Mathematics Part I Solutions Solutions for Class 10 Math Chapter 2 Quadratic Equations are provided here with simple step-by-step explanations. These solutions for Quadratic Equations are extremely popular among Class 10 students for Math Quadratic Equations Solutions come handy for quickly completing your homework and preparing for exams. All questions and answers from the Mathematics Part I Solutions Book of Class 10 Math Chapter 2 are provided here for you for free. You will also love the ad-free experience on Meritnation’s Mathematics Part I Solutions Solutions. All Mathematics Part I Solutions Solutions for class Class 10 Math are prepared by experts and are 100% accurate.

#### Question 1:

${x}^{2}+10x-200=0$ and ${x}^{2}+5x-6=0$.

#### Question 2:

Decide which of the following are quadratic equations.
(1) x2 + 5x – 2 = 0
(2) y2 = 5y – 10
(3) ${y}^{2}+\frac{1}{y}=2$
(4) $x+\frac{1}{x}=-2$
(5) (m + 2) (m – 5) = 0
(6) m3 + 3m2 – 2 = 3m3

(1) x2 + 5– 2 = 0
Only one variable x.
Maximum index = 2
So, it is a quadratic equation.

(2) y= 5y – 10
Only one variable y.
Maximum index = 2
So, it is a quadratic equation.

(3) ${y}^{2}+\frac{1}{y}=2$
$⇒{y}^{3}+1=2y\phantom{\rule{0ex}{0ex}}$
Only one variable y.
Maximum index = 3
So, it is not a quadratic equation.

(4) $x+\frac{1}{x}=-2$
$⇒{x}^{2}+1=-2x$
Only one variable x.
Maximum index = 2
So, it is a quadratic equation.

(5) (m + 2) (– 5) = 0
$⇒{m}^{2}-3m-10=0$
Only one variable m.
Maximum index = 2
So, it is a quadratic equation.

(6) m+ 3m2 – 2 = 3m3
Only one variable m.
Maximum index = 3
So, it is not a quadratic equation.

#### Question 3:

Write the following equations in the form ax2 + bx + c= 0, then write the values of a, b, c for each equation.
(1) 2y = 10 – y2
(2) (x – 1)2 = 2x + 3
(3) x2 + 5x = –(3 – x)
(4) 3m2 = 2m2 – 9
(5) P(3+6p) = –5
(6) x2 – 9 = 13

(1) 2y = 10 – y2
$⇒{y}^{2}+2y=10\phantom{\rule{0ex}{0ex}}⇒{y}^{2}+2y-10=0$

So, it is of the form ax2 + bx + = 0 where a = 1, b = 2 and c = −10.

(2) (x – 1)2 = 2x + 3
$⇒\left({x}^{2}-2x+1\right)=2x+3\phantom{\rule{0ex}{0ex}}⇒{x}^{2}-4x-2=0\phantom{\rule{0ex}{0ex}}$
So, it is of the form ax2 + bx + = 0 where a = 1, b = −4 and c = −2.

(3) x2 + 5x = –(3 – x)
$⇒{x}^{2}+5x+3-x=0\phantom{\rule{0ex}{0ex}}⇒{x}^{2}+4x+3=0$
So, it is of the form ax2 + bx + = 0 where a = 1, b = 4 and c = 3.

(4) 3m2 = 2m2 – 9
$⇒3{m}^{2}-2{m}^{2}+9=0\phantom{\rule{0ex}{0ex}}⇒{m}^{2}+0m+9=0$
So, it is of the form ax2 + bx + = 0 where a = 1, b = 0 and c = 9.

(5) p(3+6p) = –5
$⇒3p+6{p}^{2}+5=0\phantom{\rule{0ex}{0ex}}⇒6{p}^{2}+3p+5=0$
So, it is of the form ax2 + bx + = 0 where a = 6, b = 3 and c = 5.

(6) x2 – 9 = 13
$⇒{x}^{2}-9-13=0\phantom{\rule{0ex}{0ex}}⇒{x}^{2}-22=0\phantom{\rule{0ex}{0ex}}⇒{x}^{2}+0x-22=0$
So, it is of the form ax2 + bx + = 0 where a = 1, b = 0 and c =  −22.

#### Question 4:

Determine whether the values given against each of the quadratic equation are the roots of the equation.
(1) x2 + 4x – 5 = 0 , x = 1, –1
(2) 2m2 – 5m = 0,

(1) x2 + 4x – 5 = 0 , x = 1, –1

So, x = 1 is a solution of the given equation.
For x = –1
${\left(-1\right)}^{2}+4\left(-1\right)-5=0\phantom{\rule{0ex}{0ex}}⇒1-4-5=0\phantom{\rule{0ex}{0ex}}⇒-3-5=0\phantom{\rule{0ex}{0ex}}⇒-8\ne 0$
So, x = –1 is not a solution of the given equation.
Thus, only x = 1 is root of the given equation.

(2) 2m2 – 5m = 0,

So, = 2 is not a solution of the given equation.

So, $m=\frac{5}{2}$ is a solution of the given quadratic equation.
Thus, only $m=\frac{5}{2}$ is a root of the given quadratic equation.

#### Question 5:

Find k if x = 3 is a root of equation kx2 – 10x + 3 = 0

Given x = 3 is a root of equation kx2 – 10x + 3 = 0
So,
x = 3 must satisfy the given quadration equation.
$k{\left(3\right)}^{2}-10\left(3\right)+3=0\phantom{\rule{0ex}{0ex}}⇒9k-30+3=0\phantom{\rule{0ex}{0ex}}⇒9k-27=0\phantom{\rule{0ex}{0ex}}⇒9k=27\phantom{\rule{0ex}{0ex}}⇒k=\frac{27}{9}=3$
Thus, = 3.

#### Question 6:

One of the roots of equation 5m2 + 2m + k = 0 is $\frac{-7}{5}$ . Complete the following activity to find the value of 'k'.

$\overline{)\frac{-7}{5}}$ is a root of the quadratic equation 5m2 + 2m + = 0.
∴ Put m$\frac{-7}{5}$ in the equation.
$5×{\overline{)\frac{-7}{5}}}^{2}+2×\overline{)\frac{-7}{5}}+k=0\phantom{\rule{0ex}{0ex}}\overline{)\frac{49}{5}}+\overline{)\frac{-14}{5}}+k=0\phantom{\rule{0ex}{0ex}}\overline{)7}+k=0\phantom{\rule{0ex}{0ex}}k=\overline{)-7}$

#### Question 1:

Solve the following quadratic equations by factorisation.
(1) x2 – 15x + 54 = 0
(2) x2 + x – 20 = 0
(3) 2y2 + 27y + 13 = 0
(4) 5m2 = 22m + 15
(5) 2x2 – 2x + $\frac{1}{2}$ = 0
(6) $6x-\frac{2}{x}=1$
(7) $\sqrt{2}{x}^{2}+7x+5\sqrt{2}=0$
to solve this quadratic equation by factorisation, complete the following activity.
(8) $3{x}^{2}-2\sqrt{6}x+2=0$
(9) $2m\left(m-24\right)=50$
(10) $25{m}^{2}=9$
(11) $7{m}^{2}=21m$
(12) ${m}^{2}-11=0$

(1) x2 – 15x + 54 = 0

So, 6 and 9 are the roots of the given quadratic equation.

(2) x2 + x – 20 = 0

So, 4 and $-$5 are the roots of the given quadratic equation.

(3) 2y2 + 27y + 13 = 0

So,  are the roots of the given quadratic equation.

