^{4}+1=5{x}^{2}\phantom{\rule{0ex}{0ex}}\Rightarrow {x}^{4}-5{x}^{2}+1=0$

Now, the above equation clearly does not represent a quadratic equation of the form , because ${x}^{4}-5{x}^{2}+1$ is a polynomial of degree 4.

Hence, the above equation is not a quadratic equation.

(iv) Here it has been given that,

Now, solving the above equation further we get,

Now, the above equation clearly does not represent a quadratic equation of the form , because is a polynomial of degree 3.

Hence, the above equation is not a quadratic equation.

(v) Here it has been given that,

Now, the above equation clearly does not represent a quadratic equation of the form, because contains a term , where is not an integer.

Hence, the above equation is not a quadratic equation.

(vi) Here it has been given that,

Now, as we can see the above equation clearly does not represent a quadratic equation of the form, because contains an extra term, where is not an integer.

Hence, the above equation is not a quadratic equation.

(vii) Here it has been given that,

Now, after solving the above equation further we get,

Now as we can see, the above equation clearly represents a quadratic equation of the form , where , and .

Hence, the above equation is a quadratic equation.

(viii) Here it has been given that,

Now, solving the above equation further we get,

Now as we can see, the above equation clearly represents a quadratic equation of the form , where , and .

Hence, the above equation is a quadratic equation.

(ix) Here it has been given that,

Now as we can see, the above equation clearly represents a quadratic equation of the form, where, and.

Hence, the above equation is a quadratic equation.

(x) Here it has been given that,

Now, solving the above equation further we get,

Now as we can see, the above equation clearly does not represent a quadratic equation of the form , because is a polynomial having a degree of 4 which is never present in a quadratic polynomial.

Hence, the above equation is not a quadratic equation.

(xi) Here it has been given that,

Now, after solving the above equation further we get,

Now as we can see, the above equation clearly does not represent a quadratic equation of the form , because is a linear equation.

Hence, the above equation is not a quadratic equation.

(xii) Here it has been given that,

Now, solving the above equation further we get,

Now as we can see, the above equation clearly does not represent a quadratic equation of the form , because is a polynomial having a degree of 3 which is never present in a quadratic polynomial.

Hence, the above equation is not a quadratic equation.

(xiii) Here it has been given that,

Now, after solving the above equation further we get,

Now as we can see, the above equation clearly represents a quadratic equation of the form , where , and.

Hence, the above equation is a quadratic equation.

(xiv) Here it has been given that,

Now, after solving the above equation further we get,

Now as we can see, the above equation clearly represents a quadratic equation of the form , where , and .

Hence, the above equation is a quadratic equation.

(xv) Here it has been given that,

Now, solving the above equation further we get,

Now as we can see, the above equation clearly does not represent a quadratic equation of the form , because is a linear equation which does not have a term in it.

Hence, the above equation is not a quadratic equation.

#### Page No 4.4:

#### Question 2:

In each of the following, determine whether the given values are solutions of the given equation or not:

(i) ${x}^{2}-3x+2=0,x=2,x=-1$

(ii) ${x}^{2}+x+1=0,x=0,x=1$

(iii) ${x}^{2}-3\sqrt{3}x+6=0,x=\sqrt{3},x=-2\sqrt{3}$

(iv) $x+\frac{1}{x}=\frac{13}{6},x=\frac{5}{6},x=\frac{4}{3}$

(v) $2{x}^{2}-x+9={x}^{2}+4x+3,x=2,x=3$

(vi) ${x}^{2}-\sqrt{2}x-4=0,x=-\sqrt{2},x=-2\sqrt{2}$

(vii) ${a}^{2}{x}^{2}-3abx+2{b}^{2}=0,x=\frac{a}{b},x=\frac{b}{a}$

#### Answer:

We are given the following quadratic equations and we are asked to find whether the given values are solutions or not

(i)

We have been given that,

Now if is a solution of the equation then it should satisfy the equation

So, substituting in the equation we get

Hence, is a solution of the given quadratic equation.

Also, if is a solution of the equation then it should satisfy the equation

So, substituting in the equation, we get

Hence is not a solution of the quadratic equation

Therefore, from the above results we find out that is a solution and is not a solution of the given quadratic equation.

(ii) We have been given that,

Now if is a solution of the equation then it should satisfy the equation.

So, substituting in the equation, we get

Hence is not a solution of the given quadratic equation.

Also, if is a solution of the equation then it should satisfy the equation.

So, substituting in the equation, we get

Hence is not a solution of the quadratic equation.

Therefore, from the above results we find out that both and are not a solution of the given quadratic equation.

(iii) We have been given that,

Now if is a solution of the equation then it should satisfy the equation.

So, substituting in the equation, we get

Hence is a solution of the quadratic equation.

Also, if is a solution of the equation then it should satisfy the equation

So, substituting in the equation, we get

Hence is a solution of the quadratic equation.

Therefore, from the above results we find out that and are the solutions of the given quadratic equation.

(iv) We have been given that,

Now if is a solution of the equation then it should satisfy the equation.

So, substituting in the equation, we get

Hence is not a solution of the quadratic equation.

Also, if is a solution of the equation then it should satisfy the equation.

So, substituting in the equation, we get

Hence, is not a solution of the quadratic equation.

Therefore, from the above results we find out that both and are not the solutions of the given quadratic equation.

(v) We have been given that,

Now if is a solution of the equation then it should satisfy the equation.

So, substituting in the equation, we get

Hence is a solution of the given quadratic equation

Also, if is a solution of the equation then it should satisfy the equation.

So, substituting in the equation, we get

Hence is a solution of the quadratic equation.

Therefore, from the above results we find out that both and are solutions of the quadratic equation.

(vi) We have been given that,

Now if is a solution of the equation then it should satisfy the equation.

So, substituting in the equation, we get

Hence is a solution of the quadratic equation.

Also, if is a solution of the equation then it should satisfy the equation.

So, substituting in the equation, we get

Hence is not a solution of the quadratic equation.

Therefore, from the above results we find out that is a solution but is not a solution of the given quadratic equation.

(vii) We have been given that,

Now if is a solution of the equation then it should satisfy the equation.

So, substituting in the equation, we get

Hence is not a solution of the quadratic equation.

Also, if is a solution of the equation then it should satisfy the equation.

So, substituting in the equation, we get

Hence is a solution of the quadratic equation.

Therefore, from the above results we find out that is not a solution and is a solution of the given quadratic equation.

#### Page No 4.4:

#### Question 3:

In each of the following, find the value of *k* for which the given value is a solution of the given equation:

(i) $7{x}^{2}+kx-3=0,x=\frac{2}{3}$

(ii) ${x}^{2}-x\left(a+b\right)+k=0,x=a$

(iii) $k{x}^{2}+\sqrt{2}x-4=0,x=\sqrt{2}$

(iv) ${x}^{2}+3ax+k=0,x=-a$

#### Answer:

In each of the following cases find *k*.

(i) We are given here that,

Now, as we know that is a solution of the quadratic equation, hence it should satisfy the equation. Therefore substituting in the above equation gives us,

Hence, the value of .

(ii) We are given here that,

Now, as we know that is a solution of the quadratic equation, hence it should satisfy the equation. Therefore substituting in the above equation gives us,

Hence, the value of .

(iii) We are given here that,

Now, as we know that is a solution of the quadratic equation, hence it should satisfy the equation. Therefore substituting in the above equation gives us,

Hence, the value of.

(iv) We are given here that,

Hence the value of.

#### Page No 4.41:

#### Question 1:

Determine the nature of the roots of the following quadratic equations:

(i) 2*x*^{2} − 3*x* + 5 = 0

(ii) 2*x*^{2} − 6*x* + 3 = 0

(iii) $\frac{3}{5}{x}^{2}-\frac{2}{3}x+1=0$

(iv) $3{x}^{2}-4\sqrt{3}x+4=0$

(v) $3{x}^{2}-2\sqrt{6}x+2=0$

(vi) $4{x}^{2}+4\sqrt{3}x+3=0$

#### Answer:

(i) The given quadric equation is

Here,

As we know that

Putting the value of

Since,

Therefore, root of the given equation are.

(ii) The given quadric equation is

Here,

As we know that

Putting the value of

Since,

Therefore, root of the given equation are.

(iii) The given quadric equation is

Here,

As we know that

Putting the value of

Since,

Therefore, root of the given equation are.

(iv) The given quadric equation is

Here,

As we know that

Putting the value of

Since,

Therefore, root of the given equation are.

(v) The given quadric equation is

Here,

As we know that

Putting the value of

Since,

Therefore, root of the given equation are.

(vi) $4{x}^{2}+4\sqrt{3}x+3=0$$4{x}^{2}+4\sqrt{3}x+3=0\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\mathrm{Discriminant}={b}^{2}-4ac\phantom{\rule{0ex}{0ex}}={\left(4\sqrt{3}\right)}^{2}-4\left(4\right)\left(3\right)\phantom{\rule{0ex}{0ex}}=48-48\phantom{\rule{0ex}{0ex}}=0$

Since, Discriminant = 0

Therefore, the roots are real and equal.

Hence, the given equation has real and equal roots.

#### Page No 4.41:

#### Question 2:

Find the values of *k* for which the roots are real and equal in each of the following equations:

(i) $k{x}^{2}+4x+1=0$

(ii) $k{x}^{2}-2\sqrt{5}x+4=0$

(iii) $3{x}^{2}-5x+2k=0$

(iv) $4{x}^{2}+kx+9=0$

(v) $2k{x}^{2}-40x+25=0$

(vi) $9{x}^{2}-24x+k=0$

(vii) $4{x}^{2}-3kx+1=0$

(viii) ${x}^{2}-2\left(5+2k\right)x+3\left(7+10k\right)=0$

(ix) $\left(3k+1\right){x}^{2}+2\left(k+1\right)x+k=0$

(x) $k{x}^{2}+kx+1=-4{x}^{2}-x$

(xi) $\left(k+1\right){x}^{2}+2\left(k+3\right)x+\left(k+8\right)=0$

(xii) ${x}^{2}-2kx+7k-12=0$

(xiii) $\left(k+1\right){x}^{2}-2\left(3k+1\right)x+8k+1=0$

(xiv) $\left(2k+1\right){x}^{2}+2\left(k+3\right)x+\left(k+5\right)=0$

(xvii) $4{x}^{2}-2\left(k+1\right)x+\left(k+4\right)=0$

(xviii) $4{x}^{2}-2\left(k+1\right)x+\left(k+1\right)=0$

#### Answer:

(i) The given quadric equation is , and roots are real and equal

Then find the value of *k.*

Here,

As we know that

Putting the value of

The given equation will have real and equal roots, if

Thus,

Therefore, the value of

(ii) The given quadric equation is , and roots are real and equal

Then find the value of *k.*

Here,

As we know that

Putting the value of

The given equation will have real and equal roots, if *D* = 0

Thus,

Therefore, the value of

(iii) The given quadric equation is , and roots are real and equal

Then find the value of *k.*

Here,

As we know that

Putting the value of

The given equation will have real and equal roots, if *D* = 0

Thus,

Therefore, the value of

(iv) The given quadric equation is , and roots are real and equal

Then find the value of *k.*

Here,

As we know that

Putting the value of

The given equation will have real and equal roots, if *D* = 0

Thus,

Therefore, the value of

(v) The given quadric equation is , and roots are real and equal

Then find the value of *k.*

Here,

As we know that

Putting the value of

The given equation will have real and equal roots, if *D* = 0

Thus,

Therefore, the value of

(vi) The given quadric equation is , and roots are real and equal

Then find the value of *k.*

Here,

As we know that

Putting the value of

The given equation will have real and equal roots, if *D* = 0

Thus,

Therefore, the value of

(vii) The given quadric equation is , and roots are real and equal

Then find the value of *k.*

Here,

As we know that

Putting the value of

The given equation will have real and equal roots, if *D* = 0

Thus,

Therefore, the value of

(viii) The given quadric equation is , and roots are real and equal

Then find the value of *k.*

Here,

As we know that

Putting the value of

The given equation will have real and equal roots, if *D* = 0

Thus,

Now factorizing of the above equation

So, either

Therefore, the value of

(ix) The given quadric equation is , and roots are real and equal

Then find the value of *k.*

Here,

As we know that

Putting the value of

The given equation will have real and equal roots, if *D* = 0

Thus,

Now factorizing of the above equation

So, either

Therefore, the value of

(x) The given quadric equation is , and roots are real and equal

Then find the value of *k.*

Here,

So,

As we know that

Putting the value of

The given equation will have real and equal roots, if *D* = 0

Thus,

Now factorizing of the above equation

So, either

Therefore, the value of

(xi) The given quadric equation is , and roots are real and equal

Then find the value of *k.*

Here,

As we know that

Putting the value of

The given equation will have real and equal roots, if *D* = 0

Therefore, the value of

(xii) The given quadric equation is , and roots are real and equal

Then find the value of *k.*

Here,

As we know that

Putting the value of

The given equation will have real and equal roots, if *D* = 0

Now factorizing of the above equation

So, either

Therefore, the value of

(xiii) The given quadric equation is , and roots are real and equal

Then find the value of *k.*

Here,

As we know that

Putting the value of

The given equation will have real and equal roots, if *D* = 0

Now factorizing of the above equation

So, either

Therefore, the value of

(xiv) The given quadric equation is , and roots are real and equal

Then find the value of *k.*

Here,

So,

As we know that

Putting the value of

The given equation will have real and equal roots, if *D* = 0

Thus,

Now factorizing of the above equation

So, either

Therefore, the value of

(xv) The given quadric equation is , and roots are real and equal

Then find the value of *k.*

Here,

As we know that

Putting the value of

The given equation will have real and equal roots, if *D* = 0

Now factorizing of the above equation

So, either

Therefore, the value of

(xvi) The given quadric equation is , and roots are real and equal

Then find the value of *k.*

Here,

As we know that

Putting the value of

The given equation will have real and equal roots, if *D* = 0

Now factorizing the above equation

So, either

Therefore, the value of

(xvii) The given quadric equation is , and roots are real and equal

Then find the value of *k.*

Here,

As we know that

Putting the value of

The given equation will have real and equal roots, if *D* = 0

Now factorizing of the above equation

So, either

Therefore, the value of

(xviii) The given equation is $4{x}^{2}-2(k+1)x+(k+1)=0$ where a = 4, b = -2(k+1), c = (k+1)

As we know that

Putting the value of a = 4, b = -2(k+1), c = (k+1)

${\left\{-2(k+1)\right\}}^{2}-4\times 4\times (K+1)\phantom{\rule{0ex}{0ex}}4(K+1{)}^{2}-16(K+1\left)\phantom{\rule{0ex}{0ex}}\right(K+1\left)\left\{4(K+1)-16\right\}\phantom{\rule{0ex}{0ex}}\right(K+1\left)\right(4K-12\left)\phantom{\rule{0ex}{0ex}}4\right(K+1\left)\right(K-3)$

For real and equal roots D = 0

$4\left(K+1\right)\left(K-3\right)=0\phantom{\rule{0ex}{0ex}}K=-1\mathrm{or}k=3$

Therefore, the value of

#### Page No 4.42:

#### Question 3:

In the following , determine the set of values of k for which the given quadratic equation has real roots:

(i) (ii) $2{x}^{2}+x+k=0$ (iii)

(iv) (v)

#### Answer:

(i) The given quadric equation is , and roots are real.

Then find the value of *k.*

Here,

As we know that

Putting the value of

The given equation will have real roots, if

Therefore, the value of

(ii)

The given quadric equation is $2{x}^{2}+x+k=0$, and roots are real.

Then find the value of *k.*

Here, $a=2,b=1,c=k$

As we know that

Putting the value of $a=2,b=1,c=k$

$D=1-8k$

The given equation will have real roots, if

$D=1-8k\ge 0\phantom{\rule{0ex}{0ex}}\Rightarrow 8k\le 1\phantom{\rule{0ex}{0ex}}\Rightarrow k\le \frac{1}{8}$

Therefore, the value of $k\le \frac{1}{8}$.

