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#### Question 1:

Which of the following are quadratic equation in x ?
(i) ${x}^{2}-x+3=0$
(ii) $2{x}^{2}+\frac{5}{2}x-\sqrt{3}=0$
(iii) $\sqrt{2}{x}^{2}+7x+5\sqrt{2}=0$
(iv) $\frac{1}{3}{x}^{2}+\frac{1}{5}x-2=0$
(v) ${x}^{2}-3x-\sqrt{x}+4=0$
(vi) $x-\frac{6}{x}=3$
(vii) $x+\frac{2}{x}={x}^{2}$
(viii) ${x}^{2}-\frac{1}{{x}^{2}}=5$
(ix) ${\left(x+2\right)}^{3}={x}^{3}-8$
(x) $\left(2x+3\right)\left(3x+2\right)=6\left(x-1\right)\left(x-2\right)$
(xi) ${\left(x+\frac{1}{x}\right)}^{2}=2\left(x+\frac{1}{x}\right)+3$

#### Answer:

(ix) ${\left(x+2\right)}^{3}={x}^{3}-8$
$⇒{x}^{3}+6{x}^{2}+12x+8={x}^{3}-8\phantom{\rule{0ex}{0ex}}⇒6{x}^{2}+12x+16=0$
This is of the form ax2 + bx + c = 0.
Hence, the given equation is a quadratic equation.
(x) $\left(2x+3\right)\left(3x+2\right)=6\left(x-1\right)\left(x-2\right)$
$⇒6{x}^{2}+4x+9x+6=6\left({x}^{2}-3x+2\right)\phantom{\rule{0ex}{0ex}}⇒6{x}^{2}+13x+6=6{x}^{2}-18x+12\phantom{\rule{0ex}{0ex}}⇒31x-6=0$
This is not of the form ax2 + bx + c = 0.
Hence, the given equation is not a quadratic equation.
(xi) ${\left(x+\frac{1}{x}\right)}^{2}=2\left(x+\frac{1}{x}\right)+3$
$⇒{\left(\frac{{x}^{2}+1}{x}\right)}^{2}=2\left(\frac{{x}^{2}+1}{x}\right)+3\phantom{\rule{0ex}{0ex}}⇒{\left({x}^{2}+1\right)}^{2}=2x\left({x}^{2}+1\right)+3{x}^{2}\phantom{\rule{0ex}{0ex}}⇒{x}^{4}+2{x}^{2}+1=2{x}^{3}+2x+3{x}^{2}\phantom{\rule{0ex}{0ex}}⇒{x}^{4}-2{x}^{3}-{x}^{2}-2x+1=0$
This is not of the form ax2 + bx + c = 0.
Hence, the given equation is not a quadratic equation.

#### Question 2:

Which of the following are the roots of $3{x}^{2}+2x-1=0$ ?
(i) −1
(ii) $\frac{1}{3}$
(iii) $-\frac{1}{2}$

#### Question 3:

Find the value of k for which x = 1 is root of the equation ${x}^{2}+kx+3=0$.

#### Question 4:

Find the values of a and b for which are the roots of the equation $a{x}^{2}+bx-6=0.$

#### Question 5:

Solve each of the following quadratic equations:

(2x − 3)(3x + 1) = 0

#### Answer:

(2x − 3)(3x + 1) = 0

⇒ 2x − 3 = 0 or 3x + 1 = 0

⇒ 2x = 3 or 3x = −1

x$\frac{3}{2}$ or x = $-\frac{1}{3}$

Hence, the roots of the given equation are $\frac{3}{2}$ and $-\frac{1}{3}$.

#### Question 6:

Solve each of the following quadratic equations:

4x2 + 5x = 0

#### Answer:

4x2 + 5x = 0

x(4x + 5) = 0

x = 0 or 4x + 5 = 0

x = 0 or x = $-\frac{5}{4}$

Hence, the roots of the given equation are 0 and $-\frac{5}{4}$.

#### Question 7:

$3{x}^{2}-243=0$

#### Question 8:

Solve each of the following quadratic equations:

$2{x}^{2}+x-6=0$

#### Answer:

We write, $x=4x-3x$ as $2{x}^{2}×\left(-6\right)=-12{x}^{2}=4x×\left(-3x\right)$

Hence, the roots of the given equation are $-2$ and $\frac{3}{2}$.