(4) 5m2 = 22m + 15

So,  are the roots of the given quadratic equation.

(5) 2x2 – 2x + $\frac{1}{2}$ = 0

So,  are the roots of the given quadratic equation.

(6) $6x-\frac{2}{x}=1$

are the roots of the given quadratic equation.

(7) $\sqrt{2}{x}^{2}+7x+5\sqrt{2}=0$

(8) $3{x}^{2}-2\sqrt{6}x+2=0$
$⇒3{x}^{2}-\sqrt{6}x-\sqrt{6}x+2=0\phantom{\rule{0ex}{0ex}}⇒\sqrt{3}x\left(\sqrt{3}x-\sqrt{2}\right)-\sqrt{2}\left(\sqrt{3}x-\sqrt{2}\right)=0\phantom{\rule{0ex}{0ex}}⇒\left(\sqrt{3}x-\sqrt{2}\right)\left(\sqrt{3}x-\sqrt{2}\right)=0\phantom{\rule{0ex}{0ex}}⇒x=\frac{\sqrt{2}}{\sqrt{3}},\frac{\sqrt{2}}{\sqrt{3}}$

(9) $2m\left(m-24\right)=50$
$⇒m\left(m-24\right)=25\phantom{\rule{0ex}{0ex}}⇒{m}^{2}-24m=25\phantom{\rule{0ex}{0ex}}⇒{m}^{2}-24m-25=0\phantom{\rule{0ex}{0ex}}⇒{m}^{2}-25m+m-25=0\phantom{\rule{0ex}{0ex}}⇒m\left(m-25\right)+1\left(m-25\right)=0\phantom{\rule{0ex}{0ex}}⇒\left(m+1\right)\left(m-25\right)=0\phantom{\rule{0ex}{0ex}}⇒m=-1,25$

(10) $25{m}^{2}=9$

(11) $7{m}^{2}=21m$

(12) ${m}^{2}-11=0$

#### Question 1:

Solve the following quadratic equations by completing the square method.
(1) x2 + x – 20 = 0
(2) x2 + 2x – 5 = 0
(3) m2 – 5m = –3
(4) 9y2 – 12y + 2 = 0
(5) 2y2 + 9y +10 = 0
(6) 5x2 = 4x + 7

(1) x2 + x – 20 = 0

(2) x2 + 2x – 5 = 0

(3) m2 – 5m = –3

(4) 9y2 – 12y + 2 = 0
$⇒{y}^{2}-\frac{12}{9}y+\frac{2}{9}=0\phantom{\rule{0ex}{0ex}}⇒{y}^{2}-\frac{4}{3}y+\frac{2}{9}=0\phantom{\rule{0ex}{0ex}}⇒{y}^{2}-\frac{4}{3}y+{\left(\frac{\frac{4}{3}}{2}\right)}^{2}-{\left(\frac{\frac{4}{3}}{2}\right)}^{2}+\frac{2}{9}=0\phantom{\rule{0ex}{0ex}}⇒{y}^{2}-\frac{4}{3}y+{\left(\frac{2}{3}\right)}^{2}-{\left(\frac{2}{3}\right)}^{2}+\frac{2}{9}=0\phantom{\rule{0ex}{0ex}}$
$⇒\left[{y}^{2}-\frac{4}{3}y+\left(\frac{4}{9}\right)\right]-\left(\frac{4}{9}\right)+\frac{2}{9}=0\phantom{\rule{0ex}{0ex}}⇒{\left(y-\frac{2}{3}\right)}^{2}-\frac{2}{9}=0\phantom{\rule{0ex}{0ex}}⇒{\left(y-\frac{2}{3}\right)}^{2}=\frac{2}{9}\phantom{\rule{0ex}{0ex}}⇒{\left(y-\frac{2}{3}\right)}^{2}={\left(\frac{\sqrt{2}}{3}\right)}^{2}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}$

(5) 2y2 + 9y +10 = 0
$⇒{y}^{2}+\frac{9}{2}y+5=0\phantom{\rule{0ex}{0ex}}⇒{y}^{2}+\frac{9}{2}y+{\left(\frac{\frac{9}{2}}{2}\right)}^{2}-{\left(\frac{\frac{9}{2}}{2}\right)}^{2}+5=0\phantom{\rule{0ex}{0ex}}⇒\left[{y}^{2}+\frac{9}{2}y+{\left(\frac{9}{4}\right)}^{2}\right]-{\left(\frac{9}{4}\right)}^{2}+5=0\phantom{\rule{0ex}{0ex}}⇒{\left(y+\frac{9}{4}\right)}^{2}=\frac{81}{16}-5\phantom{\rule{0ex}{0ex}}⇒{\left(y+\frac{9}{4}\right)}^{2}=\frac{1}{16}$

(6) 5x2 = 4+ 7
$⇒5{x}^{2}-4x-7=0\phantom{\rule{0ex}{0ex}}⇒{x}^{2}-\frac{4}{5}x-\frac{7}{5}=0\phantom{\rule{0ex}{0ex}}⇒{x}^{2}-\frac{4}{5}x+{\left(\frac{\frac{4}{5}}{2}\right)}^{2}-{\left(\frac{\frac{4}{5}}{2}\right)}^{2}-\frac{7}{5}=0\phantom{\rule{0ex}{0ex}}⇒\left[{x}^{2}-\frac{4}{5}x+{\left(\frac{2}{5}\right)}^{2}\right]-{\left(\frac{2}{5}\right)}^{2}-\frac{7}{5}=0\phantom{\rule{0ex}{0ex}}⇒{\left(x-\frac{2}{5}\right)}^{2}=\frac{7}{5}+\frac{4}{25}=\frac{35+4}{25}=\frac{39}{25}\phantom{\rule{0ex}{0ex}}⇒{\left(x-\frac{2}{5}\right)}^{2}=\frac{39}{25}\phantom{\rule{0ex}{0ex}}$

#### Question 1:

Compare the given quadratic equations to the general form and write values of a,b, c.
(1) x2 – 7x + 5 = 0
(2) 2m2 = 5m – 5
(3) y2 = 7y

(1) x2 – 7x + 5 = 0
General form of the quadratic equation is $a{x}^{2}+bx+c=0$
Comparing x2 – 7x + 5 = 0 with the general form we have a = 1, b$-$7 and c = 5.

(2) 2m2 = 5m – 5
$⇒2{m}^{2}-5m+5=0$
General form of the quadratic equation is $a{x}^{2}+bx+c=0$
Comparing 2m2 = 5m – 5 with the general form we have a = 2, b = $-$5 and c = 5.

(3) y2 = 7y
$⇒{y}^{2}-7y+0=0$

General form of the quadratic equation is $a{x}^{2}+bx+c=0$
Comparing ${y}^{2}-7y+0=0$ with the general form we have a = 1, b = $-$7 and c = 0.