(iii) The given quadric equation is , and roots are real

Then find the value of *k.*

Here,

As we know that

Putting the value of

The given equation will have real roots, if

Therefore, the value of

(iv) The given quadric equation is , and roots are real

Then find the value of *k.*

Here,

As we know that

Putting the value of

The given equation will have real roots, if

Therefore, the value of

(v) The given quadric equation is , and roots are real.

Then find the value of *k.*

Here,

As we know that

Putting the value of

The given equation will have real roots, if

Therefore, the value of

#### Page No 4.42:

#### Question 4:

Find the values of *k* for which the following equations have real and equal roots:

(i) ${x}^{2}-2\left(k+1\right)x+{k}^{2}=0$

(ii) ${k}^{2}{x}^{2}-2\left(2k-1\right)x+4=0$

(iii) $\left(k+1\right){x}^{2}-2\left(k-1\right)x+1=0$

(iv) ${x}^{2}+k\left(2x+k-1\right)+2=0$

#### Answer:

(i) The given quadric equation is , and roots are real and equal

Then find the value of *k.*

Here,

As we know that

Putting the value of

The given equation will have real and equal roots, if *D* = 0

Therefore, the value of

(ii) The given quadric equation is , and roots are real and equal

Then find the value of *k.*

Here,

As we know that

Putting the value of

The given equation will have real and equal roots, if *D* = 0

Therefore, the value of

(iii) The given quadric equation is , and roots are real and equal

Then find the value of *k.*

Here,

As we know that

Putting the value of

The given equation will have real and equal roots, if *D* = 0

Now factorizing of the above equation

So, either

Therefore, the value of

(iv) The given equation is ${x}^{2}+k\left(2x+k-1\right)+2=0$.

$\Rightarrow {x}^{2}+2kx+k\left(k-1\right)+2=0$

So, *a* = 1, *b* = 2*k*, *c* = *k*(*k* − 1) + 2

We know $D={b}^{2}-4ac$

$\Rightarrow D={\left(2k\right)}^{2}-4\times 1\times \left[k\left(k-1\right)+2\right]\phantom{\rule{0ex}{0ex}}\Rightarrow D=4{k}^{2}-4\left[{k}^{2}-k+2\right]\phantom{\rule{0ex}{0ex}}\Rightarrow D=4{k}^{2}-4{k}^{2}+4k-8\phantom{\rule{0ex}{0ex}}\Rightarrow D=4k-8=4\left(k-2\right)$

For equal roots, *D* = 0

Thus, 4(*k* − 2) = 0

So, *k* = 2.

#### Page No 4.42:

#### Question 5:

Find the values of *k* for which the following equations have real roots

(i) (ii)

(iii) ${x}^{2}-4kx+k=0$ (iv) $kx\left(x-2\sqrt{5}\right)+10=0$

(v) $kx(x-3)+9=0$ (vi) $4{x}^{2}+kx+3=0$

#### Answer:

(i) The given quadric equation is , and roots are real and equal

Then find the value of *k.*

Here,

As we know that

Putting the value of

The given equation will have real and equal roots, if *D* = 0

Therefore, the value of

(ii) The given quadric equation is , and roots are real and equal

Then find the value of *k.*

Here,

So,

As we know that

Putting the value of

The given equation will have real and equal roots, if *D* = 0

Now factorizing of the above equation

So, either

Therefore, the value of

(iii) The given quadratic equation is ${x}^{2}-4kx+k=0$, and roots are real and equal.

Then find the value of *k.*

Here,

${x}^{2}-4kx+k=0$

So,

$a=1,b=-4k\mathrm{and}c=k.$

As we know that $D={b}^{2}-4ac$

Putting the value of $a=1,b=-4k\mathrm{and}c=k.$

$D={\left(-4k\right)}^{2}-4\left(1\right)\left(k\right)\phantom{\rule{0ex}{0ex}}=16{k}^{2}-4k$

The given equation will have real and equal roots, if *D* = 0.

So, $16{k}^{2}-4k=0$

Now factorizing the above equation,

$16{k}^{2}-4k=0\phantom{\rule{0ex}{0ex}}\Rightarrow 4k\left(4k-1\right)=0\phantom{\rule{0ex}{0ex}}\Rightarrow 4k=0\mathrm{or}4k-1=0\phantom{\rule{0ex}{0ex}}\Rightarrow k=0\mathrm{or}k=\frac{1}{4}$

Therefore, the value of $k=0,\frac{1}{4}$.

(iv) The given quadratic equation is $kx\left(x-2\sqrt{5}\right)+10=0$, and roots are real and equal.

Then find the value of *k.*

Here,

$kx\left(x-2\sqrt{5}\right)+10=0\phantom{\rule{0ex}{0ex}}\Rightarrow k{x}^{2}-2\sqrt{5}kx+10=0$

So,

$a=k,b=-2\sqrt{5}k\mathrm{and}c=10.$

As we know that $D={b}^{2}-4ac$

Putting the value of $a=k,b=-2\sqrt{5}k\mathrm{and}c=10.$

$D={\left(-2\sqrt{5}k\right)}^{2}-4\left(k\right)\left(10\right)\phantom{\rule{0ex}{0ex}}=20{k}^{2}-40k$

The given equation will have real and equal roots, if *D* = 0.

So, $20{k}^{2}-40k=0$

Now factorizing the above equation,

$20{k}^{2}-40k=0\phantom{\rule{0ex}{0ex}}\Rightarrow 20k\left(k-2\right)=0\phantom{\rule{0ex}{0ex}}\Rightarrow 20k=0\mathrm{or}k-2=0\phantom{\rule{0ex}{0ex}}\Rightarrow k=0\mathrm{or}k=2$

Therefore, the value of $k=0,2$.(v) The given quadratic equation is $px(x-3)+9=0$, and roots are real and equal.

Then find the value of *p.*

Here,

$px(x-3)+9=0\phantom{\rule{0ex}{0ex}}\Rightarrow p{x}^{2}-3px+9=0$

So,

$a=p,b=-3p\mathrm{and}c=9.$

As we know that $D={b}^{2}-4ac$

Putting the value of $a=p,b=-3p\mathrm{and}c=9.$

$D={\left(-3p\right)}^{2}-4\left(p\right)\left(9\right)\phantom{\rule{0ex}{0ex}}=9{p}^{2}-36p$

The given equation will have real and equal roots, if *D* = 0.

So, $9{p}^{2}-36p=0$

Now factorizing the above equation,

$9{p}^{2}-36p=0\phantom{\rule{0ex}{0ex}}\Rightarrow 9p\left(p-4\right)=0\phantom{\rule{0ex}{0ex}}\Rightarrow 9p=0\mathrm{or}p-4=0\phantom{\rule{0ex}{0ex}}\Rightarrow p=0\mathrm{or}p=4$

Therefore, the value of $p=0,4.$

(vi) The given quadratic equation is $4{x}^{2}+px+3=0$, and roots are real and equal.

Then find the value of *p.*

Here,

$4{x}^{2}+px+3=0$

So,

$a=4,b=p\mathrm{and}c=3.$

As we know that $D={b}^{2}-4ac$

Putting the value of $a=4,b=p\mathrm{and}c=3.$

$D={\left(p\right)}^{2}-4\left(4\right)\left(3\right)\phantom{\rule{0ex}{0ex}}={p}^{2}-48$

The given equation will have real and equal roots, if *D* = 0.

So, ${p}^{2}-48=0$

Now factorizing the above equation,

${p}^{2}-48=0\phantom{\rule{0ex}{0ex}}\Rightarrow {p}^{2}-{\left(4\sqrt{3}\right)}^{2}=0\phantom{\rule{0ex}{0ex}}\Rightarrow \left(p-4\sqrt{3}\right)\left(p+4\sqrt{3}\right)=0\phantom{\rule{0ex}{0ex}}\Rightarrow p-4\sqrt{3}=0\mathrm{or}p+4\sqrt{3}=0\phantom{\rule{0ex}{0ex}}\Rightarrow p=4\sqrt{3}\mathrm{or}p=-4\sqrt{3}$

Therefore, the value of $p=\pm 4\sqrt{3}.$

#### Page No 4.42:

#### Question 6:

Find the values of *k* for which the given quadratic equation has real and distinct roots:

(a) *kx*^{2} + 2*x* + 1 = 0

(b) *kx*^{2} + 6*x* + 1 = 0

(c) *x*^{2} − *kx* + 9 = 0

#### Answer:

(i) The given quadric equation is , and roots are real and distinct

Then find the value of *k.*

Here,

As we know that

Putting the value of

The given equation will have real and distinct roots, if

Now factorizing of the above equation

Now according to question, the value of *k* less than 1

Therefore, the value of

(ii) The given quadric equation is , and roots are real and distinct.

Then find the value of *k.*

Here,

As we know that

Putting the value of

The given equation will have real and distinct roots, if

Now factorizing of the above equation

Now according to question, the value of *k* less than 9

Therefore, the value of

(iii) The given quadric equation is , and roots are real and distinct

Then find the value of *k.*

Here,

As we know that

Putting the value of

The given equation will have real and distinct roots, if

Now factorizing of the above equation

Therefore, the value of

#### Page No 4.42:

#### Question 7:

For what value of *k*, (4 − *k*)*x*^{2} + (2*k* + 4) *x* + (8*k* + 1) = 0, is a perfect square.

#### Answer:

The given quadric equation is, and roots are real and equal

Then find the value of *k.*

Here,

As we know that

Putting the value of

The given equation will have real and equal roots, if

Thus,

Now factorizing of the above equation

So, either

Therefore, the value of

#### Page No 4.42:

#### Question 8:

Find the least positive value of *k* for which the equation *x*^{2} + *kx* + 4 = 0 has real roots.

#### Answer:

The given quadric equation is , and roots are real.

Then find the value of *k.*

Here,

As we know that

Putting the value of

The given equation will have real and equal roots, if

Now factorizing of the above equation

Now according to question, the value of *k* is positive.

Therefore, the value of

#### Page No 4.42:

#### Question 9:

(i) Find the values of *k* for which the quadratic equation $\left(3k+1\right){x}^{2}+2\left(k+1\right)x+1=0$ has equal roots. Also, find the roots.

(ii) Write all the values of *k* for which the quadratic equation *x*^{2} + *kx *+ 16 = 0 has equal roots. Find the roots of the equation so obtained.

#### Answer:

(i) The given quadric equation is $\left(3k+1\right){x}^{2}+2\left(k+1\right)x+1=0$, and roots are real and equal.

Then, find the value of *k.*

Here, $a=3k+1,b=2(k+1)\mathrm{and}c=1$.

As we know that $D={b}^{2}-4ac$

Putting the values of $a=3k+1,b=2(k+1)\mathrm{and}c=1$.

$D={\left[2\left(k+1\right)\right]}^{2}-4\left(3k+1\right)\left(1\right)\phantom{\rule{0ex}{0ex}}=4({k}^{2}+2k+1)-12k-4\phantom{\rule{0ex}{0ex}}=4{k}^{2}+8k+4-12k-4\phantom{\rule{0ex}{0ex}}=4{k}^{2}-4k$

The given equation will have real and equal roots, if *D* = 0

Thus, $4{k}^{2}-4k=0$

$\Rightarrow 4k(k-1)=0\phantom{\rule{0ex}{0ex}}\Rightarrow k=0\mathrm{or}k-1=0\phantom{\rule{0ex}{0ex}}\Rightarrow k=0\mathrm{or}k=1$

Therefore, the value of *k* is 0 or 1.

Now, for *k* = 0, the equation becomes

${x}^{2}+2x+1=0\phantom{\rule{0ex}{0ex}}\Rightarrow {x}^{2}+x+x+1=0\phantom{\rule{0ex}{0ex}}\Rightarrow x(x+1)+1(x+1)=0\phantom{\rule{0ex}{0ex}}\Rightarrow (x+1{)}^{2}=0\phantom{\rule{0ex}{0ex}}\Rightarrow x=-1,-1$

for *k* = 1, the equation becomes

$4{x}^{2}+4x+1=0\phantom{\rule{0ex}{0ex}}\Rightarrow 4{x}^{2}+2x+2x+1=0\phantom{\rule{0ex}{0ex}}\Rightarrow 2x(2x+1)+1(2x+1)=0\phantom{\rule{0ex}{0ex}}\Rightarrow (2x+1{)}^{2}=0\phantom{\rule{0ex}{0ex}}\Rightarrow x=-\frac{1}{2},-\frac{1}{2}$

Hence, the roots of the equation are $-1\mathrm{and}-\frac{1}{2}$.

*x*

^{2}+

*kx*+ 16 = 0

It is given that the quadratic equation has equal roots.

Therefore, Discriminant is equal to zero.

${x}^{2}+kx+16=0\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\mathrm{Discriminant}=0\phantom{\rule{0ex}{0ex}}\Rightarrow {b}^{2}-4ac=0\phantom{\rule{0ex}{0ex}}\Rightarrow {\left(k\right)}^{2}-4\left(1\right)\left(16\right)=0\phantom{\rule{0ex}{0ex}}\Rightarrow {k}^{2}-64=0\phantom{\rule{0ex}{0ex}}\Rightarrow {k}^{2}=64\phantom{\rule{0ex}{0ex}}\Rightarrow k=\pm 8$

Hence, the values of

*k*is ±8.

Now,

For

*k*= 8,

The equation becomes

*x*

^{2}+ 8

*x*+ 16 = 0

⇒ (

*x*+ 4)

^{2}= 0

⇒

*x*= −4

For

*k*= −8,

The equation becomes

*x*

^{2}− 8

*x*+ 16 = 0

⇒ (

*x*− 4)

^{2}= 0

⇒

*x*= 4

Hence, the roots of the equation so obtained are 4 and −4.

#### Page No 4.42:

#### Question 10:

Find the values of *p* for which the quadratic equation $\left(2p+1\right){x}^{2}-\left(7p+2\right)x+\left(7p-3\right)=0$ has equal roots. Also, find these roots.

#### Answer:

The given quadric equation is $\left(2p+1\right){x}^{2}-\left(7p+2\right)x+\left(7p-3\right)=0$, and roots are real and equal.

Then, find the value of *p.*

Here, $a=2p+1,b=-7p-2\mathrm{and}c=7p-3$.

As we know that $D={b}^{2}-4ac$

Putting the values of $a=2p+1,b=-7p-2\mathrm{and}c=7p-3$.

$D={\left[-\left(7p+2\right)\right]}^{2}-4\left(2p+1\right)\left(7p-3\right)\phantom{\rule{0ex}{0ex}}=(49{p}^{2}+28p+4)-4\left(14{p}^{2}-6p+7p-3\right)\phantom{\rule{0ex}{0ex}}=49{p}^{2}+28p+4-56{p}^{2}-4p+12\phantom{\rule{0ex}{0ex}}=-7{p}^{2}+24p+16$

The given equation will have real and equal roots, if *D* = 0

Thus, $-7{p}^{2}+24p+16=0$

$\Rightarrow 7{p}^{2}-24p-16=0\phantom{\rule{0ex}{0ex}}\Rightarrow 7{p}^{2}-28p+4p-16=0\phantom{\rule{0ex}{0ex}}\Rightarrow 7p(p-4)+4(p-4)=0\phantom{\rule{0ex}{0ex}}\Rightarrow (7p+4)(p-4)=0\phantom{\rule{0ex}{0ex}}\Rightarrow 7p+4=0\mathrm{or}p-4=0\phantom{\rule{0ex}{0ex}}\Rightarrow p=-\frac{4}{7}\mathrm{or}p=4$

Therefore, the value of *p* is 4 or $-\frac{4}{7}$.