#### Question 2:

Solve using formula.
(1) x2 + 6x + 5 = 0
(2) x2 – 3x – 2 = 0
(3) 3m2 + 2m – 7 = 0
(4) 5m2 – 4m – 2 = 0
(5) ${y}^{2}+\frac{1}{3}y=2$
(6) 5x2 + 13x + 8 = 0

(1) x2 + 6x + 5 = 0
On comparing with the equation $a{x}^{2}+bx+c=0$
a = 1, = 6 and c = 5
Now ${b}^{2}-4ac={6}^{2}-4×1×5=36-20=16$
$x=\frac{-b±\sqrt{{b}^{2}-4ac}}{2a}$

(2) x2 – 3x – 2 = 0
On comparing with the equation $a{x}^{2}+bx+c=0$
a = 1, $-$3 and c = $-$2
Now ${b}^{2}-4ac={\left(-3\right)}^{2}-4×1×\left(-2\right)=9+8=17$
$x=\frac{-b±\sqrt{{b}^{2}-4ac}}{2a}$

(3) 3m2 + 2m – 7 = 0
On comparing with the equation $a{x}^{2}+bx+c=0$
a = 3, = 2 and c = $-$7
Now ${b}^{2}-4ac={\left(2\right)}^{2}-4×3×\left(-7\right)=4+84=88$
$x=\frac{-b±\sqrt{{b}^{2}-4ac}}{2a}$

(4) 5m2 – 4m – 2 = 0
On comparing with the equation $a{x}^{2}+bx+c=0$
a = 5, = – 4and c = $-$2
Now ${b}^{2}-4ac={\left(-4\right)}^{2}-4×5×\left(-2\right)=16+40=56$
$x=\frac{-b±\sqrt{{b}^{2}-4ac}}{2a}$

(5) ${y}^{2}+\frac{1}{3}y=2$
Multiplying the equation by 3
$3{y}^{2}+y=6\phantom{\rule{0ex}{0ex}}⇒3{y}^{2}+y-6=0$
On comparing with the equation $a{x}^{2}+bx+c=0$
a = 3, = 1 and c = $-$6
Now ${b}^{2}-4ac={\left(1\right)}^{2}-4×3×\left(-6\right)=1+72=73$
$x=\frac{-b±\sqrt{{b}^{2}-4ac}}{2a}$

(6) 5x2 + 13x + 8 = 0
On comparing with the equation $a{x}^{2}+bx+c=0$
a = 5, = 13 and c = 8
Now ${b}^{2}-4ac={\left(13\right)}^{2}-4×5×8=169-160=9$
$x=\frac{-b±\sqrt{{b}^{2}-4ac}}{2a}$

#### Question 3:

With the help of the flow chart given below solve the equation ${x}^{2}+2\sqrt{3}x+3=0$ using the formula.

comparing ${x}^{2}+2\sqrt{3}x+3=0$ with $a{x}^{2}+bx+c=0$ we get

${b}^{2}-4ac={\left(2\sqrt{3}\right)}^{2}-4×1×3=12-12=0$
Formula to solve a quadratic equation will be
$x=\frac{-b±\sqrt{{b}^{2}-4ac}}{2a}\phantom{\rule{0ex}{0ex}}⇒x=\frac{-2\sqrt{3}±\sqrt{0}}{2×1}=\frac{-2\sqrt{3}}{2}=-\sqrt{3}$
Thus, $x=-\sqrt{3},-\sqrt{3}$.

#### Question 1:

Fill in the gaps and complete.

(3)If α, β are roots of quadratic equation,

If  are the roots of the quadratic equation then the quadratic equation can be written as
${x}^{2}-\left(\alpha +\beta \right)x+\alpha \beta =0$
So, the quadratic equation with sum of roots as $\alpha +\beta =-7$ and product of roots as $\alpha \beta =5$ will be
${x}^{2}+7x+5=0$

Sum of roots = $\frac{-b}{a}=\frac{-\left(-4\right)}{2}=2$
Product of roots = $\frac{c}{a}=\frac{-3}{2}$

#### Question 2:

Find the value of discriminant.
(1) x2 + 7x – 1 = 0
(2) 2y2 – 5y + 10 = 0
(3) $\sqrt{2}{x}^{2}+4x+2\sqrt{2}=0$

(1) x2 + 7x – 1 = 0
Comparing the given equation with $a{x}^{2}+bx+c=0$,

So, the discriminant ${b}^{2}-4ac={\left(7\right)}^{2}-4×1×\left(-1\right)=49+4=53$

(2) 2y2 – 5y + 10 = 0
Comparing the given equation with $a{x}^{2}+bx+c=0$,

So, the discriminant ${b}^{2}-4ac={\left(-5\right)}^{2}-4×2×10=25-80=-55$

(3) $\sqrt{2}{x}^{2}+4x+2\sqrt{2}=0$
Comparing the given equation with $a{x}^{2}+bx+c=0$,

So, the discriminant ${b}^{2}-4ac={\left(4\right)}^{2}-4×\sqrt{2}×2\sqrt{2}=16-16=0$

#### Question 3:

Determine the nature of roots of the following quadratic equations.
(1) x2 – 4x + 4 = 0
(2) 2y2 – 7y +2 = 0
(3) m2 + 2m + 9 = 0

(1) x2 – 4x + 4 = 0
Comparing the given equation with the quadratic equation $a{x}^{2}+bx+c=0$,
$a=1,b=-4,c=4$
Discriminant, $△={b}^{2}-4ac={\left(-4\right)}^{2}-4×1×4=16-16=0$
Since the discriminant = 0 so, the roots of the given quadratic equation are real and equal.

(2) 2y– 7y +2 = 0
Comparing the given equation with the quadratic equation $a{x}^{2}+bx+c=0$,
$a=2,b=-7,c=2$
Discriminant, $△={b}^{2}-4ac={\left(-7\right)}^{2}-4×2×2=49-16=33$
Since the discriminant > 0 so, the roots of the given quadratic equation are real and unequal.

(3) m2 + 2m + 9 = 0
Comparing the given equation with the quadratic equation $a{x}^{2}+bx+c=0$,
$a=1,b=2,c=9$
Discriminant, $△={b}^{2}-4ac={\left(2\right)}^{2}-4×1×9=4-36=-32$
Since the discriminant < 0 so, the roots of the given quadratic equation are not real.

#### Question 4:

Form the quadratic equation from the roots given below.
(1) 0 and 4
(2) 3 and –10
(3) $\frac{1}{2},-\frac{1}{2}$
(4) $2-\sqrt{5},2+\sqrt{5}$

(1) 0 and 4
Sum of roots = 0 + 4 = 4
Product of roots = 0 $×$ 4 = 0
The general form of the quadratic equation is
So, the quadratic equation obtained is ${x}^{2}-4x+0=0$
$⇒{x}^{2}-4x=0\phantom{\rule{0ex}{0ex}}⇒x\left(x-4\right)=0$

(2) 3 and –10
Sum of roots = 3 + (–10) = −7
Product of roots = 3 $×$ –10 = –30
The general form of the quadratic equation is
So, the quadratic equation obtained is ${x}^{2}-\left(-7\right)x+\left(-30\right)=0$
${x}^{2}+7x-30=0$

(3) $\frac{1}{2},-\frac{1}{2}$
Sum of roots = $\frac{1}{2}+\left(-\frac{1}{2}\right)=0$
Product of roots = $\frac{1}{2}×\left(\frac{-1}{2}\right)=\frac{-1}{4}$
The general form of the quadratic equation is
So, the quadratic equation obtained is ${x}^{2}-\left(0\right)x+\left(\frac{-1}{4}\right)=0$
$⇒{x}^{2}-\frac{1}{4}=0$

(4) $2-\sqrt{5},2+\sqrt{5}$
Sum of roots = $2-\sqrt{5}+2+\sqrt{5}=4$
Product of roots = $\left(2-\sqrt{5}\right)\left(2+\sqrt{5}\right)=4-5=-1$
The general form of the quadratic equation is
So, the quadratic equation obtained is ${x}^{2}-\left(4\right)x+\left(-1\right)=0$
$⇒{x}^{2}-4x-1=0$