Now, for *p* = 4, the equation becomes

$9{x}^{2}-30x+25=0\phantom{\rule{0ex}{0ex}}\Rightarrow 9{x}^{2}-15x-15x+25=0\phantom{\rule{0ex}{0ex}}\Rightarrow 3x(3x-5)-5(3x-5)=0\phantom{\rule{0ex}{0ex}}\Rightarrow (3x-5{)}^{2}=0\phantom{\rule{0ex}{0ex}}\Rightarrow x=\frac{5}{3},\frac{5}{3}$

for *p* = $-\frac{4}{7}$, the equation becomes

$\left(-\frac{8}{7}+1\right){x}^{2}-\left(-4+2\right)x+\left(-4-3\right)=0\phantom{\rule{0ex}{0ex}}\Rightarrow \left(\frac{-8+7}{7}\right){x}^{2}+2x-7=0\phantom{\rule{0ex}{0ex}}\Rightarrow -\frac{1}{7}{x}^{2}+2x-7=0\phantom{\rule{0ex}{0ex}}\Rightarrow -{x}^{2}+14x-49=0\phantom{\rule{0ex}{0ex}}\Rightarrow {x}^{2}-14x+49=0\phantom{\rule{0ex}{0ex}}\Rightarrow {x}^{2}-7x-7x+49=0\phantom{\rule{0ex}{0ex}}\Rightarrow x(x-7)-7(x-7)=0\phantom{\rule{0ex}{0ex}}\Rightarrow (x-7{)}^{2}=0\phantom{\rule{0ex}{0ex}}\Rightarrow x=7,7$

Hence, the roots of the equation are $\frac{5}{3}\mathrm{and}7$.

#### Page No 4.42:

#### Question 11:

If −5 is a root of the quadratic equation $2{x}^{2}+px-15=0$ and the quadratic equation $p({x}^{2}+x)+k=0$ has equal roots, find the value of *k*.

#### Answer:

The given quadratic equation is $2{x}^{2}+px-15=0$, and one root is −5.

Then, it satisfies the given equation.

$2{\left(-5\right)}^{2}+p\left(-5\right)-15=0\phantom{\rule{0ex}{0ex}}\Rightarrow 50-5p-15=0\phantom{\rule{0ex}{0ex}}\Rightarrow -5p=-35\phantom{\rule{0ex}{0ex}}\Rightarrow p=7$

The quadratic equation $p({x}^{2}+x)+k=0$, has equal roots.

Putting the value of *p*, we get

$7\left({x}^{2}+x\right)+k=0\phantom{\rule{0ex}{0ex}}\Rightarrow 7{x}^{2}+7x+k=0$

Here, $a=7,b=7\mathrm{and}c=k$.

As we know that $D={b}^{2}-4ac$

Putting the values of $a=7,b=7\mathrm{and}c=k$.

$D={\left(7\right)}^{2}-4\left(7\right)\left(k\right)\phantom{\rule{0ex}{0ex}}=49-28k$

The given equation will have real and equal roots, if *D* = 0

Thus, $49-28k=0$

$\Rightarrow 28k=49\phantom{\rule{0ex}{0ex}}\Rightarrow k=\frac{49}{28}\phantom{\rule{0ex}{0ex}}\Rightarrow k=\frac{7}{4}$

Therefore, the value of *k* is $\frac{7}{4}$.

#### Page No 4.42:

#### Question 12:

If 2 is a root of the quadratic equation $3{x}^{2}+px-8=0$ and the quadratic equation $4{x}^{2}-2px+k=0$ has equal roots, find the value of *k*.

#### Answer:

The given quadratic equation is $3{x}^{2}+px-8=0$, and one root is 2.

Then, it satisfies the given equation.

$3{\left(2\right)}^{2}+p\left(2\right)-8=0\phantom{\rule{0ex}{0ex}}\Rightarrow 12+2p-8=0\phantom{\rule{0ex}{0ex}}\Rightarrow 2p=-4\phantom{\rule{0ex}{0ex}}\Rightarrow p=-2$

The quadratic equation $4{x}^{2}-2px+k=0$, has equal roots.

Putting the value of *p*, we get

$4{x}^{2}-2(-2)x+k=0\phantom{\rule{0ex}{0ex}}\Rightarrow 4{x}^{2}+4x+k=0$

Here, $a=4,b=4\mathrm{and}c=k$.

As we know that $D={b}^{2}-4ac$

Putting the values of $a=4,b=4\mathrm{and}c=k$.

$D={\left(4\right)}^{2}-4\left(4\right)\left(k\right)\phantom{\rule{0ex}{0ex}}=16-16k$

The given equation will have real and equal roots, if *D* = 0

Thus, $16-16k=0$

$\Rightarrow 16k=16\phantom{\rule{0ex}{0ex}}\Rightarrow k=1$

Therefore, the value of *k* is 1.

#### Page No 4.42:

#### Question 13:

If 1 is a root of the quadratic equation $3{x}^{2}+ax-2=0$ and the quadratic equation $a({x}^{2}+6x)-b=0$ has equal roots, find the value of *b*.

#### Answer:

The given quadratic equation is $3{x}^{2}+ax-2=0$, and one root is 1.

Then, it satisfies the given equation.

$3{\left(1\right)}^{2}+a\left(1\right)-2=0\phantom{\rule{0ex}{0ex}}\Rightarrow 3+a-2=0\phantom{\rule{0ex}{0ex}}\Rightarrow 1+a=0\phantom{\rule{0ex}{0ex}}\Rightarrow a=-1$

The quadratic equation $a({x}^{2}+6x)-b=0$, has equal roots.

Putting the value of *a*, we get

$-1\left({x}^{2}+6x\right)-b=0\phantom{\rule{0ex}{0ex}}\Rightarrow {x}^{2}+6x+b=0$

Here, $A=1,B=6\mathrm{and}C=b$.

As we know that $D={B}^{2}-4AC$

Putting the values of $A=1,B=6\mathrm{and}C=b$.

$D={\left(6\right)}^{2}-4\left(1\right)\left(b\right)\phantom{\rule{0ex}{0ex}}=36-4b$

The given equation will have real and equal roots, if *D* = 0

Thus, $36-4b=0$

$\Rightarrow 4b=36\phantom{\rule{0ex}{0ex}}\Rightarrow b=9$

Therefore, the value of *b* is $9$.

#### Page No 4.42:

#### Question 14:

Find the value of *p* for which the quadratic equation $\left(p+1\right){x}^{2}-6(p+1)x+3(p+9)=0,p\ne -1$ has equal roots. Hence, find the roots of the equation.

**Disclaimer:** There is a misprinting in the given question. In the question '*q*' is printed instead of 9.

#### Answer:

The given quadratic equation $\left(p+1\right){x}^{2}-6(p+1)x+3(p+9)=0$, has equal roots.

Here, $a=p+1,b=-6p-6\mathrm{and}c=3p+27$.

As we know that $D={b}^{2}-4ac$

Putting the values of $a=p+1,b=-6p-6\mathrm{and}c=3p+27$.

$D={\left[-6(p+1)\right]}^{2}-4\left(p+1\right)\left[3\left(p+9\right)\right]\phantom{\rule{0ex}{0ex}}=36({p}^{2}+2p+1)-12({p}^{2}+10p+9)\phantom{\rule{0ex}{0ex}}=36{p}^{2}-12{p}^{2}+72p-120p+36-108\phantom{\rule{0ex}{0ex}}=24{p}^{2}-48p-72$

The given equation will have real and equal roots, if *D* = 0

Thus, $24{p}^{2}-48p-72=0$

$\Rightarrow {p}^{2}-2p-3=0\phantom{\rule{0ex}{0ex}}\Rightarrow {p}^{2}-3p+p-3=0\phantom{\rule{0ex}{0ex}}\Rightarrow p(p-3)+1(p-3)=0\phantom{\rule{0ex}{0ex}}\Rightarrow (p+1)(p-3)=0\phantom{\rule{0ex}{0ex}}\Rightarrow p+1=0\mathrm{or}p-3=0\phantom{\rule{0ex}{0ex}}\Rightarrow p=-1\mathrm{or}p=3$

Therefore, the value of *p* is −1, 3.

It is given that *p* ≠ −1, thus *p* = 3 only.

Now the equation becomes

$4{x}^{2}-24x+36=0\phantom{\rule{0ex}{0ex}}\Rightarrow {x}^{2}-6x+9=0\phantom{\rule{0ex}{0ex}}\Rightarrow {x}^{2}-3x-3x+9=0\phantom{\rule{0ex}{0ex}}\Rightarrow x(x-3)-3(x-3)=0\phantom{\rule{0ex}{0ex}}\Rightarrow (x-3{)}^{2}=0\phantom{\rule{0ex}{0ex}}\Rightarrow x=3,3$

â€‹Hence, the root of the equation is 3.

#### Page No 4.42:

#### Question 15:

Determine the nature of the roots of the following quadratic equations:

(i) (ii) ,$a\ne 0,b\ne 0$

(iii) (iv)

#### Answer:

(i) The given quadric equation is

Here,

As we know that

Putting the value of

Since,

Therefore, root of the given equation are.

(ii) The given quadric equation is

Here,

As we know that

Putting the value of

Since,

Therefore, root of the given equation are.

(iii) The given quadric equation is

Here,

As we know that

Putting the value of

Since,

Therefore, root of the given equation are.

(iv) The given quadric equation is

Here,

As we know that

Putting the value of

Since,

Therefore, root of the given equation are.

#### Page No 4.42:

#### Question 16:

Determine the set of values of k for which the following quadratic equation have real roots:

(i) (ii)

(iii) (iv)

#### Answer:

(i) The given quadric equation is , and roots are real

Then find the value of *k.*

Here,

As we know that

Putting the value of

The given equation will have real roots, if

Therefore, the value of

(ii) The given quadric equation is , and roots are real.

Then find the value of *k.*

Here,

As we know that

Putting the value of

The given equation will have real roots, if

Therefore, the value of

(iii) The given quadric equation is , and roots are real.

Then find the value of *k.*

Here,

As we know that

Putting the value of

The given equation will have real roots, if

Therefore, the value of

(iv) The given quadric equation is , and roots are real

Then find the value of *k.*

Here,

As we know that

Putting the value of

The given equation will have real roots, if

Since left hand side is always positive. So

Therefore, the value of

#### Page No 4.42:

#### Question 17:

If the roots of the equation $\left(b-c\right){x}^{2}+\left(c-a\right)x+\left(a-b\right)=0$ are equal, then prove that 2*b* = *a* + *c*.

#### Answer:

The given quadric equation is, and roots are real

Then prove that*.*

Here,

As we know that

Putting the value of

As we know that

The given equation will have real roots, if

Square root both side we get

Hence

#### Page No 4.43:

#### Question 18:

If the roots of the equation (*a*^{2} +* **b*^{2})*x*^{2} − 2 (*ac* + *bd*)*x* + (*c*^{2} + *d*^{2}) = 0 are equal, prove that $\frac{a}{b}=\frac{c}{d}$.

#### Answer:

The given quadric equation is, and roots are real

Then prove that*.*

Here,

As we know that

Putting the value of

The given equation will have real roots, if

Square root both sides we get,

Hence

#### Page No 4.43:

#### Question 19:

If the roots of the equations *ax*^{2} + 2*bx* + *c* = 0 and $b{x}^{2}-2\sqrt{ac}x+b=0$ are simultaneously real, then prove that *b*^{2} = *ac*.

#### Answer:

The given equations are

…... (1)

…… (2)

Roots are simultaneously real

Then prove that*.*

Let be the discriminants of equation (1) and (2) respectively,

Then,

And

Both the given equation will have real roots, if

…… (3)

…... (4)

From equations (3) and (4) we get

Hence,

#### Page No 4.43:

#### Question 20:

If *p*,* q* are real and *p* ≠ *q*, then show that the roots of the equation (*p* −* q*) *x*^{2} + 5(*p* + *q*) *x* − 2(*p* − *q*) = 0 are real and unequal.

#### Answer:

The quadric equation is

Here,

As we know that

Putting the value of

$D={\left\{5\left(p+q\right)\right\}}^{2}-4\left(p-q\right)\left(-2\left(p-q\right)\right)\phantom{\rule{0ex}{0ex}}=25\left({p}^{2}+2pq+{q}^{2}\right)+8\left({p}^{2}-2pq+{q}^{2}\right)\phantom{\rule{0ex}{0ex}}=25{p}^{2}+50pq+25{q}^{2}+8{p}^{2}-16pq+8{q}^{2}\phantom{\rule{0ex}{0ex}}=33{p}^{2}+34pq+33{q}^{2}$

Since*, P* and *q* are real and * *, therefore, the value of .

Thus, the roots of the given equation are real and unequal.

Hence, proved

#### Page No 4.43:

#### Question 21:

If the roots of the equation $\left({c}^{2}-ab\right){x}^{2}-2\left({a}^{2}-bc\right)x+{b}^{2}-ac=0$ are equal, prove that either *a* = 0 or *a*^{3} + *b*^{3} + *c*^{3} = 3*abc*.

#### Answer:

The given quadric equation is, and roots are equal.

Then prove that either or

Here,

As we know that

Putting the value of

The given equation will have real roots, if

So, either

Hence

#### Page No 4.43:

#### Question 22:

Show that the equation $2\left({a}^{2}+{b}^{2}\right){x}^{2}+2\left(a+b\right)x+1=0$ has no real roots, when *a* ≠ *b*.

#### Answer:

The quadric equation is

Here,

As we know that

Putting the value of

We have,

Thus, the value of

Therefore, the roots of the given equation are not real

Hence, proved

#### Page No 4.43:

#### Question 23:

Prove that both the roots of the equation $\left(x-a\right)\left(x-b\right)+\left(x-b\right)\left(x-c\right)+\left(x-c\right)\left(x-a\right)=0$ are real but they are equal only when *a* = *b* = *c*.

#### Answer:

The quadric equation is

Here,

After simplifying the equation

As we know that

Putting the value of

Since, . So the solutions are real

Let * *

Then

Thus, the value of

Therefore, the roots of the given equation are real and but they are equal only when,

Hence proved

#### Page No 4.43:

#### Question 24:

If *a*, *b*, *c* are real numbers such that *ac* ≠ 0, then show that at least one of the equations *ax*^{2} + *bx* + *c* = 0 and −*ax*^{2} + *bx* + *c* = 0 has real roots.

#### Answer:

The given equations are

…... (1)

…… (2)

Roots are simultaneously real

Let be the discriminants of equation (1) and (2) respectively,

Then,

And

Both the given equation will have real roots, if.

Thus,

${b}^{2}-4ac\ge 0$

${b}^{2}\ge 4ac$ ...... (3)

And,

…... (4)

Now given that are real number and as well as from equations (3) and (4) we get

At least one of the given equation has real roots

Hence, proved

#### Page No 4.43:

#### Question 25:

If the equation $\left(1+{m}^{2}\right){x}^{2}+2mcx+\left({c}^{2}-{a}^{2}\right)=0$ has equal roots, prove that *c*^{2} = *a*^{2}(1 + *m*^{2}).

#### Answer:

The given equation, has equal roots

Then prove that*.*

Here,

As we know that

Putting the value of

The given equation will have real roots, if

Hence,

#### Page No 4.5:

#### Question 4:

Determine. if 3 is a root of the equation given below:

$\sqrt{{x}^{2}-4x+3}+\sqrt{{x}^{2}-9}=\sqrt{4{x}^{2}-14x+16}$

#### Answer:

We have been given that,

We have to check whether *x* = 3 is the solution of the given equation or not.

Now, if is a root of the above quadratic equation, then it should satisfy the whole. So substituting in the above equation, we have,

Now since, we can see from above that left hand side and right hand side are not equal. Therefore is not the solution of the given quadratic equation.

#### Page No 4.5:

#### Question 5:

If *x* = 2/3 and *x* = −3 are the roots of the equation *ax*^{2} + 7*x* + *b* = 0, find the values of *a* and *b*.

#### Answer:

We have been given that,

We have to find *a *and *b*

Now, if is a root of the equation, then it should satisfy the equation completely. Therefore we substitute in the above equation. We get,

…… (1)

Also, if is a root of the equation, then it should satisfy the equation completely. Therefore we substitute in the above equation. We get,

…… (2)

Now, we multiply equation (2) by 9 and then subtract equation (1) from it. So we have,

Now, put this value of ‘*a*’ in equation (2) in order to get the value of ‘*b*’. So,

Therefore, we have and .

#### Page No 4.51:

#### Question 1:

Find two consecutive numbers whose squares have the sum 85.

#### Answer:

Let two consecutive numbers be *x *and

Then according to question

Or

Since, *x *being a number,

Therefore,

When* *then

And when* *then

Thus, two consecutive number be either

#### Page No 4.51:

#### Question 2:

Divide 29 into two parts so that the sum of the squares of the parts is 425.