#### Question 5:

Sum of the roots of a quadratic equation is double their product. Find k if equation x2 – 4kx + k +3 = 0

Let the roots of the quadratic equation be .
Sum of roots = $\alpha +\beta$
Product of roots = $\alpha \beta$
Given that sum of roots = double the product
$⇒\alpha +\beta =2\alpha \beta$                               .....(I)
Given: x2 – 4kx + +3 = 0           .....(II)
General form of the quadratic equation is

On comparing (II) and (III) we have

Solving these equations we have
$4k=2\left(k+3\right)\phantom{\rule{0ex}{0ex}}⇒2k=k+3\phantom{\rule{0ex}{0ex}}⇒2k-k=3\phantom{\rule{0ex}{0ex}}⇒k=3$

#### Question 6:

α, β are roots of y2 – 2y –7 = 0 find,
(1) α2 + β2
(2) α3 + β3

α, β are roots of y2 – 2y –7 = 0
$a=1,b=-2,c=-7$$a=1,b=-2,c=-7$
(1)
$\alpha +\beta =\frac{-b}{a}=\frac{-\left(-2\right)}{1}=2\phantom{\rule{0ex}{0ex}}\alpha \beta =\frac{c}{a}=\frac{-7}{1}=-7$
${\alpha }^{2}+{\beta }^{2}={\left(\mathrm{\alpha }+\mathrm{\beta }\right)}^{2}-2\alpha \beta \phantom{\rule{0ex}{0ex}}={\left(2\right)}^{2}-2\left(-7\right)\phantom{\rule{0ex}{0ex}}=4+14\phantom{\rule{0ex}{0ex}}=18$

(2)
${\alpha }^{3}+{\beta }^{3}={\left(\alpha +\beta \right)}^{3}-3\alpha \beta \left(\alpha +\beta \right)\phantom{\rule{0ex}{0ex}}={\left(2\right)}^{3}-3\left(-7\right)\left(2\right)\phantom{\rule{0ex}{0ex}}=8+42\phantom{\rule{0ex}{0ex}}=50$

#### Question 7:

The roots of each of the following quadratic equations are real and equal, find k.
(1) 3y2 + ky +12 = 0
(2) kx (x – 2) + 6 = 0

(1) 3y+ ky +12 = 0
The roots of the given quadratic equation are real and equal. So, the discriminant will be 0.
${b}^{2}-4ac=0\phantom{\rule{0ex}{0ex}}⇒{k}^{2}-4×3×12=0\phantom{\rule{0ex}{0ex}}⇒{k}^{2}-144=0\phantom{\rule{0ex}{0ex}}⇒{k}^{2}=144\phantom{\rule{0ex}{0ex}}⇒k=±12$

(2) kx (x – 2) + 6 = 0
$⇒k{x}^{2}-2kx+6=0$
The roots of the given quadratic equation are real and equal. So, the discriminant will be 0.

But k cannot be equal to 0 since then there will not be any quadratic equation.
So, = 6

#### Question 1:

Product of Pragati’s age 2 years ago and 3 years hence is 84. Find her present age.

Let Pragati's present age be years.
Her age 2 years ago = $x-2$
Her age 3 years hence = $x+3$
Product of Pragati’s age 2 years ago and 3 years hence is 84.

But age cannot be negative so, = 9.
Thus, Pragati's present age is 9 years.

#### Question 2:

The sum of squares of two consecutive natural numbers is 244; find the numbers.

Disclaimer: There is an error in the given question. Here it should be two consecutive even natural numbers.

Let the two consecutive natural numbers be
Sum of squares of these two consecutive natural numbers is 244
${x}^{2}+{\left(x+2\right)}^{2}=244\phantom{\rule{0ex}{0ex}}⇒{x}^{2}+{x}^{2}+4+4x=244\phantom{\rule{0ex}{0ex}}⇒2{x}^{2}+4x-240=0\phantom{\rule{0ex}{0ex}}⇒{x}^{2}+2x-120=0\phantom{\rule{0ex}{0ex}}⇒x=\frac{-b±\sqrt{{b}^{2}-4ac}}{2a}\phantom{\rule{0ex}{0ex}}⇒x=\frac{-2±\sqrt{{2}^{2}-4×1×\left(-120\right)}}{2}\phantom{\rule{0ex}{0ex}}⇒x=\frac{-2±\sqrt{4+480}}{2}$
$⇒x=\frac{-2±\sqrt{484}}{2}\phantom{\rule{0ex}{0ex}}⇒x=\frac{-2±22}{2}\phantom{\rule{0ex}{0ex}}⇒x=\frac{-2+22}{2},\frac{-2-22}{2}\phantom{\rule{0ex}{0ex}}⇒x=\frac{20}{2},\frac{-24}{2}\phantom{\rule{0ex}{0ex}}⇒x=10,-12$
But the natural number cannot be negative so,
The two numbers are 10 and 10 + 2 = 12.

#### Question 3:

In the orange garden of Mr. Madhusudan there are 150 orange trees. The number of trees in each row is 5 more than that in each column. Find the number of trees in each row and each column with the help of following flow chart.

Let the number of trees in each column be x
Number of trees in a row = + 5
Total trees = Number of rows $×$ Number of columns = $x\left(x+5\right)$ = 150
$⇒x\left(x+5\right)=150\phantom{\rule{0ex}{0ex}}⇒{x}^{2}+5x-150=0\phantom{\rule{0ex}{0ex}}⇒{x}^{2}+15x-10x-150=0\phantom{\rule{0ex}{0ex}}⇒x\left(x+15\right)-10\left(x+15\right)=0\phantom{\rule{0ex}{0ex}}⇒\left(x+15\right)\left(x-10\right)=0\phantom{\rule{0ex}{0ex}}⇒x=10,-15\phantom{\rule{0ex}{0ex}}$
But number of columns cannot be negative so, number of columns = 10.
Number of rows = 10 + 5 = 15

#### Question 4:

Vivek is older than Kishor by 5 years. The sum of the reciprocals of their ages is $\frac{1}{6}$. Find their present ages.

Let the present age of Kishor be years.
Vivek is older than Kishor by 5 years.
So, Vivek's age will be + 5.
The sum of the reciprocals of their ages is $\frac{1}{6}$.
$\frac{1}{x}+\frac{1}{x+5}=\frac{1}{6}\phantom{\rule{0ex}{0ex}}⇒\frac{x+5+x}{x\left(x+5\right)}=\frac{1}{6}\phantom{\rule{0ex}{0ex}}⇒\frac{5+2x}{{x}^{2}+5x}=\frac{1}{6}\phantom{\rule{0ex}{0ex}}⇒30+12x={x}^{2}+5x\phantom{\rule{0ex}{0ex}}⇒{x}^{2}-7x-30=0\phantom{\rule{0ex}{0ex}}⇒{x}^{2}-10x+3x-30=0\phantom{\rule{0ex}{0ex}}⇒x\left(x-10\right)+3\left(x-10\right)=0\phantom{\rule{0ex}{0ex}}⇒\left(x+3\right)\left(x-10\right)=0\phantom{\rule{0ex}{0ex}}⇒x=-3,10$
But age cannot be negative so, age of Kishore is 10 years and that of Vivek is 10 + 5 = 15 years.

#### Question 5:

Suyash scored 10 marks more in second test than that in the first. 5 times the score of the second test is the same as square of the score in the first test. Find his score in the first test.