#### Answer:

Let first numbers be *x *and other

Then according to question

Or

Since, 29* *being a positive number, so *x* cannot be negative.

Therefore,

When* *then

Thus, two consecutive number be

#### Page No 4.51:

#### Question 3:

Two squares have sides *x* cm and (*x* + 4) cm. The sum of their areas is 656 cm^{2}. Find the sides of the squares.

#### Answer:

Given that the sides of two square be *x *cm* *and

Then according to question

Or

Since, sides of the squares* *being a positive, so *x* cannot be negative.

Therefore,

When* *then

Thus, sides of the squares* *be

#### Page No 4.51:

#### Question 4:

The sum of two numbers is 48 and their product is 432. Find the numbers.

#### Answer:

Let first numbers be *x *and other

Then according to question

Or

Thus, two number be

#### Page No 4.51:

#### Question 5:

If an integer is added to its square, the sum is 90. Find the integer with the help of quadratic equation.

#### Answer:

Let an integer be *x*.

Then according to question

Or

Thus, an integer be

#### Page No 4.51:

#### Question 6:

Find the whole number which when decreased by 20 is equal to 69 times the reciprocal of the number.

#### Answer:

Let the whole numbers be *x*.

Then according to question

Or

Since, whole numbers being a positive, so *x* cannot be negative.

Thus, whole numbers be

#### Page No 4.51:

#### Question 7:

Find two consecutive natural numbers whose product is 20.

#### Answer:

Let two consecutive numbers be *x *and

Then according to question

Or

Since, *x *being a natural number,

Therefore negative value is not possible

So when* *then

Thus, two consecutive numbers are

#### Page No 4.51:

#### Question 8:

The sum of squares of two consecutive odd positive integers is 394. Find them.

#### Answer:

Let two consecutive odd positive integer be * *and other

Then according to question

Since, *x *being a positive number, so *x* cannot be negative.

Therefore,

When* *then odd positive

And

Thus, two consecutive odd positive integer be

#### Page No 4.51:

#### Question 9:

The sum of two numbers is 8 and 15 times the sum of their reciprocals is also 8. Find the numbers.

#### Answer:

Let first numbers be *x *and other

Then according to question

Or

Thus, two consecutive number be

#### Page No 4.51:

#### Question 10:

The sum of a number and its positive square root is 6/25. Find the number.

#### Answer:

Let first numbers be *x *

Then according to question

Let then

Or

Since, being a positive number, so *y *cannot be negative.

Therefore,

Thus, the required number be

#### Page No 4.51:

#### Question 11:

The sum of a number and its square is 63/4, find the numbers.

#### Answer:

Let first numbers be *x*

Then according to question

Let then

Or

Thus, the required number be

#### Page No 4.52:

#### Question 12:

There are three consecutive integers such that the square of the first increased by the product of the other two gives 154. What are the integers?

#### Answer:

Let three consecutive integer be * *and

Then according to question

Or

Since, *x *being a positive number, so *x* cannot be negative.

Therefore,

When* *then other positive integer

And

Thus, three consecutive positive integer be

#### Page No 4.52:

#### Question 13:

The product of two successive integral multiples of 5 is 300. Determine the multiples.

#### Answer:

Let the successive integer multiples of 5 be

Then according to question

Therefore,

Or

When* * then integer

And when then integer

Thus, three consecutive positive integer be

#### Page No 4.52:

#### Question 14:

The sum of the squares of two numbers is 233 and one of the number is 3 less than twice the other number. Find the numbers.

#### Answer:

Let the numbers be integers. One of the numbers be *x*. So, the other will be .

Then according to question,

Or

Since, we have assumed the numbers to be integers, so *x* cannot be a rational number/fraction.

Therefore, for *x* = 8

Other number =

Thus, whole numbers be .

#### Page No 4.52:

#### Question 15:

Find the consecutive even integers whose squares have the sum 340.

#### Answer:

Let two consecutive even integer be * *and other

Then according to question

Since, *x *being a positive number, so *x* cannot be negative.

Therefore,

When* *then even integer

And

Thus, two consecutive odd positive integer be

#### Page No 4.52:

#### Question 16:

The difference of two numbers is 4. If the difference of their reciprocal is $\frac{4}{21}$, find the numbers.

#### Answer:

Let one numbers be *x *then other.

Then according to question

By cross multiplication

Or

Since, *x *being a number,

Therefore,

When* *then

And when* *then

Thus, two consecutive number be either

#### Page No 4.52:

#### Question 17:

Find two natural numbers which differ by 3 and whose squares have the sum 17.

#### Answer:

Let one natural number be * *and other.

Then according to question

Since, *x *being a natural number, so *x* cannot be negative.

Therefore,

When* *then even integer

Thus, two natural number be

#### Page No 4.52:

#### Question 18:

The sum of the squares of three consecutive natural number is 149. Find the numbers.

#### Answer:

Let three consecutive integer be * *and

Then according to question

Or

Since, *x *being a positive number, so *x* cannot be negative.

Therefore,

When* *then other positive integer

And

Thus, three consecutive positive integer be

#### Page No 4.52:

#### Question 19:

The sum of two numbers is 16. The sum of their reciprocals is 1/3. Find the numbers.

#### Answer:

Let one numbers be *x *then other.

Then according to question

By cross multiplication

Or

Since, *x *being a number,

Therefore,

When* *then

Thus, two consecutive number be either

#### Page No 4.52:

#### Question 20:

Determine two consecutive multiples of 3 whose product is 270.

#### Answer:

Let the required number be * *and

Then according to question

Or

Since, *x *being a positive number, so *x* cannot be negative.

Therefore,

When* *then positive integer

And

Thus, three consecutive positive integer be

#### Page No 4.52:

#### Question 21:

The sum of a number and its reciprocal is 17/4. Find the number.

#### Answer:

Let a numbers be *x *and its reciprocal is

Then according to question

By cross multiplication

Or

Thus, two consecutive number be either

#### Page No 4.52:

#### Question 22:

A two-digit number is such that the product of its digits is 8. When 18 is subtracted from the number, the digits interchange their places. Find the number.

#### Answer:

Let the tens digit be , then, the unit digits

Therefore, number

And number obtained by interchanging the digits

Then according to question

Or

So the digit can never be negative.

Therefore,

When* *then the unit digits is

And therefore the number is

Thus, the required number be

#### Page No 4.52:

#### Question 23:

A two-digit number is such that the product of digit is 12. When 36 is added to the number the digits interchange their places. Determine the number.

#### Answer:

Let the tens digit be then, the unit digits

Therefore, number

And number obtained by interchanging the digits

Then according to question

Or

So, the digit can never be negative.

Therefore,

When* *then the unit digits

.

And number

Thus, the required number be

#### Page No 4.52:

#### Question 24:

A natural number when increased by 12 equals 160 times its reciprocal. Find the number.

#### Answer:

Let the natural number be N.

Now, $\left(N+12\right)=\frac{160}{N}$

$\Rightarrow {N}^{2}+12N-160=0\phantom{\rule{0ex}{0ex}}\Rightarrow N\left(N+20\right)-8\left(N+20\right)=0\phantom{\rule{0ex}{0ex}}\Rightarrow N=8(N=-20;\mathrm{Not}\mathrm{a}\mathrm{natural}\mathrm{number})$

Hence, required natural number is 8.

#### Page No 4.52:

#### Question 25:

Two number differ by 3 and their product is 504. Find the numbers.

#### Answer:

Let two required numbers be *x *and

Then according to question

Or

Since, *x *being a number,

Therefore,

When* *then

And when* *then

Thus, two consecutive number be either

#### Page No 4.52:

#### Question 26:

Two number differ by 4 and their product is 192. Find the numbers.

#### Answer:

Let two required numbers be *x *and

Then according to question

Or

Since, *x *being a number,

Therefore,

When* *then

And when* *then

Thus, two consecutive number be either

#### Page No 4.52:

#### Question 27:

A two digit number is 4 times the sum of its digits and twice the product of its digits. Find the number.

#### Answer:

Let the require digit be

Then according to question

.….. (1)

And, .…..(2)

Now putting the value of *y* in equation (2) from (1)

So, either

Or

So, the digit can never be negative.

When* *then

Therefore, number

Thus, the required number be

#### Page No 4.52:

#### Question 28:

The difference of the squares of two positive integers is 180. The square of the smaller number is 8 times the larger, find the numbers.

#### Answer:

Let the larger number be *x*.

Then according to the question,

Square of the smaller number = 8*x*, then

${x}^{2}-8x=180\phantom{\rule{0ex}{0ex}}\Rightarrow {x}^{2}-8x-180=0\phantom{\rule{0ex}{0ex}}\Rightarrow {x}^{2}-18x+10x-180=0\phantom{\rule{0ex}{0ex}}\Rightarrow x(x-18)+10(x-18)=0\phantom{\rule{0ex}{0ex}}\Rightarrow (x+10)(x-18)=0\phantom{\rule{0ex}{0ex}}\Rightarrow x+10=0\mathrm{or}x-18=0\phantom{\rule{0ex}{0ex}}\Rightarrow x=-10\mathrm{or}x=18$

Since, *x *being a positive integer so, *x* cannot be negative,

Therefore, larger number = 18.

then the smaller number = $\sqrt{8\times 18}=12$

Thus, the two positive numbers are 12 and 18.

#### Page No 4.52:

#### Question 29:

The sum of two numbers is 18. The sum of their reciprocals is 1/4. Find the numbers.

#### Answer:

Let one of the number be *x *then other number is (18 − *x*).

Then according to question,

$\frac{1}{x}+\frac{1}{18-x}=\frac{1}{4}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{18-x+x}{x(18-x)}=\frac{1}{4}\phantom{\rule{0ex}{0ex}}\Rightarrow 18\times 4=18x-{x}^{2}\phantom{\rule{0ex}{0ex}}\Rightarrow 72=18x-{x}^{2}\phantom{\rule{0ex}{0ex}}\Rightarrow {x}^{2}-18x+72=0\phantom{\rule{0ex}{0ex}}\Rightarrow {x}^{2}-12x-6x+72=0\phantom{\rule{0ex}{0ex}}\Rightarrow x(x-12)-6(x-12)=0\phantom{\rule{0ex}{0ex}}\Rightarrow (x-6)(x-12)=0\phantom{\rule{0ex}{0ex}}\Rightarrow x-6=0\mathrm{or}x-12=0\phantom{\rule{0ex}{0ex}}\Rightarrow x=6\mathrm{or}x=12$

Since, *x *being a number,

Therefore,

When* $x=12$ *then another number will be

$18-x=18-12=6$

And when* $x=6$ *then another number will be

$18-x=18-6=12$

Thus, the two numbers are 6 and 12.

#### Page No 4.52:

#### Question 30:

The sum of two number *a* and *b* is 15, and the sum of their reciprocals $\frac{1}{a}\mathrm{and}\frac{1}{b}$ is 3/10. Find the numbers *a* and *b*.

#### Answer:

Given that be two numbers in such a way that.

Then according to question

By cross multiplication

…. (1)

Now putting the value of *b* in equation (1)

Or

Therefore,

When* *then

And when* *then

Thus, two consecutive number be either

#### Page No 4.52:

#### Question 31:

The sum of two numbers is 9. The sum of their reciprocals is 1/2. Find the numbers.

#### Answer:

Let one numbers be *x *then other (9 − *x*).

Then according to question

$\frac{1}{x}+\frac{1}{\left(9-x\right)}=\frac{1}{2}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{\left(9-x\right)+x}{x\left(9-x\right)}=\frac{1}{2}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{9}{x\left(9-x\right)}=\frac{1}{2}$

By cross multiplication

$\Rightarrow 18=x\left(9-x\right)\phantom{\rule{0ex}{0ex}}\Rightarrow {x}^{2}-9x+18=0\phantom{\rule{0ex}{0ex}}\Rightarrow {x}^{2}-6x-3x+18=0\phantom{\rule{0ex}{0ex}}\Rightarrow \left(x-6\right)x-3\left(x-6\right)=0\phantom{\rule{0ex}{0ex}}\Rightarrow \left(x-6\right)\left(x-3\right)=0\phantom{\rule{0ex}{0ex}}\Rightarrow x=6,3$

Since, *x *being a number,

Therefore,

When x = 6 then

(9 − *x*) = (9 − 6) = 3

When x = 3 then

(9 − *x*) = (9 − 3) = 6

Thus, two consecutive number be either 3, 6 or 6, 3.

#### Page No 4.52:

#### Question 32:

Three consecutive positive integers are such that the sum of the square of the first and the product of other two is 46, find the integers.

#### Answer:

Let three consecutive integer be * *and

Then according to question

Or

Since, *x *being a positive number, so *x* cannot be negative.

Therefore,

When* *then other positive integer

And

Thus, three consecutive positive integer be

#### Page No 4.52:

#### Question 33:

The difference of squares of two number is 88. If the larger number is 5 less than twice the smaller number, then find the two numbers.

#### Answer:

Let the smaller numbers be *x*

Then according to question,

The larger number be, then

Or

Since, *x *being a positive integer so, *x* cannot be negative,

Therefore,

When* *then larger number be

Thus, two consecutive number be either

#### Page No 4.52:

#### Question 34:

Find two consecutive odd positive integers, sum of whose squares is 970.

#### Answer:

Let one of the number be *x** *then other number is *x* + 2.

Then according to question,

${x}^{2}+{\left(x+2\right)}^{2}=970\phantom{\rule{0ex}{0ex}}\Rightarrow {x}^{2}+{x}^{2}+4x+4=970\phantom{\rule{0ex}{0ex}}\Rightarrow 2{x}^{2}+4x-966=0\phantom{\rule{0ex}{0ex}}\Rightarrow {x}^{2}+2x-483=0\phantom{\rule{0ex}{0ex}}\Rightarrow {x}^{2}+23x-21x-483=0\phantom{\rule{0ex}{0ex}}\Rightarrow x(x+23)-21(x+23)=0\phantom{\rule{0ex}{0ex}}\Rightarrow (x-21)(x+23)=0\phantom{\rule{0ex}{0ex}}\Rightarrow x-21=0\mathrm{or}x+23=0\phantom{\rule{0ex}{0ex}}\Rightarrow x=21\mathrm{or}x=-23$

Since, *x *being an odd positive integer,

Therefore, *x *= 21.

Then another number will be

$x+2=21+2=23$

Thus, the two consecutive odd positive integers are 21 and 23.

#### Page No 4.52:

#### Question 35:

The difference of two natural numbers is 3 and the difference of their reciprocals is $\frac{3}{28}$. Find the numbers.

#### Answer:

Let the smaller number be *x** *then the other number be 3 + *x*.

Then according to question,

$\frac{1}{x}-\frac{1}{3+x}=\frac{3}{28}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{3+x-x}{x(3+x)}=\frac{3}{28}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{3}{3x+{x}^{2}}=\frac{3}{28}\phantom{\rule{0ex}{0ex}}\Rightarrow 28=3x+{x}^{2}\phantom{\rule{0ex}{0ex}}\Rightarrow {x}^{2}+3x-28=0\phantom{\rule{0ex}{0ex}}\Rightarrow {x}^{2}+7x-4x-28=0\phantom{\rule{0ex}{0ex}}\Rightarrow x(x+7)-4(x+7)=0\phantom{\rule{0ex}{0ex}}\Rightarrow (x-4)(x+7)=0\phantom{\rule{0ex}{0ex}}\Rightarrow x-4=0\mathrm{or}x+7=0\phantom{\rule{0ex}{0ex}}\Rightarrow x=4\mathrm{or}x=-7$

Since, *x *being a natural number,

Therefore, *x *= 4.

Then another number will be

$3+x=3+4=7$

Thus, the two natural numbers are 7 and 4.

#### Page No 4.52:

#### Question 36:

The sum of the squares of two consecutive odd numbers is 394. Find the numbers.

#### Answer:

Let one of the number be *x** *then the other number be *x* + 2.