Let the score in the first test be x
Suyash's score in second test is 10 more than that in the first.
So, his score in the second test = + 10
5 times the score of the second test is the same as square of the score in the first test.
$5\left(x+10\right)={x}^{2}\phantom{\rule{0ex}{0ex}}⇒5x+50={x}^{2}\phantom{\rule{0ex}{0ex}}⇒{x}^{2}-5x-50=0\phantom{\rule{0ex}{0ex}}⇒{x}^{2}-10x+5x-50=0\phantom{\rule{0ex}{0ex}}⇒x\left(x-10\right)+5\left(x-10\right)=0\phantom{\rule{0ex}{0ex}}⇒\left(x+5\right)\left(x-10\right)=0\phantom{\rule{0ex}{0ex}}⇒x=10,-5$
The score is not negative so, Suyash scored 10 marks in first test.

#### Question 6:

Mr. Kasam runs a small business of making earthen pots. He makes certain number of pots on daily basis. Production cost of each pot is Rs 40 more than 10 times total number of pots, he makes in one day. If production cost of all pots per day is Rs 600, find production cost of one pot and number of pots he makes per day.

Let the number of pots Mr. Kasam makes in one day be x.
Production cost of each pot is Rs 40 more than 10 times total number of pots, he makes in one day = 10+ 40
Production cost of all pots per day is Rs 600
(10+ 40)= 600
$10{x}^{2}+40x=600\phantom{\rule{0ex}{0ex}}⇒{x}^{2}+4x-60=0\phantom{\rule{0ex}{0ex}}⇒{x}^{2}+10x-6x-60=0\phantom{\rule{0ex}{0ex}}⇒x\left(x+10\right)-6\left(x+10\right)=0\phantom{\rule{0ex}{0ex}}⇒\left(x-6\right)\left(x+10\right)=0\phantom{\rule{0ex}{0ex}}⇒x=6,-10$
But the number of pots cannot be negative so,
$x\ne -10\phantom{\rule{0ex}{0ex}}⇒x=6$
Production cost of 1 pot =

#### Question 7:

Pratik takes 8 hours to travel 36 km down stream and return to the same spot. The speed of boat in still water is 12 km. per hour. Find the speed of water current.

Let the speed of water current = y km/h
Speed of boat in still water = 12 km/h
Speed upstream = $12-y$
Speed downstream = 12 +
Distance travelled = 36 km downstream and 36 km upstream
$\mathrm{Speed}=\frac{\mathrm{Distance}}{\mathrm{Time}}\phantom{\rule{0ex}{0ex}}⇒\mathrm{Time}=\frac{\mathrm{Distance}}{\mathrm{Speed}}$
$\frac{36}{12-y}+\frac{36}{12+y}=8\phantom{\rule{0ex}{0ex}}⇒36\left[\frac{1}{12-y}+\frac{1}{12+y}\right]=8\phantom{\rule{0ex}{0ex}}⇒\frac{12+y+12-y}{144-{y}^{2}}=\frac{1}{4}\phantom{\rule{0ex}{0ex}}⇒\frac{24}{144-{y}^{2}}=\frac{1}{4}\phantom{\rule{0ex}{0ex}}⇒144-{y}^{2}=24×4\phantom{\rule{0ex}{0ex}}⇒{y}^{2}=36\phantom{\rule{0ex}{0ex}}⇒{y}^{2}-36=0\phantom{\rule{0ex}{0ex}}⇒y=±6$
Speed cannot be negative so, the speed of water current = 6 km/h.

#### Question 8:

Pintu takes 6 days more than those of Nishu to complete certain work. If they work together they finish it in 4 days. How many days would it take to complete the work if they work alone.

Let the number of days taken by Nishu be x
So, number of days taken by Pintu will be + 6
Nishu's 1 day work = $\frac{1}{x}$
Pintu's one day work = $\frac{1}{x+6}$
Work done together in 4 days = $\frac{1}{4}$
$\frac{1}{x}+\frac{1}{x+6}=\frac{1}{4}\phantom{\rule{0ex}{0ex}}⇒\frac{x+6+x}{x\left(x+6\right)}=\frac{1}{4}\phantom{\rule{0ex}{0ex}}⇒\frac{6+2x}{{x}^{2}+6x}=\frac{1}{4}\phantom{\rule{0ex}{0ex}}⇒24+8x={x}^{2}+6x\phantom{\rule{0ex}{0ex}}⇒{x}^{2}-2x-24=0\phantom{\rule{0ex}{0ex}}⇒{x}^{2}-6x+4x-24=0\phantom{\rule{0ex}{0ex}}⇒\left(x-6\right)\left(x+4\right)=0\phantom{\rule{0ex}{0ex}}⇒x=6,-4$
Number of days cannot be negative so, x = 6.
Thus, number of days taken by Nishu = 6 and that by Pintu = 6 + 6 = 12.

#### Question 9:

If 460 is divided by a natural number, quotient is 6 more than five times the divisor and remainder is 1. Find quotient and diviser.

Let the divisor be x
Dividend = 460
Quotient = 5x + 6
Remainder = 1
We know
$\mathrm{Divident}=\mathrm{Divisor}×\mathrm{Quotient}+\mathrm{Remainder}$
$460=x\left(5x+6\right)+1\phantom{\rule{0ex}{0ex}}⇒460=5{x}^{2}+6x+1\phantom{\rule{0ex}{0ex}}⇒5{x}^{2}+6x-459=0\phantom{\rule{0ex}{0ex}}⇒x=\frac{-b±\sqrt{{b}^{2}-4ac}}{2a}\phantom{\rule{0ex}{0ex}}⇒x=\frac{-6±\sqrt{{\left(6\right)}^{2}-4×5×\left(-459\right)}}{2×5}\phantom{\rule{0ex}{0ex}}⇒x=\frac{-6±\sqrt{36+9180}}{10}\phantom{\rule{0ex}{0ex}}⇒x=\frac{-6±\sqrt{9216}}{10}$
$⇒x=\frac{-6±96}{10}\phantom{\rule{0ex}{0ex}}⇒x=\frac{-6+96}{10},\frac{-6-96}{10}\phantom{\rule{0ex}{0ex}}⇒x=9,-10.2$
Thus, divisor = 9, quotient = 5x + 6 = $5×9+6=45+6=51$

#### Question 10:

In the adjoining fig. $\square$ABCD is a trapezium AB || CD and its area is 33 cm2 . From the information given in the figure find the lengths of all sides of the  $\square$ABCD. Fill in the empty boxes to get the solution.

$\square$ABCD is a trapezium.
AB || CD
A($\square$ABCD) = $\frac{1}{2}\left(\mathrm{AB}+\mathrm{CD}\right)×\overline{)\mathrm{AM}}$

But length is never negative.

AB = 7, CD = 15, BC = 5

#### Question 1:

Choose the correct answers for the following questions.
(1) Which one is the quadratic equation ?

 (A) $\frac{5}{x}-3={x}^{2}$ (B) x(x + 5) = 2 (C) n – 1 = 2n (D) $\frac{1}{{x}^{2}}\left(x+2\right)=x$

(2) Out of the following equations which one is not a quadratic equation ?
 (A) x2 + 4x = 11 + x2 (B) x2 = 4x (C) 5x2 = 90 (D) 2x – x2 = x2 + 5

(3) The roots of x2 + kx + k = 0 are real and equal, find k.
 (A) 0 (B) 4 (C) 0 or 4 (D) 2

(4) For $\sqrt{2}{x}^{2}-5x+\sqrt{2}=0$ find the value of the discriminant.
 (A) –5 (B) 17 (C) $\sqrt{2}$ (D) $2\sqrt{2}-5$

(5) Which of the following quadratic equations has roots 3,5 ?
 (A) x2 – 15x + 8 = 0 (B) x2 – 8x + 15 = 0 (C) x2 + 3x + 5 = 0 (D) x2 + 8x – 15 = 0

(6) Out of the following equations, find the equation having the sum of its roots –5.
 (A) 3x2 – 15x + 3 = 0 (B) x2 – 5x + 3 = 0 (C) x2 + 3x – 5 = 0 (D) 3x2 + 15x + 3 = 0

(7) $\sqrt{5}{m}^{2}-\sqrt{5}m+\sqrt{5}=0$ which of the following statement is true for this given equation ?
 (A) Real and uneual roots (B) Real and equal roots (C) Roots are not real (D) Three roots.

(8) One of the roots of equation x2 + mx – 5 = 0 is 2; find m.
 (A) –2 (B) $-\frac{1}{2}$ (C) $\frac{1}{2}$ (D) 2

(1) The general form of a quadratic equation is $a{x}^{2}+bx+c=0$.
For (A), $\frac{5}{x}-3={x}^{2}$
$5-3x={x}^{3}$
Thus, (A) is not quadratic equation.
For (B) x(x + 5) = 2
${x}^{2}+5x-2=0$
Thus, (B) is a qudratic equation.
(C) is also not a quadratic equation.
(D) $\frac{1}{{x}^{2}}\left(x+2\right)=x$
$x+2={x}^{3}$ which is also not quadratic.
Hence, the correct answer is option (B).

(2) (A) x2 + 4x = 11 + x2
$⇒4x=11$
Thus, (A) is not a quadratic equation.
(B) and (C) can be written as ${x}^{2}-4x+0=0$ and $5{x}^{2}-90+0=0$ respectively.
So, B and C are quadratic equations.
(D) 2x – x2 = x2 + 5 can be written as $2{x}^{2}-2x+5=0$
So, it also forms a quadratic equation.
Hence, the correct answer is option (A).

(3) Given quadratic equation is x2 + kx + k = 0
For real and equal roots, D = 0
${b}^{2}-4ac=0\phantom{\rule{0ex}{0ex}}⇒{k}^{2}-4×1×k=0\phantom{\rule{0ex}{0ex}}⇒{k}^{2}-4k=0\phantom{\rule{0ex}{0ex}}⇒k\left(k-4\right)=0\phantom{\rule{0ex}{0ex}}⇒k=0,k=4$
Hence, the correct answer is option C.

(4) $\sqrt{2}{x}^{2}-5x+\sqrt{2}=0$
$a=\sqrt{2},b=-5,c=\sqrt{2}\phantom{\rule{0ex}{0ex}}D={b}^{2}-4ac\phantom{\rule{0ex}{0ex}}={\left(-5\right)}^{2}-4×\sqrt{2}×\sqrt{2}\phantom{\rule{0ex}{0ex}}=25-8\phantom{\rule{0ex}{0ex}}=17$
Hence, the correct answer is option B.

(5) The roots of the quadratic equation x2 – 8x + 15 = 0 are
${x}^{2}-5x-3x+15=0\phantom{\rule{0ex}{0ex}}⇒x\left(x-5\right)-3\left(x-5\right)=0\phantom{\rule{0ex}{0ex}}⇒\left(x-3\right)\left(x-5\right)=0\phantom{\rule{0ex}{0ex}}⇒x=3,5$
Hence, the correct answer is option B.

(6) For the equation 3x2 + 15x + 3 = 0

Hence, the correct answer is option D.

(7) For the given quadratic equation $\sqrt{5}{m}^{2}-\sqrt{5}m+\sqrt{5}=0$
$D={b}^{2}-4ac={\left(-\sqrt{5}\right)}^{2}-4×\sqrt{5}×\sqrt{5}=5-20=-15$
Since D < 0 so, the roots are not real.
Hence, the correct answer is option C.

(8) For the equation, x2 + mx – 5 = 0
one of the roots is 2. So, 2 should satisfy the given equation.
${\left(2\right)}^{2}+m×\left(2\right)-5=0\phantom{\rule{0ex}{0ex}}⇒4+2m-5=0\phantom{\rule{0ex}{0ex}}⇒2m-1=0\phantom{\rule{0ex}{0ex}}⇒m=\frac{1}{2}$
Hence, the correct answer is option C.

#### Question 2:

Which of the following equations is quadratic ?
(1) ${x}^{2}+2x+11=0$
(2) ${x}^{2}-2x+5={x}^{2}$
(3) ${\left(x+2\right)}^{2}=2{x}^{2}$

General form of a quadratic equation is
$a{x}^{2}+bx+c=0$
(1)  ${x}^{2}+2x+11=0$ is in the form of the general quadratic equation.
(2) ${x}^{2}-2x+5={x}^{2}$
$⇒-2x+5=0$
It is not in the form of quadratic equation.
(3) ${\left(x+2\right)}^{2}=2{x}^{2}$
$⇒{x}^{2}+4+4x=2{x}^{2}\phantom{\rule{0ex}{0ex}}⇒{x}^{2}-4x-4=0$
It is also in the form of a quadratic equation.
Hence, (1) and (3) are quadratic equations.

#### Question 3:

Find the value of discriminant for each of the following equations.
(1) $2{y}^{2}-y+2=0$
(2) $5{m}^{2}-m=0$
(3) $\sqrt{5}{x}^{2}-x-\sqrt{5}=0$

(1) $2{y}^{2}-y+2=0$
For the quadratic equation, $a{x}^{2}+bx+c=0$
$D={b}^{2}-4ac$
Here,
$a=2,b=-1,c=2\phantom{\rule{0ex}{0ex}}D={\left(-1\right)}^{2}-4×2×2=1-16=-15$

(2) $5{m}^{2}-m=0$
For the quadratic equation, $a{x}^{2}+bx+c=0$
$D={b}^{2}-4ac$
Here,
$m=5,b=-1,c=0\phantom{\rule{0ex}{0ex}}D={\left(-1\right)}^{2}-4×5×0=1$

(3) $\sqrt{5}{x}^{2}-x-\sqrt{5}=0$
For the quadratic equation, $a{x}^{2}+bx+c=0$
$D={b}^{2}-4ac$
Here,
$a=\sqrt{5},b=-1,c=-\sqrt{5}\phantom{\rule{0ex}{0ex}}D={\left(-1\right)}^{2}-4×\left(\sqrt{5}\right)×\left(-\sqrt{5}\right)\phantom{\rule{0ex}{0ex}}=1+20=21$

#### Question 4:

One of the roots of quadratic equation $2{x}^{2}+kx-2=0$ is –2. find k.