Then according to question,

${x}^{2}+{\left(x+2\right)}^{2}=394\phantom{\rule{0ex}{0ex}}\Rightarrow {x}^{2}+{x}^{2}+4x+4=394\phantom{\rule{0ex}{0ex}}\Rightarrow 2{x}^{2}+4x-390=0\phantom{\rule{0ex}{0ex}}\Rightarrow {x}^{2}+2x-195=0\phantom{\rule{0ex}{0ex}}\Rightarrow {x}^{2}+15x-13x-195=0\phantom{\rule{0ex}{0ex}}\Rightarrow x(x+15)-13(x+15)=0\phantom{\rule{0ex}{0ex}}\Rightarrow (x-13)(x+15)=0\phantom{\rule{0ex}{0ex}}\Rightarrow x-13=0\mathrm{or}x+15=0\phantom{\rule{0ex}{0ex}}\Rightarrow x=13\mathrm{or}x=-15$

Since, *x *being an odd number,

Therefore, *x *= 13.

Then another number will be

$x+2=13+2=15$

Thus, the two consecutive odd numbers are 13 and 15.

#### Page No 4.52:

#### Question 37:

The sum of the squares of two consecutive multiples of 7 is 637. Find the multiples.

#### Answer:

Let one of the number be 7*x** *then the other number be 7(*x* + 1).

Then according to question,

${\left(7x\right)}^{2}+{\left[7\left(x+1\right)\right]}^{2}=637\phantom{\rule{0ex}{0ex}}\Rightarrow 49{x}^{2}+49({x}^{2}+2x+1)=637\phantom{\rule{0ex}{0ex}}\Rightarrow 49{x}^{2}+49{x}^{2}+98x+49-637=0\phantom{\rule{0ex}{0ex}}\Rightarrow 98{x}^{2}+98x-588=0\phantom{\rule{0ex}{0ex}}\Rightarrow {x}^{2}+x-6=0\phantom{\rule{0ex}{0ex}}\Rightarrow {x}^{2}+3x-2x-6=0\phantom{\rule{0ex}{0ex}}\Rightarrow x(x+3)-2(x+3)=0\phantom{\rule{0ex}{0ex}}\Rightarrow (x-2)(x+3)=0\phantom{\rule{0ex}{0ex}}\Rightarrow x-2=0\mathrm{or}x+3=0\phantom{\rule{0ex}{0ex}}\Rightarrow x=2\mathrm{or}x=-3$

Since, the numbers are multiples of 7,

Therefore, one number = 7 × 2 =14.

Then another number will be

$7(x+1)=7\times 3=21$

Thus, the two consecutive multiples of 7 are 14 and 21.

#### Page No 4.53:

#### Question 38:

The sum of the squares of two consecutive even numbers is 340. Find the numbers.

#### Answer:

Let one of the number be *x** *then the other number be *x* + 2.

Then according to question,

${x}^{2}+{\left(x+2\right)}^{2}=340\phantom{\rule{0ex}{0ex}}\Rightarrow {x}^{2}+{x}^{2}+4x+4=340\phantom{\rule{0ex}{0ex}}\Rightarrow 2{x}^{2}+4x-336=0\phantom{\rule{0ex}{0ex}}\Rightarrow {x}^{2}+2x-168=0\phantom{\rule{0ex}{0ex}}\Rightarrow {x}^{2}+14x-12x-168=0\phantom{\rule{0ex}{0ex}}\Rightarrow x(x+14)-12(x+14)=0\phantom{\rule{0ex}{0ex}}\Rightarrow (x-12)(x+14)=0\phantom{\rule{0ex}{0ex}}\Rightarrow x-12=0\mathrm{or}x+14=0\phantom{\rule{0ex}{0ex}}\Rightarrow x=12\mathrm{or}x=-14$

Since, *x *being an even number,

Therefore, *x *= 12.

Then another number will be

$x+2=12+2=14$

Thus, the two consecutive even numbers are 12 and 14.

#### Page No 4.53:

#### Question 39:

The numerator of a fraction is 3 less than the denominator. If 2 is added to both the numerator and the denominator, then the sum of the new fraction and the original fraction is $\frac{29}{20}$. Find the original fraction.

#### Answer:

Let the denominator of the original fraction be *x** *then the numerator be *x* − 3.

Then according to question,

$\frac{x-3}{x}+\frac{x-3+2}{x+2}=\frac{29}{20}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{x-3}{x}+\frac{x-1}{x+2}=\frac{29}{20}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{\left(x-3\right)\left(x+2\right)+\left(x-1\right)x}{x\left(x+2\right)}=\frac{29}{20}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{{x}^{2}-3x+2x-6+{x}^{2}-x}{{x}^{2}+2x}=\frac{29}{20}\phantom{\rule{0ex}{0ex}}\Rightarrow 20(2{x}^{2}-2x-6)=29({x}^{2}+2x)\phantom{\rule{0ex}{0ex}}\Rightarrow 40{x}^{2}-40x-120-29{x}^{2}-58x=0\phantom{\rule{0ex}{0ex}}\Rightarrow 11{x}^{2}-98x-120=0\phantom{\rule{0ex}{0ex}}\Rightarrow 11{x}^{2}-110x+12x-120=0\phantom{\rule{0ex}{0ex}}\Rightarrow 11x(x-10)+12(x-10)=0\phantom{\rule{0ex}{0ex}}\Rightarrow (11x+12)(x-10)=0\phantom{\rule{0ex}{0ex}}\Rightarrow 11x+12=0\mathrm{or}x-10=0\phantom{\rule{0ex}{0ex}}\Rightarrow x=-\frac{12}{11}\mathrm{or}x=10$

Since, *x *being an integer,

Therefore, *x *= 10.

Then the numerator will be

$x-3=10-3=7$

Thus, the original fraction is $\frac{7}{10}$.

#### Page No 4.53:

#### Question 40:

Find a natural number whose square diminished by 84 is equal to thrice of 8 more than the given number.

#### Answer:

Let the given number be *x.*

Given that the square of a natural number diminished by 84 is equal to thrice of 8 more than the given number.

$\Rightarrow {x}^{2}-84=3\left(8+x\right)\phantom{\rule{0ex}{0ex}}\Rightarrow {x}^{2}-84=24+3x\phantom{\rule{0ex}{0ex}}\Rightarrow {x}^{2}-3x-108=0\phantom{\rule{0ex}{0ex}}\Rightarrow {x}^{2}-12x+9x-108=0\phantom{\rule{0ex}{0ex}}\Rightarrow x(x-12)+9(x-12)=0\phantom{\rule{0ex}{0ex}}\Rightarrow (x+9)(x-12)=0\phantom{\rule{0ex}{0ex}}\Rightarrow x=-9\mathrm{or}x=12$

We ignore the negative value as we have taken natural numbers under consideration.

Hence, *x* = 12 is the required number.

#### Page No 4.58:

#### Question 1:

The speed of a boat in still water is 8 km/hr. It can go 15 km upstream and 22 km downstream in 5 hours. Find the speed of the stream.

#### Answer:

Let the speed of stream be then

Speed downstream.

Therefore, Speed upstream

Time taken by the boat to go upstream

Time taken by the boat to returns downstream

Now it is given that the boat returns to the same point in 5 hr.

So,

$\frac{296-7x}{64-{x}^{2}}=5$

$5{x}^{2}-7x+296-320=0$

$5{x}^{2}-7x-24=0$

$5{x}^{2}-15x+8x-24=0$

$5x\left(x-3\right)+8\left(x-3\right)=0$

$\left(x-3\right)\left(5x+8\right)=0$

$x=3,x=-\frac{8}{5}$

But, the speed of the stream can never be negative.

Hence, the speed of the stream is

#### Page No 4.58:

#### Question 2:

A train , travelling at a uniform speed for 360 km , would have taken 48 minutes less to travel the same distance if its speed were 5 km/hr more. Find the original speed of the train .

#### Answer:

*Let the speed of train be *x* km/h*

Distance to be travelled = 360 km

We know that,

Time take by the train intially when it was travelling with uniform speed of *x* km/h =

If the speed was increased by 5 km/h

Time taken by train =

With increased speed, the time taken is 48 min less. So, difference in time will be

Hence, the speed of train = 45 km/hr

#### Page No 4.58:

#### Question 3:

A fast train takes one hour less than a slow train for a journey of 200 km. If the speed of the slow train is 10 km/hr less than that of the fast train, find the speed of the two trains.

#### Answer:

Let the speed of the fast train bethen

the speed of the slow train be

Time taken by the fast train to cover

Time taken by the slow train to cover

Therefore,

$\frac{200}{\left(x-10\right)}-\frac{200}{x}=1\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{200x-200\left(x-10\right)}{x\left(x-10\right)}=1\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{200x-220x+2000}{{x}^{2}-10x}=1\phantom{\rule{0ex}{0ex}}\Rightarrow {x}^{2}-10x=2000\phantom{\rule{0ex}{0ex}}\Rightarrow {x}^{2}-10x-2000=0$

$\Rightarrow {x}^{2}-50x+40x-2000=0\phantom{\rule{0ex}{0ex}}\Rightarrow x\left(x-50\right)+40\left(x-50\right)=0\phantom{\rule{0ex}{0ex}}\Rightarrow \left(x-50\right)\left(x+40\right)=0$

So, either

Or

But, the speed of the train can never be negative.

Thus, when then

Hence, the speed of the fast train is

and the speed of the slow train is respectively.

#### Page No 4.58:

#### Question 4:

A passenger train takes one hour less for a journey of 150 km if its speed is increased by 5 km/hr from its usual speed. Find the usual speed of the train.

#### Answer:

Let the usual speed of train bethen

Increased speed of the train

Time taken by the train under usual speed to cover

Time taken by the train under increased speed to cover

Therefore,

So, either

Or

But, the speed of the train can never be negative.

Hence, the usual speed of train is

#### Page No 4.58:

#### Question 5:

The time taken by a person to cover 150 km was 2.5 hrs more than the time taken in the return journey. If he returned at a speed of 10 km/hr more than the speed of going, what was the speed per hour in each direction?

#### Answer:

Let the ongoing speed of person be. Then,

Returning speed of the person is.

Time taken by the person in going direction to cover

Time taken by the person in returning direction to cover

Therefore,

So, either

Or

But, the speed of the train can never be negative.

Thus, when then

Hence, ongoing speed of person is

and returning speed of the person is respectively.

#### Page No 4.58:

#### Question 6:

A plane left 40 minutes late due to bad weather and in order to reach its destination, 1600 km away in time, it had to increase its speed by 400 km/hr from its usual speed. Find the usual speed of the plane.

#### Answer:

Let the usual speed of plane be. Then,

Increased speed of the plane

Time taken by the plane under usual speed to cover

Time taken by the plane under increased speed to cover

Therefore,

So, either

Or

But, the speed of the plane can never be negative.

Hence, the usual speed of train is

#### Page No 4.58:

#### Question 7:

An aeroplane take 1 hour less for a journey of 1200 km if its speed is increased by 100 km/hr from its usual speed. Find its usual speed.

#### Answer:

Let the usual speed of aero plane be. Then,

Increased speed of the aero plane

Time taken by the aero plane under usual speed to cover

Time taken by the aero plane under increased speed to cover

Therefore,

So, either

Or

But, the speed of the aero plane can never be negative.

Hence, the usual speed of train is

#### Page No 4.58:

#### Question 8:

A train travels at a certain average speed for a distance 63 km and then travels a distance of 72 km at an average speed of 6 km/hr more than the original speed . If it takes 3 hours to complete total journey , what is its original average speed ?

#### Answer:

*Let the original speed of the train be *x.

For a distance of 63 km, let the speed be x km/h.

So, time = $\frac{63}{x}$

For a distance of 72 km, speed = 6 + x km/h

Time = $\frac{72}{x+6}$

Total time = 3 hours

$\frac{63}{x}+\frac{72}{x+6}=3\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{63x+378+72x}{x\left(x+6\right)}=3\phantom{\rule{0ex}{0ex}}\Rightarrow 63x+378+72x=3{x}^{2}+18x\phantom{\rule{0ex}{0ex}}\Rightarrow 3{x}^{2}-117x-378=0\phantom{\rule{0ex}{0ex}}\Rightarrow \left(3x-126\right)\left(x+3\right)=0\phantom{\rule{0ex}{0ex}}\Rightarrow x=-3,42$

So, the original speed = 42 km/h.

#### Page No 4.59:

#### Question 9:

A train covers a distance of 90 km at a uniform speed. Had the speed been 15 km/hour more, it would have taken 30 minutes less for a journey. Find the original speed of the train.

#### Answer:

Let the original speed of train be. Then,

Increased speed of the train

Time taken by the train under usual speed to cover

Time taken by the train under increased speed to cover

Therefore,

So, either

Or

But, the speed of the train can never be negative.

Hence, the original speed of train is

#### Page No 4.59:

#### Question 10:

A train travels 360 km at a uniform speed. If the speed had been 5 km/hr more, it would have taken 1 hour less for the same journey. Find the speed of the train.

#### Answer:

Let the original speed of train be. Then,

Increased speed of the train

Time taken by the train under usual speed to cover

Time taken by the train under increased speed to cover

Therefore,

So, either

Or

But, the speed of the train can never be negative.

Hence, the original speed of train is

#### Page No 4.59:

#### Question 11:

An express train takes 1 hour less than a passenger train to travel 132 km between Mysore and Bangalore (without taking into consideration the time they stop at intermediate stations). If the average speed of the express train is 11 km/hr more than that of the passenger train, find the average speed of two trains.

#### Answer:

Let the speed of the passenger train be. Then,

Speed of the express train

Time taken by the passenger train to cover between Mysore to Bangalore

Time taken by the express train to cover between Mysore to Bangalore

Therefore,

So, either

Or

But, the speed of the train can never be negative.

Thus, when then speed of express train

Hence, the speed of the passenger train is

and the speed of the express train is respectively.

#### Page No 4.59:

#### Question 12:

An aeroplane left 50 minutes later than its scheduled time, and in order to reach the destination, 1250 km away, in time, it had to increase its speed by 250 km/hr from its usual speed. Find its usual speed.

#### Answer:

Let the usual speed of aero plane be. Then,

Increased speed of the aero plane

Time taken by the aero plane under usual speed to cover

Time taken by the aero plane under increased speed to cover

Therefore,

So, either

Or

But, the speed of the aero plane can never be negative.

Hence, the usual speed of train is

#### Page No 4.59:

#### Question 13:

While boarding an aeroplane, a passenger got hurt. The pilot showing promptness and concern, made arrangements to hospitalise the injured and so the plane started late by 30 minutes to reach the destination, 1500 km away in time, the pilot increased the speed by 100 km/hr. Find the original speed/hour of the plane.

#### Answer:

Let the original speed of the plane be *x* km/hr.

Increased speed of the plane = (*x* + 100) km/hr.

Total Distance = 1500 km.

We know that, $\mathrm{Time}=\frac{\mathrm{Distance}}{\mathrm{Speed}}$

Time taken to reach the destination at original speed = *t*_{1} = $\frac{1500}{x}$ hr

Time taken to reach the destination at increasing speed = *t*_{2} = $\frac{1500}{x+100}$ hr

According to the question,

*t*_{1} − *t*_{2} = 30 min

$\Rightarrow \frac{1500}{x}-\frac{1500}{x+100}=\frac{30}{60}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{1500(x+100)-1500x}{x(x+100)}=\frac{1}{2}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{1500x+150000-1500x}{{x}^{2}+100x}=\frac{1}{2}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{150000}{{x}^{2}+100x}=\frac{1}{2}\phantom{\rule{0ex}{0ex}}\Rightarrow 300000={x}^{2}+100x\phantom{\rule{0ex}{0ex}}\Rightarrow {x}^{2}+100x-300000=0\phantom{\rule{0ex}{0ex}}\Rightarrow {x}^{2}+600x-500x-300000=0\phantom{\rule{0ex}{0ex}}\Rightarrow x(x+600)-500(x+600)=0\phantom{\rule{0ex}{0ex}}\Rightarrow (x-500)(x+600)=0\phantom{\rule{0ex}{0ex}}\Rightarrow x-500=0\mathrm{or}x+600=0\phantom{\rule{0ex}{0ex}}\Rightarrow x=500\mathrm{or}x=-600$

Since, speed cannot be negative.

Thus, the original speed/hour of the plane is 500 km/hr.

#### Page No 4.59:

#### Question 14:

A motor boat whose speed in still water is 18 km/hr takes 1 hour more to go 24 km up stream that to return down stream to the same spot. Find the speed of the stream.

#### Answer:

Let the speed of the stream be *x* km/hr.

speed of the boat in still water = 18 km/hr.

Total Distance = 24 km.

We know that,

Speed of the boat up stream = speed of the boat in still water − speed of the stream

= (18 − *x*) km/hr

Speed of the boat down stream = speed of the boat in still water + speed of the stream

= (18 + *x*) km/hr

Time of up stream journey = *t*_{1} = $\frac{24}{18-x}$ hr

Time of down stream journey = *t*_{2} = $\frac{24}{18+x}$ hr

According to the question,

*t*_{1} − *t*_{2} = 1 hr

$\Rightarrow \frac{24}{18-x}-\frac{24}{18+x}=1\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{24(18+x-18+x)}{(18-x)(18+x)}=1\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{24\left(2x\right)}{(18{)}^{2}-{x}^{2}}=1\phantom{\rule{0ex}{0ex}}\Rightarrow 48x=324-{x}^{2}\phantom{\rule{0ex}{0ex}}\Rightarrow {x}^{2}+48x-324=0\phantom{\rule{0ex}{0ex}}\Rightarrow {x}^{2}+54x-6x-324=0\phantom{\rule{0ex}{0ex}}\Rightarrow x(x+54)-6(x+54)=0\phantom{\rule{0ex}{0ex}}\Rightarrow (x-6)(x+54)=0\phantom{\rule{0ex}{0ex}}\Rightarrow x-6=0\mathrm{or}x+54=0\phantom{\rule{0ex}{0ex}}\Rightarrow x=6\mathrm{or}x=-54$

Since, speed cannot be negative.

Thus, speed of the stream is 6 km/hr.

#### Page No 4.59:

#### Question 15:

A car moves a distance of 2592 km with uniform speed. The number of hours taken for the journey is one-half the number representing the speed, in km/hour. Find the time taken to cover the distance.

#### Answer:

Distance covered = 2592 km

Let the speed of the car be *x km/h*

According to the question, Time = $\frac{\mathrm{speed}}{2}=\frac{x}{2}h\phantom{\rule{0ex}{0ex}}$

we know that

$\mathrm{speed}=\frac{\mathrm{distance}}{\mathrm{time}}$

$x=\frac{2592}{x/2}\phantom{\rule{0ex}{0ex}}{x}^{2}=5184\phantom{\rule{0ex}{0ex}}x=\sqrt{5184}\phantom{\rule{0ex}{0ex}}x=72\phantom{\rule{0ex}{0ex}}\mathrm{Speed}\mathrm{of}\mathrm{the}\mathrm{car}=72\mathrm{km}/\mathrm{h}\phantom{\rule{0ex}{0ex}}\mathrm{Time}\mathrm{taken}\mathrm{to}\mathrm{cover}\mathrm{the}\mathrm{given}\mathrm{distance}=36\mathrm{h}$

#### Page No 4.59:

#### Question 16:

A motor boat whose speed in still water is 9 km/hr, goes 15 km downstream and comes back to the same spot, in a total time of 3 hours 45 minutes. Find the speed of the stream.

#### Answer:

Let speed of stream be *x* km/h.

Given:

Speed of boat = 9 km/h

Distance covered upstream = 15 km

Distance covered downstream = 15 km

Total time taken = 3 hours 45 minutes = $\frac{15}{4}\mathrm{hours}$

Now, Speed of boat upstream = 9 − *x* km/h

Speed of boat downstream = 9 + *x* km/h

$\frac{\mathrm{Distance}}{\mathrm{Speed}}=\mathrm{Time}$

According to the question,

$\frac{15}{9+x}+\frac{15}{9-x}=\frac{15}{4}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{15\left(9-x\right)+15\left(9+x\right)}{\left(9+x\right)\left(9-x\right)}=\frac{15}{4}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{135-15x+135+15x}{81-{x}^{2}}=\frac{15}{4}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{135+135}{81-{x}^{2}}=\frac{15}{4}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{270}{81-{x}^{2}}=\frac{15}{4}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{18}{81-{x}^{2}}=\frac{1}{4}\phantom{\rule{0ex}{0ex}}\Rightarrow 18\left(4\right)=81-{x}^{2}\phantom{\rule{0ex}{0ex}}\Rightarrow 72=81-{x}^{2}\phantom{\rule{0ex}{0ex}}\Rightarrow {x}^{2}=9\phantom{\rule{0ex}{0ex}}\Rightarrow x=\pm 3\phantom{\rule{0ex}{0ex}}\mathrm{But}x\mathrm{is}\mathrm{the}\mathrm{speed}\mathrm{of}\mathrm{stream}\mathrm{which}\mathrm{is}\mathrm{always}\mathrm{positive}.\phantom{\rule{0ex}{0ex}}\mathrm{Thus},x=3\mathrm{km}/\mathrm{h}$

Hence, the speed of the stream is 3 km/h.

#### Page No 4.61:

#### Question 1:

Ashu is* x* years old while his mother Mrs Veena is *x*^{2} years old. Five years hence Mrs Veena will be three times old as Ashu. Find their present ages.

#### Answer:

Given that Ashu’s present age isand his mother Mrs. Veena is

Then according to question,

Five years later, Ashu’s is

And his mother Mrs. Veena is

Thus

So, either

Or

But, the age can never be negative.

Therefore, when then

Hence, Ashu’s present age isand his mother Mrs. Veena is

#### Page No 4.61:

#### Question 2:

The sum of ages of a man and his son is 45 years. Five years ago, the product of their ages was four times the man's age at the time. Find their present ages.

#### Answer:

Let the present age of the man be

Then present age of his son is

Five years ago, man’s age

And his son’s age

Then according to question,

So, either

Or

But, the father’s age never be 5 years

Therefore, when then

Hence, man’s present age isand his son’s age is

#### Page No 4.61:

#### Question 3:

The product of Shikha's age five years ago and her age 8 years later is 30, her age at both times being given in years. Find her present age.

#### Answer:

Let the present age of Shikha be

Then, 8 years later, age of her

Five years ago, her age

Then according to question,

So, either

Or

But the age never be negative

Hence, the present age of Shikha be

#### Page No 4.61:

#### Question 4:

The product of Ramu's age (in years) five years ago and his age (in years) nice years later is 15. Determine Ramu's present age.

#### Answer:

Let the present age of Ramu be

Then, 9 years later, age of her

Five years ago, her age

Then according to question,

So, either

Or

But the age never be negative

Hence, the present age of Ramu be

#### Page No 4.61:

#### Question 5:

Is the following situation possible? If so, determine their present ages. The sum of the ages of two friends is 20 years ago, the product of their ages in years was 48.

#### Answer:

Let the present age of two friends be *x* years and (20 − *x*) years respectively.

Then, 4 years later, the age of two friends will be (*x* − 4) years and (20 − *x* − 4) years respectively

Then according to question,

$-{x}^{2}+20x-64-48=0$

${x}^{2}-20x+112=0$

Let *D* be the discriminant of the above quadratic equation.

Then,

Putting the value of *a* = 1, *b* = − 20 and *c* = 112

$D={\left(-20\right)}^{2}-4\times 1\times 112$

= 400 − 448

= − 48

Thus,

So, the above equation does not have real roots.

Hence, the given situation is not possible.

#### Page No 4.61:

#### Question 6:

A girls is twice as old as her sister. Four years hence, the product of their ages (in years) will be 160. Find their present ages.

#### Answer:

Let the present age of girl bethen, age of her sister

Then, 4 years later, age of girl and her sister’s age be

Then according to question,

So, either

Or

But the age never be negative

Therefore, when then

Hence, the present age of girl be and her sister’s age be

#### Page No 4.62:

#### Question 7:

The sum of the reciprocals of Rehman's ages (in years) 3 years ago and 5 years from now is 1/3. Find his present age.

#### Answer:

Let the present age of Rehman be

Then, 8 years later, age of her

Five years ago, her age

Then according to question,

So, either

Or

But the age never be negative

Hence, the present age of Rehman be

#### Page No 4.62:

#### Question 8:

If Zeba were younger by 5 years than what she really is, then the square of her age (in years) would have been 11 more than 5 times her actual age. What is her age now?

#### Answer:

Let, the present age of Zeba be *x* years.

If she were 5 years younger, then the square of her age would have been 11 more than 5 times her actual age

$\Rightarrow {\left(x-5\right)}^{2}=11+5x\phantom{\rule{0ex}{0ex}}\Rightarrow {x}^{2}+25-10x=11+5x\phantom{\rule{0ex}{0ex}}\Rightarrow {x}^{2}-15x+14=0\phantom{\rule{0ex}{0ex}}\Rightarrow {x}^{2}-14x-x+14=0\phantom{\rule{0ex}{0ex}}\Rightarrow x\left(x-14\right)-1\left(x-14\right)=0\phantom{\rule{0ex}{0ex}}\Rightarrow \left(x-14\right)\left(x-1\right)=0\phantom{\rule{0ex}{0ex}}\Rightarrow x=14\mathrm{or}x=1$

The age cannot be 1 year as we have also talked about the age 5 years ago.

Therefore, Zeba's present age is $x=14$ years.

#### Page No 4.62:

#### Question 9:

At present Asha's age (in years) is 2 more than the square of her daughter Nisha's age. When Nisha grows to her mother's present age, Asha's age would be one year less than 10 times the present age of Nisha. Find the present ages of both Asha and Nisha.

#### Answer:

Let, the present age of Asha be *a* years

Present age of Nisha be *n *years.

Given that at present Asha's age is 2 more than the square of her daughter Nisha's age.

$\Rightarrow a=2+{n}^{2}$

So, we can say that Nisha will take $(2+{n}^{2})-n$ years to reach her mother's age.

So, after $(2+{n}^{2})-n$ years, her mother's age will be $(2+{n}^{2})+(2+{n}^{2}-n)$ years.

It is also given that when Nisha grows to her mother's present age, Asha's age would be one year less than 10 times the present age of Nisha.

Therefore, after $2+{n}^{2}-n$ years, Asha's age will be $10n-1$ years

$\Rightarrow 2+{n}^{2}+2+{n}^{2}-n=10n-1\phantom{\rule{0ex}{0ex}}\Rightarrow 2{n}^{2}-10n-n+5=0\phantom{\rule{0ex}{0ex}}\Rightarrow 2n(n-5)-1(n-5)=0\phantom{\rule{0ex}{0ex}}\Rightarrow (2n-1)(n-5)=0\phantom{\rule{0ex}{0ex}}\Rightarrow n=\frac{1}{2}\mathrm{or}n=5$

We ignore the fractional value.

Therefore, Nisha's age is *n* = 5 years.

Asha's age = $2+{n}^{2}=2+25=27$ years

#### Page No 4.64:

#### Question 1:

The hypotenuse of a right triangle is 25 cm. The difference between the lengths of the other two sides of the triangle is 5 cm. Find the lengths of these sides.

#### Answer:

Let the length of one side of right triangle be then other side be

And given that hypotenuse

As we know that by Pythagoras theorem,

So, either

Or

But the side of right triangle can never be negative

Therefore, when then

Hence, length of one side of right triangle be then other side be

#### Page No 4.64:

#### Question 2:

The diagonal of a rectangular field is 60 meters more than the shorter side. If the longer side is 30 meters more than the shorter side, find the sides of the field.

#### Answer:

Let the length of smaller side of rectangle be then larger side be and their diagonal be

Then, as we know that Pythagoras theorem

But, the side of rectangle can never be negative.

Therefore, when then

Hence, length of smaller side of rectangle be and larger side be

#### Page No 4.64:

#### Question 3:

The hypotenuse of a right triangle is $3\sqrt{10}\mathrm{cm}$. If the smaller leg is tripled and the longer leg doubled, new hypotenuse wll be $9\sqrt{5}\mathrm{cm}$. How long are the legs of the triangle?

#### Answer:

Let the length of smaller side of right triangle be then larger side be

Then, as we know that by Pythagoras theorem

….. (1)

If the smaller side is triple and the larger side be doubled, the new hypotenuse is

Therefore,

….. (2)

From equation (1) we get

Now putting the value of in equation (2)

But, the side of right triangle can never be negative

Therefore, when then

Hence, length of smaller side of right triangle be then larger side be

#### Page No 4.64:

#### Question 4:

A pole has to be erected at a point on the boundary of a circular park of diameter 13 meters in such a way that the difference of its distances from two diametrically opposite fixed gates A and B on the boundary is 7 meters. Is it the possible to do so? If yes, at what distances from the two gates should the pole be erected?

#### Answer:

Let *P *be the required location on the boundary of a circular park such that its distance from gate *B *is that is *BP*

Then, *AP *

In the right triangle *ABP *we have by using Pythagoras theorem

But the side of right triangle can never be negative

Therefore,

Hence, *P *is at a distance of from the gate *B*.

#### Page No 4.70:

#### Question 1:

The perimeter of a rectangular field is 82 m and its area is 400 m^{2}. Find the breadth of the rectangle.

#### Answer:

Let the breadth of the rectangle be. Then

Perimeter

And area of the rectangle

or

Since perimeter is 82 meter. So breadth can’t be 25 meter.

Hence, breadth

#### Page No 4.70:

#### Question 2:

The length of a hall is 5 m more than its breadth. If the area of the floor of the hall is 84 m^{2}, what are the length and breadth of the hall?

#### Answer:

Let the breadth of the rectangular hall beand the lengthThen

And area of the rectangle

or

Sides of the rectangular hall never are negative.

Therefore, length

Hence, breadth of the hall be be

#### Page No 4.70:

#### Question 3:

Two squares have sides *x* cm and (*x* + 4) cm. The sum of their areas is 656 cm^{2}. Find the sides of the squares.

#### Answer:

Given that sides of the squares are and .Then

According to question,

Sum of the areas of square

So,

or

Sides of the square never are negative.

Therefore the side of the other square is

Hence, sides of the square be and respectively.

#### Page No 4.71:

#### Question 4:

The area of a right angled triangle is 165 m^{2}. Determine its base and altitude if the latter exceeds the former by 7 m.

#### Answer:

Let the base of the right triangle beand the altitudeThen

According to question,

Areas of the right triangle

And as we know that the area of the right triangle

or

Since negative value is not possible. So *x *=15 m

Therefore the altitude is

Hence, base of the right triangle be and altitude be

#### Page No 4.71:

#### Question 5:

Is it possible to design a rectangular mango grove whose length is twice its breadth and the area is 800 m^{2}? If so, find its length and breadth.

#### Answer:

Let the breadth of the rectangular mango grove be *x *meter and the length. Then

Area of the rectangle

Sides of the rectangular hall never be negative

Therefore, length

, it is possible.

Hence, breadth of the hall be be

#### Page No 4.71:

#### Question 6:

Is it possible to design a rectangular park of perimeter 80 m and area 400 m^{2}. If so, find its length and breadth.

#### Answer:

Let the breadth of the rectangle be. Then

Perimeter

And area of the rectangle

, it is possible.

Hence, breadth of the rectangular park be and length be

#### Page No 4.71:

#### Question 7:

Sum of the areas of two squares is 640 m^{2}. If the difference of their perimeters is 64 m. Find the sides of the two squares.

#### Answer:

Let the sides of the squares are *x *m and .Then

According to question,

Sum of the difference of their perimeter=64 m

….. (1)

And sum of the areas of square

….. (2)

Putting the value of *x* in equation (2) from equation (1)

or

Sides of the square never are negative.

Therefore, putting the value of *x* in equation (1)

Hence, sides of the square be and respectively.

#### Page No 4.71:

#### Question 8:

Sum of the area of two squares is 400 cm^{2}. If the difference of their perimeters is 16 cm, find the sides of two squares.

#### Answer:

Let the side of the smaller square be *x *cm.

Perimeter of any square = (4 × side of the square)* *cm.

It is given that the difference of the perimeters of two squares is 16 cm.

Then side of the bigger square = $\frac{16+4x}{4}=\left(4+x\right)$ cm.