$2{x}^{2}+kx-2=0$
One of the roots is –2 so, it will satisfy the given equation.
$2{\left(-2\right)}^{2}+k\left(-2\right)-2=0\phantom{\rule{0ex}{0ex}}⇒8-2k-2=0\phantom{\rule{0ex}{0ex}}⇒6-2k=0\phantom{\rule{0ex}{0ex}}⇒6=2k\phantom{\rule{0ex}{0ex}}⇒k=3$

#### Question 5:

Two roots of quadratic equations are given ; frame the equation.
(1) 10 and –10
(2)
(3) 0 and 7

(1) 10 and –10
Sum of roots = 10 + (–10) = 0
Product of roots = $10×\left(-10\right)=-100$
The general form of the quadratic equation is

So, the quadratic equation will be
${x}^{2}-0x-100=0\phantom{\rule{0ex}{0ex}}⇒{x}^{2}-100=0$

(2)
Sum of roots = $1-3\sqrt{5}+1+3\sqrt{5}=2$
Product of roots = $\left(1-3\sqrt{5}\right)\left(1+3\sqrt{5}\right)=1-45=-44$
The general form of the quadratic equation is

So, the quadratic equation will be
${x}^{2}-2x-44=0$

(3) 0 and 7
Sum of roots = 0 + 7 = 7
Product of roots = $0×7=0$
The general form of the quadratic equation is

So, the quadratic equation will be
${x}^{2}-7x+0=0\phantom{\rule{0ex}{0ex}}⇒{x}^{2}-7x=0$

#### Question 6:

Determine the nature of roots for each of the quadratic equation.
(1) $3{x}^{2}-5x+7=0$
(2) $\sqrt{3}{x}^{2}+\sqrt{2}x-2\sqrt{3}=0$
(3) ${m}^{2}-4x-3=0$

(1) $3{x}^{2}-5x+7=0$
In order to find the nature of the roots we need to find the discriminant.
$D={b}^{2}-4ac\phantom{\rule{0ex}{0ex}}a=3,b=-5,c=7\phantom{\rule{0ex}{0ex}}D={\left(-5\right)}^{2}-4×3×7=25-84=-59$
Since, the D < 0 so, there is no real root.

(2) $\sqrt{3}{x}^{2}+\sqrt{2}x-2\sqrt{3}=0$
In order to find the nature of the roots we need to find the discriminant.
$D={b}^{2}-4ac\phantom{\rule{0ex}{0ex}}a=\sqrt{3},b=\sqrt{2},c=-2\sqrt{3}\phantom{\rule{0ex}{0ex}}D={\left(\sqrt{2}\right)}^{2}-4×\sqrt{3}×\left(-2\sqrt{3}\right)=2+24=26>0$
Since, the D > 0 so, real and unequal roots.

(3) ${m}^{2}-4x-3=0$
In order to find the nature of the roots we need to find the discriminant.
$D={b}^{2}-4ac\phantom{\rule{0ex}{0ex}}a=1,b=-4,c=-3\phantom{\rule{0ex}{0ex}}D={\left(-4\right)}^{2}-4×\left(1\right)×\left(-3\right)=16+12=28$
Since, the D > 0 so, real and unequal roots.

#### Question 7:

(1) $\frac{1}{x+5}=\frac{1}{{x}^{2}}$
(2) ${x}^{2}-\frac{3x}{10}-\frac{1}{10}=0$
(3) ${\left(2x+3\right)}^{2}=25$
(4) ${m}^{2}+5m+5=0$
(5) $5{m}^{2}+2m+1=0$
(6) ${x}^{2}-4x-3=0$

(1) $\frac{1}{x+5}=\frac{1}{{x}^{2}}$
$⇒{x}^{2}=x+5\phantom{\rule{0ex}{0ex}}⇒{x}^{2}-x-5=0\phantom{\rule{0ex}{0ex}}⇒x=\frac{-\left(-1\right)±\sqrt{{\left(-1\right)}^{2}-4×1×\left(-5\right)}}{2×1}\phantom{\rule{0ex}{0ex}}⇒x=\frac{1±\sqrt{1+20}}{2}=\frac{1±\sqrt{21}}{2}$
(2) ${x}^{2}-\frac{3x}{10}-\frac{1}{10}=0$
$⇒10{x}^{2}-3x-1=0\phantom{\rule{0ex}{0ex}}⇒x=\frac{-\left(-3\right)±\sqrt{{\left(-3\right)}^{2}-4×10×\left(-1\right)}}{2×10}\phantom{\rule{0ex}{0ex}}⇒x=\frac{3±\sqrt{9+40}}{20}=\frac{3±\sqrt{49}}{20}\phantom{\rule{0ex}{0ex}}⇒x=\frac{3±7}{20}\phantom{\rule{0ex}{0ex}}⇒x=\frac{3+7}{20},\frac{3-7}{20}\phantom{\rule{0ex}{0ex}}⇒x=\frac{10}{20},\frac{-4}{20}\phantom{\rule{0ex}{0ex}}⇒x=\frac{1}{2},\frac{-1}{5}$

(3) ${\left(2x+3\right)}^{2}=25$
$⇒4{x}^{2}+9+12x=25\phantom{\rule{0ex}{0ex}}⇒4{x}^{2}+12x-16=0\phantom{\rule{0ex}{0ex}}⇒{x}^{2}+3x-4=0\phantom{\rule{0ex}{0ex}}⇒{x}^{2}+4x-x-4=0\phantom{\rule{0ex}{0ex}}⇒x\left(x+4\right)-1\left(x+4\right)=0\phantom{\rule{0ex}{0ex}}⇒\left(x-1\right)\left(x+4\right)=0\phantom{\rule{0ex}{0ex}}⇒x=1,-4$

(4) ${m}^{2}+5m+5=0$
$⇒m=\frac{-5±\sqrt{{\left(5\right)}^{2}-4×1×5}}{2}\phantom{\rule{0ex}{0ex}}⇒m=\frac{-5±\sqrt{25-20}}{2}\phantom{\rule{0ex}{0ex}}⇒m=\frac{-5±\sqrt{5}}{2}\phantom{\rule{0ex}{0ex}}$

(5) $5{m}^{2}+2m+1=0$
$⇒m=\frac{-2±\sqrt{{\left(2\right)}^{2}-4×1×5}}{2}\phantom{\rule{0ex}{0ex}}⇒m=\frac{-2±\sqrt{4-20}}{2}\phantom{\rule{0ex}{0ex}}⇒m=\frac{-2±\sqrt{-16}}{2}\phantom{\rule{0ex}{0ex}}$
So, the roots are not real as the discriminant is negative.

(6) ${x}^{2}-4x-3=0$
$⇒x=\frac{-\left(-4\right)±\sqrt{{\left(-4\right)}^{2}-4×1×\left(-3\right)}}{2×1}\phantom{\rule{0ex}{0ex}}⇒x=\frac{4±\sqrt{16+12}}{2}\phantom{\rule{0ex}{0ex}}⇒x=\frac{4±\sqrt{28}}{2}\phantom{\rule{0ex}{0ex}}⇒x=\frac{4±2\sqrt{7}}{2}\phantom{\rule{0ex}{0ex}}⇒x=2±\sqrt{7}$

#### Question 8:

Find m if (m – 12) x2 + 2(m –12) x + 2 = 0 has real and equal roots.

For real and equal roots, the Discriminant should be 0.
(m – 12) x2 + 2(–12) + 2 = 0
$∆={b}^{2}-4ac=0\phantom{\rule{0ex}{0ex}}a=m-12,b=2\left(m-12\right),c=2$
${\left[2\left(m-12\right)\right]}^{2}-4\left(m-12\right)×2=0\phantom{\rule{0ex}{0ex}}⇒4{m}^{2}+576-96m-8m+96=0\phantom{\rule{0ex}{0ex}}⇒4{m}^{2}-104m+672=0\phantom{\rule{0ex}{0ex}}⇒{m}^{2}-26m+168=0\phantom{\rule{0ex}{0ex}}⇒{m}^{2}-14m-12m+168=0\phantom{\rule{0ex}{0ex}}⇒m\left(m-14\right)-12\left(m-14\right)=0\phantom{\rule{0ex}{0ex}}⇒\left(m-14\right)\left(m-12\right)=0\phantom{\rule{0ex}{0ex}}m=12,14$
But if m = 12, the quadratic equation will be formed hence, the vaalue of is 14.