According to the question,

${x}^{2}+{\left(4+x\right)}^{2}=400\phantom{\rule{0ex}{0ex}}\Rightarrow {x}^{2}+16+{x}^{2}+8x=400\phantom{\rule{0ex}{0ex}}\Rightarrow 2{x}^{2}+8x-384=0\phantom{\rule{0ex}{0ex}}\Rightarrow {x}^{2}+4x-192=0\phantom{\rule{0ex}{0ex}}\Rightarrow {x}^{2}+16x-12x-192=0\phantom{\rule{0ex}{0ex}}\Rightarrow x(x+16)-12(x+16)=0\phantom{\rule{0ex}{0ex}}\Rightarrow (x-12)(x+16)=0\phantom{\rule{0ex}{0ex}}\Rightarrow x-12=0\mathrm{or}x+16=0\phantom{\rule{0ex}{0ex}}\Rightarrow x=12\mathrm{or}x=-16$

Since, side of the square cannot be negative.

Thus, the side of the smaller square is 12 cm.

and the side of the bigger square is (4 + 12) = 16 cm.

#### Page No 4.71:

#### Question 9:

The area of a rectangular plot is 528 m^{2}. The length of the plot (in metres) is one metre more then twice its breadth. Find the length and the breadth of the plot.

#### Answer:

Let the breadth of the rectangular plot be *x *m.

Then, the length of the rectangular plot = (1 + 2*x*) m.

According to the question,

Length × Breadth = Area

$x(1+2x)=528\phantom{\rule{0ex}{0ex}}\Rightarrow 2{x}^{2}+x-528=0\phantom{\rule{0ex}{0ex}}\Rightarrow 2{x}^{2}+33x-32x-528=0\phantom{\rule{0ex}{0ex}}\Rightarrow x(2x+33)-16(2x+33)=0\phantom{\rule{0ex}{0ex}}\Rightarrow (x-16)(2x+33)=0\phantom{\rule{0ex}{0ex}}\Rightarrow x-16=0\mathrm{or}2x+33=0\phantom{\rule{0ex}{0ex}}\Rightarrow x=16\mathrm{or}x=-\frac{33}{2}$

Since, length and breadth of the rectangle cannot be negative.

Thus, the breadth of the rectangular plot is 16 m.

and the length of the rectangular plot is (1 + 2×16) = 33 m.

#### Page No 4.71:

#### Question 10:

In the centre of a rectangular lawn of dimensions 50 m × 40 m, a rectangular pond has to be constructed so that the area of the grass surrounding the pond would be 1184 m^{2}. Find the length and breadth of the pond.

#### Answer:

It is given that the dimensions of rectangular park is 50 m × 40 m.

∴ Area of the rectangular park = 50 × 40 = 2000 m^{2}

Area of the grass surrounding the pond = 1184 m^{2}

Now,

Area of the rectangular pond

= Area of the rectangular park − Area of the grass surrounding the rectangular pond

= 2000 − 1184

= 816 m^{2}

Let the uniform width of the surrounding grass be *x*.

∴ Length of the rectangular pond = (50 − 2*x*) m

Breadth of the rectangular pond = (40 − 2*x*) m

Now,

Area of rectangular pond = 816 m^{2}

∴ (50 − 2*x*) × (40 − 2*x*) = 816

⇒ 2000 − 80*x *− 100*x* + 4*x*^{2} = 816

⇒ 4*x*^{2} − 180*x* + 2000 − 816 = 0

⇒ 4*x*^{2} − 180*x *+ 1184 = 0

⇒ *x*^{2} − 45*x *+ 296 = 0

⇒* x*^{2} − 37*x* − 8*x* + 296 = 0

⇒ *x*(*x *− 37) − 8(*x *− 37) = 0

⇒ (*x* − 8)(*x* − 37) = 0

⇒ *x* − 8 = 0 or *x* − 37 = 0

⇒* x *= 8 or *x* = 37

For *x* = 37,

Length of rectangular pond = 50 − 2 × 37 = −24 m, which is not possible

So, *x* ≠ 37

Therefore, *x* = 8.

When *x* = 8,

Length of the rectangular pond = 50 − 2 × 8 = 50 − 16 = 34 m

Breadth of the rectangular pond = 40 − 2 × 8 = 40 − 16 = 24 m.

#### Page No 4.73:

#### Question 1:

A takes 10 days less than the time taken by B to finish a piece of work. If both A and B together can finish the work in 12 days, find the time taken by B to finish the work.

#### Answer:

Let *B* alone takes *x *days to finish the work. Then,* B’*s one day’s work.

Similarly, *A *alone can finish it in days to finish the work. Then,* A’*s one day’s work.

It is given that

*A’*s one day’s work + *B’*s one day’s work’s one day’s work

But is not correct.

therefore, is correct

Hence, the time taken by *B *to finish the work in

#### Page No 4.73:

#### Question 2:

If two pipes function simultaneously, a reservoir will be filled in 12 hours. One pipe fills the reservoir 10 hours faster than the other. How many hours will the second pipe take to fill the reservoir?

#### Answer:

Let the first pipe takes *x* hours to fill the reservoir. Then the second pipe will takes hours to fill the reservoir.

Since, the faster pipe takes *x* hours to fill the reservoir.

Therefore, portion of the reservoir filled by the faster pipe in one hour

So, portion of the reservoir filled by the faster pipe in 12 hours

Similarly,

Portion of the reservoir filled by the slower pipe in 12 hours

It is given that the reservoir is filled in 12 hours.

So,

But, cannot be negative.

Therefore, when then

Hence, the second pipe will takes to fill the reservoir.

#### Page No 4.73:

#### Question 3:

Two water taps together can fill a tank in $9\frac{3}{8}\mathrm{hours}$. The tap of larger diameter takes 10 hours less than the smaller one to fill the tank separately. Find the time in which each tap can separately fill the tank.

#### Answer:

Let the first water tape takes *x* hours to fill the tank. Then the second water tape will takes hours to fill the tank.

Since, the faster water tape takes *x* hours to fill the tank.

Therefore, portion of the tank filled by the faster water tape in one hour

So, portion of the tank filled by the faster water tape in hours

Similarly,

Portion of the tank filled by the slower water tape in hours

It is given that the tank is filled in hours.

So,

But, cannot be negative.

Therefore, when then

Hence, the first water tape will takes to fill the tank, and the second water tape will takes to fill the tank.

#### Page No 4.73:

#### Question 4:

Two pipes running together can fill a tank in $11\frac{1}{9}\mathrm{minutes}$. If one pipe takes 5 minutes more than the other to fill the tank separately, find the time in which each pipe would fill the tank separately.

#### Answer:

Let the first pipe takes *x* minutes to fill the tank. Then the second pipe will takes minutes to fill the tank.

Since, the first pipe takes *x* minutes to fill the tank.

Therefore, portion of the tank filled by the first pipe in one minutes

So, portion of the tank filled by the first pipe in minutes

Similarly,

Portion of the tank filled by the second pipe in minutes

It is given that the tank is filled in minutes.

So,

But, cannot be negative.

Therefore, when *x* = 20 then

Hence, the first water tape will takes 20 min to fill the tank, and the second water tape will take 25 min to fill the tank.

#### Page No 4.74:

#### Question 5:

To fill a swimming pool two pipes are used. If the pipe of larger diameter used for 4 hours and the pipe of smaller diameter for 9 hours, only half of the pool can be filled. Find, how long it would take for each pipe to fill the pool separately, if the pipe of smaller diameter takes 10 hours more than the pipe of larger diameter to fill the pool?

#### Answer:

Let the pipe of larger diameter takes* x* hours.

Then, the pipe of smaller diameter takes *x *+ 10 hours to fill the pool.

Now, the part of the pool filled by the larger pipe in 1 hour = $\frac{1}{x}$

and the part of the pool filled by the smaller pipe in 1 hour = $\frac{1}{x+10}$

If the larger pipe is used for 4 hours and the smaller pipe is used for 9 hours, only half of the pool can be filled,

$\therefore \frac{4}{x}+\frac{9}{x+10}=\frac{1}{2}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{4(x+10)+9x}{x(x+10)}=\frac{1}{2}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{4x+40+9x}{{x}^{2}+10x}=\frac{1}{2}\phantom{\rule{0ex}{0ex}}\Rightarrow 2(13x+40)={x}^{2}+10x\phantom{\rule{0ex}{0ex}}\Rightarrow {x}^{2}+10x-26x-80=0\phantom{\rule{0ex}{0ex}}\Rightarrow {x}^{2}-16x-80=0\phantom{\rule{0ex}{0ex}}\Rightarrow {x}^{2}-20x+4x-80=0\phantom{\rule{0ex}{0ex}}\Rightarrow x(x-20)+4(x-20)=0\phantom{\rule{0ex}{0ex}}\Rightarrow (x+4)(x-20)=0\phantom{\rule{0ex}{0ex}}\Rightarrow x+4=0\mathrm{or}x-20=0\phantom{\rule{0ex}{0ex}}\Rightarrow x=-4\mathrm{or}x=20$

Since, time cannot be negative.

∴ *x* = 20

Thus, the pipe of larger diameter takes* *20 hours and the pipe of smaller diameter takes (20 + 10) = 30 hours to fill the pool separately.

#### Page No 4.74:

#### Question 6:

Two water taps together can fill a tank in $1\frac{7}{8}$ hours. The tap with longer diameter takes 2 hours less than the tap with the smaller one to fill the tank separately. Find the time in which each tap can fill the tank separately.

#### Answer:

Let the smaller tap takes *x* hours to fill the tank.

Then, the larger one takes (*x* − 2) hours to fill the tank.

Tank filled in 1 hour by smaller tap = $\frac{1}{x}$

Tank filled in 1 hour by larger tap = $\frac{1}{x-2}$

Tank filled in 1 hour by both the taps = $\frac{8}{15}$

According to the question,

$\frac{1}{x}+\frac{1}{x-2}=\frac{8}{15}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{\left(x-2\right)+\left(x\right)}{\left(x\right)\left(x-2\right)}=\frac{8}{15}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{2x-2}{{x}^{2}-2x}=\frac{8}{15}\phantom{\rule{0ex}{0ex}}\Rightarrow 15\left(2x-2\right)=8\left({x}^{2}-2x\right)\phantom{\rule{0ex}{0ex}}\Rightarrow 30x-30=8{x}^{2}-16x\phantom{\rule{0ex}{0ex}}\Rightarrow 8{x}^{2}-46x+30=0\phantom{\rule{0ex}{0ex}}\Rightarrow 4{x}^{2}-23x+15=0\phantom{\rule{0ex}{0ex}}\Rightarrow x=\frac{-\left(-23\right)\pm \sqrt{{\left(-23\right)}^{2}-4\left(4\right)\left(15\right)}}{2\left(4\right)}\phantom{\rule{0ex}{0ex}}\Rightarrow x=\frac{23\pm \sqrt{529-240}}{8}\phantom{\rule{0ex}{0ex}}\Rightarrow x=\frac{23\pm \sqrt{289}}{8}\phantom{\rule{0ex}{0ex}}\Rightarrow x=\frac{23\pm 17}{8}\phantom{\rule{0ex}{0ex}}\Rightarrow x=\frac{23+17}{8}\mathrm{or}x=\frac{23-17}{8}\phantom{\rule{0ex}{0ex}}\Rightarrow x=5\mathrm{or}x=0.75\phantom{\rule{0ex}{0ex}}\mathrm{But}x\ne 0.75\left(\because \left(x-2\right)\mathrm{becomes}\mathrm{negative}\right)\phantom{\rule{0ex}{0ex}}\mathrm{Thus},x=5\phantom{\rule{0ex}{0ex}}$

Hence, the time in which each tap can fill the tank separately is 5 hours and 3 hours respectively.

#### Page No 4.8:

#### Question 1:

The product of two consecutive positive integers is 306. Form the quadratic equation to find the integers, if *x* denotes the smaller integer.

#### Answer:

Since it is given in the question that the numbers we have to find are consecutive positive integer numbers, therefore the difference between the two numbers should be equal to 1.

For e.g. 7 and 8 or 26 and 27 are consecutive numbers.

Let us assume the first number to be ‘*x*’. So our next consecutive number should be ‘*x *+ 1’. Now the question also says that the product of these two numbers is 306.

Therefore,

Hence, this is our required quadratic equation.

#### Page No 4.8:

#### Question 2:

John and Jivanti together have 45 marbles. Both of them lost 5 marbles each, and the product of the number of marbles they now have is 128. Form the quadratic equation to find how many marbles they had to start with, if John and *x* marbles.

#### Answer:

It is given that John had ‘*x*’ marbles.

We are also given that both John and Javanti had 45 marbles together.

So, Javanti should have ’45 − *x*’ marbles with her.

Now, it is given that both of them lose 5 marbles each.

So in the new situation, John will have ‘*x *− 5’marbles and Javanti will have ’45 − *x *− 5’ marbles.

Also it is given that the product of the number of marbles both of them now is 128.

Therefore,

Hence, this is the required quadratic equation.

#### Page No 4.8:

#### Question 3:

A cottage industry produces a certain number of toys in a day. The cost of production of each toy (in rupees) was found to be 55 minus the number of articles produced in a day. On a particular day, the total cost of production was Rs. 750. If *x* denotes the number of toys produced that day, from the quadratic equation of find *x*.

#### Answer:

Now we know that ‘*x*’ denotes the total number of toys produced in that day.

But, the cost of production of a single toy is 55 minus the number of toys produced that day i.e. ‘*x*’.

So, the total production cost would be the product of the cost of a single toy and the total number of toys i.e. product of ‘55 −* x*’ and ‘*x*’. Now, it is given here that total production cost of that day was Rs.750.

Therefore,

Hence, this is the required quadratic equation.

#### Page No 4.8:

#### Question 4:

The height of a right triangle is 7 cm less than its base. If the hypotenuse is 13 cm, form the quadratic equation to find the base of the triangle.

#### Answer:

Now, since we have to find out base, let us assume the base to be ‘*x*’ cm.

Therefore the height of the triangle becomes ‘*x* −7’.

It is also given that the hypotenuse is 13 cm.

By Pythagoras Theorem,

Hence, this is our required quadratic equation.

#### Page No 4.8:

#### Question 5:

An express train takes 1 hour less than a passenger train to travel 132 km between Mysore and Banglore. If the average speed of the express train is 11 km/hr more than that of the passenger train, from the quadratic equation to find the average speed of express train.

#### Answer:

Now let us assume that the speed of the express train be ‘*x*’ km/hr. Therefore according to the question speed of the passenger train will be ‘*x *−11’ km/hr. Now we know that the total distance travelled by both the trains was 132 km.

We also know that

So the time taken by express train would be hr and the time taken by the passenger train would behr. Now, we also know that the express train took 1 hr less than the passenger train to travel the whole distance.

Therefore, we have

Therefore, this is the required equation.

#### Page No 4.8:

#### Question 6:

A train travels 360 km at a uniform speed. If the speed had been 5 km/hr more, it would have taken 1 hour less for the same journey. Form the quadratic equation to find the speed of the train.

#### Answer:

Let us assume that the speed of the train be ‘*x*’ km/hr. We are also given that the distance covered during the journey is 360 km.

Now, time taken during the journey = hr

Time taken for the new journey = hr

According to the question,

Hence, this is the required quadratic equation.

#### Page No 4.80:

#### Question 1:

A piece of cloth costs Rs. 35. If the piece were 4 m longer and each meter costs Rs. one less, the cost would remain unchanged. How long is the piece?

#### Answer:

Let the length of the piece be *x* metres.

Then, rate per metre

According to question, new length.

Since the cost remain same. Therefore, new rate per metre

It is given that

Because *x *cannot be negative.

Thus,* *is the require solution.

Therefore, the length of the piece be

#### Page No 4.80:

#### Question 2:

Some students planned a picnic. The budget for food was Rs. 480. But eight of these failed to go and thus the cost of food for each member increased by Rs. 10. How many students attended the picnic?

#### Answer:

Let *x *students planned a picnic.

Then, the share of each student

According to question, 8 students fail to go picnic, then remaining students.