#### Question 9:

The sum of two roots of a quadratic equation is 5 and sum of their cubes is 35, find the equation.

Let the roots be
Sum of roots = 5
$\alpha +\beta =5$
Also,
${\alpha }^{3}+{\beta }^{3}=35\phantom{\rule{0ex}{0ex}}⇒{\alpha }^{3}+{\beta }^{3}={\left(\alpha +\beta \right)}^{3}-3\alpha \beta \left(\alpha +\beta \right)\phantom{\rule{0ex}{0ex}}⇒35={5}^{3}-3\alpha \beta \left(5\right)\phantom{\rule{0ex}{0ex}}⇒15\alpha \beta =125-35\phantom{\rule{0ex}{0ex}}⇒\alpha \beta =6$
${x}^{2}-\left(\alpha +\beta \right)x+\alpha \beta =0\phantom{\rule{0ex}{0ex}}$
So, the required quadratic equation will be
${x}^{2}-\left(5\right)x+6=0\phantom{\rule{0ex}{0ex}}⇒{x}^{2}-5x+6=0\phantom{\rule{0ex}{0ex}}$

#### Question 10:

Find quadratic equation such that its roots are square of sum of the roots and square of difference of the roots of equation $2{x}^{2}+2\left(p+q\right)x+{p}^{2}+{q}^{2}=0$

$2{x}^{2}+2\left(p+q\right)x+{p}^{2}+{q}^{2}=0$
Let the roots of the given quadratic equation be $\alpha$ and $\beta$
Sum of roots, $\alpha$ + $\beta$$\frac{-2\left(p+q\right)}{2}=-\left(p+q\right)$
$⇒{\left(\alpha +\beta \right)}^{2}={\left(p+q\right)}^{2}$                              .....(A)
$\alpha \beta =\frac{{p}^{2}+{q}^{2}}{2}$

$AB={\left(p+q\right)}^{2}\left[-{\left(p-q\right)}^{2}\right]\phantom{\rule{0ex}{0ex}}=-\left({p}^{2}+{q}^{2}+2pq\right)\left({p}^{2}+{q}^{2}-2pq\right)\phantom{\rule{0ex}{0ex}}=-\left[{\left({p}^{2}\right)}^{2}+{\left({q}^{2}\right)}^{2}-2{p}^{2}{q}^{2}\right]\phantom{\rule{0ex}{0ex}}=-{\left({p}^{2}-{q}^{2}\right)}^{2}$
General form of quadratic equation is
${x}^{2}-\left(A+B\right)x+AB=0\phantom{\rule{0ex}{0ex}}$
Putting the value of A and B we get
${x}^{2}-4pq-{\left({p}^{2}-{q}^{2}\right)}^{2}=0$.

#### Question 11:

Mukund possesses Rs 50 more than what Sagar possesses. The product of the amount they have is 15,000. Find the amount each one has.

Let amount with Sagar be Rs x.
Amount with Mukund = Rs + 50
The product of the amount they have is 15,000.
$x\left(x+50\right)=15000\phantom{\rule{0ex}{0ex}}⇒{x}^{2}+50x=15000\phantom{\rule{0ex}{0ex}}⇒{x}^{2}+50x-15000=0\phantom{\rule{0ex}{0ex}}⇒{x}^{2}+150x-100x-15000=0\phantom{\rule{0ex}{0ex}}⇒x\left(x+150\right)-100\left(x+150\right)=0\phantom{\rule{0ex}{0ex}}⇒\left(x-100\right)\left(x+150\right)=0\phantom{\rule{0ex}{0ex}}⇒x=100,-150$
But amount cannot be negative so,
Amount with Sagar = Rs 100 and that with Mukund is Rs 150.

#### Question 12:

The difference between squares of two numbers is 120. The square of smaller number is twice the greater number. Find the numbers.

Let the smaller number be x.
Larger number be y
The square of smaller number is twice the greater number.
${x}^{2}=2y$
The difference between squares of two numbers is 120.
${y}^{2}-{x}^{2}=120\phantom{\rule{0ex}{0ex}}⇒{y}^{2}-2y-120=0\phantom{\rule{0ex}{0ex}}⇒{y}^{2}-12y+10y-120=0\phantom{\rule{0ex}{0ex}}⇒y\left(y-12\right)+10\left(y-12\right)=0\phantom{\rule{0ex}{0ex}}⇒\left(y+10\right)\left(y-12\right)=0\phantom{\rule{0ex}{0ex}}⇒y=12,-10$
So, the numbers are 12 as the larger number and the smaller number is
${x}^{2}=2×12=24\phantom{\rule{0ex}{0ex}}⇒x=±\sqrt{24}$

#### Question 13:

Ranjana wants to distribute 540 oranges among some students. If 30 students were more each would get 3 oranges less. Find the number of students.

Total oranges = 540
Let the total number of students be x
Orange each student gets will be $\frac{540}{x}$
If 30 more students are there so, the total number of students will be + 30
Number of oranges per person will be $\frac{540}{x+30}$.
If 30 students were more each would get 3 oranges less = $\frac{540}{x}-3$
$\frac{540}{x+30}$$\frac{540}{x}-3$
$⇒3=540\left(\frac{1}{x}-\frac{1}{x+30}\right)\phantom{\rule{0ex}{0ex}}⇒1=180\left(\frac{x+30-x}{x\left(x+30\right)}\right)\phantom{\rule{0ex}{0ex}}⇒1=180\left(\frac{30}{{x}^{2}+30x}\right)\phantom{\rule{0ex}{0ex}}⇒{x}^{2}+30x=5400\phantom{\rule{0ex}{0ex}}⇒{x}^{2}+30x-5400=0\phantom{\rule{0ex}{0ex}}⇒{x}^{2}+90x-60x-5400=0\phantom{\rule{0ex}{0ex}}⇒x\left(x+90\right)-60\left(x+90\right)=0\phantom{\rule{0ex}{0ex}}⇒\left(x-60\right)\left(x+90\right)=0\phantom{\rule{0ex}{0ex}}⇒x=60,-90$
But number of students cannot be negative so, the number of students is 60.

#### Question 14:

Mr. Dinesh owns an agricultural farm at village Talvel. The length of the farm is 10 meter more than twice the breadth. In order to harvest rain water, he dug a square shaped pond inside the farm. The side of pond is $\frac{1}{3}$ of the breadth of the farm. The area of the farm is 20 times the area of the pond. Find the length and breadth of the farm and of the pond.

Length of farm = $2x+10$
Area of farm = $lb=x\left(2x+10\right)=2{x}^{2}+10x$
Side of pond = $\frac{x}{3}$
Area of pond = ${s}^{2}={\left(\frac{x}{3}\right)}^{2}$
$⇒2{x}^{2}+10x=20×\frac{{x}^{2}}{9}\phantom{\rule{0ex}{0ex}}⇒18{x}^{2}+90x=20{x}^{2}\phantom{\rule{0ex}{0ex}}⇒9{x}^{2}+45x=10{x}^{2}\phantom{\rule{0ex}{0ex}}⇒{x}^{2}-45x=0\phantom{\rule{0ex}{0ex}}⇒x\left(x-45\right)=0\phantom{\rule{0ex}{0ex}}⇒x=0,45$
Side of pond = $\frac{45}{3}=15$ units