Therefore, new share of each student

It is given that

Because *x *cannot be negative.

Thus,* *the total numbers of students attend a picnic

Therefore, the total numbers of students attend a picnic be

#### Page No 4.80:

#### Question 3:

A dealer sells an article for Rs. 24 and gains as much percent as the cost price of the article. Find the cost price of the article.

#### Answer:

Let the cost price of article be Rs. *x*.

Then, gain percent = *x*

Therefore, the selling price of article

It is given that

Because *x *cannot be negative.

Thus,* *is the require solution.

Therefore, the cost price of article be

#### Page No 4.80:

#### Question 4:

Out of a group of swans, 7/2 times the square root of the total number are playing on the share of a pond. The two remaining ones are swinging in water. Find the total number of swans.

#### Answer:

Let the total number of swans be *x*.

Then, total numbers of swans are playing on the share of a pond =

It is given that

Let , then

Because is not correct.

Thus,* *is correct. Putting the value of *y*

Square root both sides, we get

Therefore, the total number of swans be

#### Page No 4.80:

#### Question 5:

If the list price of a toy is reduced by Rs. 2, a person can buy 2 toys more for Rs. 360. Find the original price of the toy.

#### Answer:

Let the original list price of the toy be Rs. *x *.

Then, the number of toys brought for Rs.360

According to question, reduced list price of the toys.

Therefore, the number of toys brought for Rs.360

It is given that

Because *x *cannot be negative.

Thus,* *is the require solution.

Therefore, the original list price of the toy be

#### Page No 4.80:

#### Question 6:

Rs. 9000 were divided equally among a certain number of persons. Had there been 20 more persons, each would have got Rs. 160 less. Find the original number of persons.

#### Answer:

Let the original number of persons be *x*.

Then, by the given information,

Thus, the original number of persons is Rs 25.

#### Page No 4.80:

#### Question 7:

Some students planned a picnic. The budget for food was Rs. 500. But, 5 of them failed to go and thus the cost of food for each member increased by Rs. 5. How many students attended the picnic?

#### Answer:

Let *x *students planned a picnic.

Then, the share of each student

According to question, 5 students fail to go picnic, then remaining students.

Therefore, new share of each student

It is given that

Because *x *cannot be negative.

Thus,* *the total numbers of students attend a picnic

Therefore, the total numbers of students attend a picnic be

#### Page No 4.80:

#### Question 8:

A pole has to be erected at a point on the boundary of a circular park of diameter 13 meters in such a way that the difference of its distance from two diametrically opposite fixed gates A and B on the boundary is 7 meters. Is it the possible to do so? If yes, at what distances from the two gates should the pole be erected?

#### Answer:

Let *P *be the required location on the boundary of a circular park such that its distance from gate *B *is that is *BP*

Then, *AP *

In the right triangle *ABP *we have by using Pythagoras theorem

But, the side of right triangle can never be negative.

Therefore,

Hence, *P *is at a distance of from the gate *B*.

#### Page No 4.80:

#### Question 9:

In a class test, the sum of the marks obtained by P in Mathematics and science is 28. Had he got 3 marks more in mathematics and 4 marks less in Science. The product of his marks would have been 180. Find his marks in two subjects.

#### Answer:

Let marks obtained by *P* in mathematics be *x*, then in science

It is given that,

Therefore, when then

Hence, marks in mathematics and marks in science .

Or, when then

Hence, marks in mathematics and marks in science .

#### Page No 4.81:

#### Question 10:

In a class test, the sum of Shefali's marks in Mathematics and English is 30. Had she got 2 marks more in mathematics and 3 marks less in English, the product of her marks would have been 210. Find her marks in two subjects.

#### Answer:

Let marks obtained by Shefali in mathematics be *x*, then in english

It is given that,

Therefore, when then

Hence, marks in mathematics and marks in science .

Or,

when then

Hence, marks in mathematics and marks in science .

#### Page No 4.81:

#### Question 11:

A cottage industry produces a certain number of pottery articles in a day. It was observed on a particular day that the cost of production on each article (in rupees) was 3 more than twice the number of articles produced on that day. If the total cost of production on that day was Rs. 90, find the number of articles produced and the cost of each article.

#### Answer:

Let the number of article produced by the cottage industry be *x.*

Then the cost of production of each article

It is given that total cost of production

Therefore,

Therefore, *x *cannot be negative.

So, when then

Hence, the number of article produced by the cottage industry be and the cost of production of each article .

#### Page No 4.81:

#### Question 12:

At *t* minutes past 2 pm the time needed by the minutes hand and a clock to show 3 pm was found to be 3 minutes less than $\frac{{t}^{2}}{4}$ minutes. Find *t*.

#### Answer:

It is given that at *t* minutes past 2 pm the time needed by the minutes hand and a clock to show 3 pm was found to be 3 minutes less than $\frac{{t}^{2}}{4}$ minutes i.e. $\frac{{t}^{2}}{4}-3$.

There are 60 minutes in 1 hour, so at *t* minutes past 2, time left to 3 will be $60-t$ minutes.

$\Rightarrow 60-t=\frac{{t}^{2}}{4}-3\phantom{\rule{0ex}{0ex}}\Rightarrow 240-4t={t}^{2}-12\phantom{\rule{0ex}{0ex}}\Rightarrow {t}^{2}+4t-252=0\phantom{\rule{0ex}{0ex}}\Rightarrow {t}^{2}+18t-14t-252=0\phantom{\rule{0ex}{0ex}}\Rightarrow t\left(t+18\right)-14\left(t+18\right)=0\phantom{\rule{0ex}{0ex}}\Rightarrow \left(t-14\right)\left(t+18\right)=0\phantom{\rule{0ex}{0ex}}\Rightarrow t=14\mathrm{or}t=-18$

We ignore the negative value because time cannot be negative.

Therefore, *t* = 14

#### Page No 4.82:

#### Question 1:

If the equation *x*^{2} + 4*x* + *k* = 0 has real and distinct roots, then

(a) *k* < 4

(b) *k* > 4

(c) *k* ≥ 4

(d) *k* ≤ 4

#### Answer:

The given quadric equation is , and roots are real and distinct.

Then find the value of *k.*

Here,

As we know that

Putting the value of

The given equation will have real and distinct roots, if

Therefore, the value of

Thus, the correct answer is

#### Page No 4.82:

#### Question 2:

If the equation *x*^{2} − *ax* + 1 = 0 has two distinct roots, then

(a) |*a*| = 2

(b) |*a*| < 2

(c) |*a*| > 2

(d) None of these

#### Answer:

The given quadric equation is , and roots are distinct.

Then find the value of *a.*

Here,

As we know that

Putting the value of

The given equation will have real and distinct roots, if

Therefore, the value of

Thus, the correct answer is

#### Page No 4.82:

#### Question 3:

If the equation 9*x*^{2} + 6*kx* + 4 = 0 has equal roots, then the roots are both equal to

(a) $\pm \frac{2}{3}$

(b) $\pm \frac{3}{2}$

(c) 0

(d) ±3

#### Answer:

The given quadric equation is , and roots are equal.

Then find roots of given equation*.*

Here,

As we know that

Putting the value of

The given equation will have equal roots, if

So, putting the value of* k *in quadratic equation

When then equation be and when then

Therefore, the value of

Thus, the correct answer is

#### Page No 4.82:

#### Question 4:

If *ax*^{2} + *bx* + *c* = 0 has equal roots, then *c* =

(a) $\frac{-b}{2a}$

(b) $\frac{b}{2a}$

(c) $\frac{-{b}^{2}}{4a}$

(d) $\frac{{b}^{2}}{4a}$

#### Answer:

The given quadric equation is , and roots are equal

Then find the value of *c.*

Let be two roots of given equation

Then, as we know that sum of the roots

And the product of the roots

Putting the value of

Therefore, the value of

Thus, the correct answer is

#### Page No 4.82:

#### Question 5:

If the equation *ax*^{2} + 2*x* + *a* = 0 has two distinct roots, if

(a) *a* = ±1

(b) *a* = 0

(c) *a* = 0, 1

(d) *a* = −1, 0

#### Answer:

The given quadric equation is , and roots are distinct.

Then find the value of *a.*

Here,

As we know that

Putting the value of

The given equation will have real and distinct roots, if

Therefore, the value of

Thus, the correct answer is

#### Page No 4.82:

#### Question 6:

The positive value of *k* for which the equation *x*^{2}^{ }+* kx* + 64 = 0 and *x*^{2} − 8*x* + *k* = 0 will both have real roots, is

(a) 4

(b) 8

(c) 12

(d) 16

#### Answer:

The given quadric equation are , and roots are real.

Then find the value of *a.*

Here, ….. (1)

….. (2)

As we know that

Putting the value of

The given equation will have real and distinct roots, if

Therefore, putting the value of in equation (2) we get

The value of satisfying to both equations

Thus, the correct answer is

#### Page No 4.82:

#### Question 7:

The value of $\sqrt{6+\sqrt{6+\sqrt{6+}}}....$ is

(a) 4

(b) 3

(c) −2

(d) 3.5

#### Answer:

Let

Squaring both sides we get

The value of *x* cannot be negative.

Thus, the value of *x* = 3

Therefore, the correct answer is

#### Page No 4.82:

#### Question 8:

If 2 is a root of the equation *x*^{2} + *bx* + 12 = 0 and the equation* **x*^{2} + *bx* + *q* = 0 has equal roots, then *q* =

(a) 8

(b) −8

(c) 16

(d) −16

#### Answer:

2 is the common roots given quadric equation are , and

Then find the value of *q.*

Here, ….. (1)

….. (2)

Putting the value of in equation (1) we get

Now, putting the value of in equation (2) we get

Then,

As we know that

Putting the value of

The given equation will have equal roots, if

Thus, the correct answer is

#### Page No 4.82:

#### Question 9:

If the equations $\left({a}^{2}+{b}^{2}\right){x}^{2}-2\left(ac+bd\right)x+{c}^{2}+{d}^{2}=0$ has equal roots, then

(a) *ab* = *cd*

(b) *ad* = *bc*

(c) $ad=\sqrt{bc}$

(d) $ab=\sqrt{cd}$

#### Answer:

The given quadric equation is , and roots are equal.

Here,

As we know that

Putting the value of

The given equation will have equal roots, if

Thus, the correct answer is

#### Page No 4.82:

#### Question 10:

If the roots of the equations $\left({a}^{2}+{b}^{2}\right){x}^{2}-2b\left(a+c\right)x+\left({b}^{2}+{c}^{2}\right)=0$ are equal, then

(a) 2*b* = *a* + *c*

(b)* **b*^{2} = *ac*

(c) $b=\frac{2ac}{a+c}$

(d) *b* = *ac*

#### Answer:

The given quadric equation is , and roots are equal.

Here,

As we know that

Putting the value of

The given equation will have equal roots, if

Thus, the correct answer is

#### Page No 4.82:

#### Question 11:

If the equation *x*^{2} − *bx* + 1 = 0 does not possess real roots, then

(a) −3 < *b* < 3

(b) −2 < *b* < 2

(c)* b* > 2

(d) *b* < −2

#### Answer:

The given quadric equation is , and does not have real roots.

Then find the value of *b.*

Here,

As we know that

Putting the value of

The given equation does not have real roots, if

Therefore, the value of

Thus, the correct answer is

#### Page No 4.82:

#### Question 12:

If *x* = 1 is a common roots of the equations *ax*^{2} + *ax* + 3 = 0 and *x*^{2} + *x* + *b* = 0, then *ab* =

(a) 3

(b) 3.5

(c) 6

(d) −3

#### Answer:

is the common roots given quadric equation are , and

Then find the value of *q.*

Here, ….. (1)

….. (2)

Putting the value of in equation (1) we get

Now, putting the value of in equation (2) we get

Then,

Thus, the correct answer is

#### Page No 4.83:

#### Question 13:

If *p *and *q* are the roots of the equation* **x*^{2} − *px* + *q* = 0, then

(a) *p *= 1, *q* = −2

(b) *b* = 0, *q* = 1

(c) *p* = −2, *q* = 0

(d) *p* = −2, *q* = 1

#### Answer:

Given that *p* and *q *be the roots of the equation

Then find the value of *p *and *q.*

Here,

*p* and *q *be the roots of the given equation

Therefore, sum of the roots

….. (1)

Product of the roots

As we know that

Putting the value of in equation (1)

Therefore, the value of

Thus, the correct answer is

#### Page No 4.83:

#### Question 14:

If *a* and *b* can take values 1, 2, 3, 4. Then the number of the equations of the form *ax*^{2} +* bx* + 1 = 0 having real roots is

(a) 10

(b) 7

(c) 6

(d) 12

#### Answer:

Given that the equation _{.}

For given equation to have real roots, discriminant (D) ≥ 0

⇒ *b*^{2} − 4*a* ≥ 0

⇒ *b*^{2} ≥ 4*a*

⇒ *b* ≥ 2√*a*

Now, it is given that *a* and *b* can take the values of 1, 2, 3 and 4.

The above condition *b* ≥ 2√*a* can be satisfied when

i) *b* = 4 and *a* = 1, 2, 3, 4

ii) *b* = 3 and *a* = 1, 2

iii) *b* = 2 and *a* = 1

So, there will be a maximum of 7 equations for the values of (*a*, *b*) = (1, 4), (2, 4), (3, 4), (4, 4), (1, 3), (2, 3) and (1, 2).

Thus, the correct option is (*b*).

#### Page No 4.83:

#### Question 15:

The number of quadratic equations having real roots and which do not change by squaring their roots is

(a) 4

(b) 3

(c) 2

(d) 1

#### Answer:

As we know that the number of quadratic equations having real roots and which do not change by squaring their roots is 2.

Thus, the correct answer is

#### Page No 4.83:

#### Question 16:

If $\left({a}^{2}+{b}^{2}\right){x}^{2}+2\left(ab+bd\right)x+{c}^{2}+{d}^{2}=0$ has no real roots, then

(a)* ab *= *bc*

(b)* ab* = *cd*

(c) *ac* = *bd*

(d)* ad* ≠ *bc*

#### Answer:

The given quadric equation is , and roots are equal.

Here,

As we know that

Putting the value of

The given equation will have no real roots, if

Thus, the correct answer is

#### Page No 4.83:

#### Question 17:

If the sum of the roots of the equation *x*^{2} − *x* = λ(2*x* − 1) is zero, then λ =

(a) −2

(b) 2

(c) $-\frac{1}{2}$

(d) $\frac{1}{2}$

#### Answer:

The given quadric equation is , and roots are zero.

Then find the value of *.*

Here,

As we know that

Putting the value of

The given equation will have zero roots, if

Therefore, the value of

Thus, the correct answer is

#### Page No 4.83:

#### Question 18:

If *x* = 1 is a common root of *ax*^{2} + *ax* + 2 = 0 and* **x*^{2} + *x* + *b *= 0, then, *ab* =

(a) 1

(b) 2

(c) 4

(d) 3

#### Answer:

is the common roots given quadric equation are , and

Then find the value of *ab.*

Here, ….. (1)

….. (2)

Putting the value of in equation (2) we get

Now, putting the value of in equation (1) we get

Then,

Thus, the correct answer is

#### Page No 4.83:

#### Question 19:

The value of* c* for which the equation *ax*^{2} + 2*bx* +* c* = 0 has equal roots is

(a) $\frac{{b}^{2}}{a}$

(b) $\frac{{b}^{2}}{4a}$

(c) $\frac{{a}^{2}}{b}$

(d) $\frac{{a}^{2}}{4b}$

#### Answer:

The given quadric equation is , and roots are equal

Then find the value of *c.*

Let be two roots of given equation

Then, as we know that sum of the roots

And the product of the roots

Putting the value of

Therefore, the value of

Thus, the correct answer is

#### Page No 4.83:

#### Question 20:

If ${x}^{2}+k\left(4x+k-1\right)+2=0$ has equal roots, then k =

(a) $-\frac{2}{3},1$

(b) $\frac{2}{3},-1$

(c) $\frac{3}{2},\frac{1}{3}$

(d) $-\frac{3}{2},-\frac{1}{3}$

#### Answer:

The given quadric equation is , and roots are equal

Then find the value of *k.*

Here,