#### Question 9:

Solve each of the following quadratic equations:

${x}^{2}+6x+5=0$

#### Answer:

We write, $6x=x+5x$ as ${x}^{2}×5=5{x}^{2}=x×5x$

Hence, the roots of the given equation are −1 and −5.

#### Question 10:

Solve each of the following quadratic equations:

$9{x}^{2}-3x-2=0$

#### Answer:

We write, $-3x=3x-6x$ as $9{x}^{2}×\left(-2\right)=-18{x}^{2}=3x×\left(-6x\right)$

Hence, the roots of the given equation are $-\frac{1}{3}$ and $\frac{2}{3}$.

#### Question 11:

${x}^{2}+12x+35=0$

#### Question 12:

${x}^{2}=18x-77$

#### Question 13:

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

#### Question 14:

$6{x}^{2}+x-12=0$

#### Question 15:

Solve each of the following quadratic equations:

$3{x}^{2}-2x-1=0$

#### Answer:

We write, $-2x=-3x+x$ as $3{x}^{2}×\left(-1\right)=-3{x}^{2}=\left(-3x\right)×x$

Hence, the roots of the given equation are $1$ and $-\frac{1}{3}$.

#### Question 16:

$4{x}^{2}-9x=100$

#### Question 17:

$15{x}^{2}-28=x$

#### Question 18:

$4-11x=3{x}^{2}\phantom{\rule{0ex}{0ex}}$

#### Question 19:

$48{x}^{2}-13x-1=0$

#### Question 20:

Solve each of the following quadratic equations:

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

#### Answer:

We write, $2\sqrt{2}x=3\sqrt{2}x-\sqrt{2}x$ as ${x}^{2}×\left(-6\right)=-6{x}^{2}=3\sqrt{2}x×\left(-\sqrt{2}x\right)$

Hence, the roots of the given equation are $-3\sqrt{2}$ and $\sqrt{2}$.

#### Question 21:

Solve each of the following quadratic equations:

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

#### Answer:

We write, $10x=3x+7x$ as $\sqrt{3}{x}^{2}×7\sqrt{3}=21{x}^{2}=3x×7x$

$\therefore \sqrt{3}{x}^{2}+10x+7\sqrt{3}=0\phantom{\rule{0ex}{0ex}}⇒\sqrt{3}{x}^{2}+3x+7x+7\sqrt{3}=0\phantom{\rule{0ex}{0ex}}⇒\sqrt{3}x\left(x+\sqrt{3}\right)+7\left(x+\sqrt{3}\right)=0\phantom{\rule{0ex}{0ex}}⇒\left(x+\sqrt{3}\right)\left(\sqrt{3}x+7\right)=0$

Hence, the roots of the given equation are $-\sqrt{3}$ and $-\frac{7\sqrt{3}}{3}$.

#### Question 22:

$\sqrt{3}{x}^{2}+11x+6\sqrt{3}=0$

#### Question 23:

$3\sqrt{7}{x}^{2}+4x-\sqrt{7}=0$

#### Question 24:

Solve each of the following quadratic equations:

$\sqrt{7}{x}^{2}-6x-13\sqrt{7}=0$

#### Answer:

We write, $-6x=7x-13x$ as $\sqrt{7}{x}^{2}×\left(-13\sqrt{7}\right)=-91{x}^{2}=7x×\left(-13x\right)$

$\therefore \sqrt{7}{x}^{2}-6x-13\sqrt{7}=0\phantom{\rule{0ex}{0ex}}⇒\sqrt{7}{x}^{2}+7x-13x-13\sqrt{7}=0\phantom{\rule{0ex}{0ex}}⇒\sqrt{7}x\left(x+\sqrt{7}\right)-13\left(x+\sqrt{7}\right)=0\phantom{\rule{0ex}{0ex}}⇒\left(x+\sqrt{7}\right)\left(\sqrt{7}x-13\right)=0$

Hence, the roots of the given equation are $-\sqrt{7}$ and $\frac{13\sqrt{7}}{7}$.

#### Question 25:

$4\sqrt{6}{x}^{2}-13x-2\sqrt{6}=0$

#### Question 26:

Solve each of the following quadratic equations:

$3{x}^{2}-2\sqrt{6}x+2=0$                             [CBSE 2010, 12]

#### Answer:

We write, $-2\sqrt{6}x=-\sqrt{6}x-\sqrt{6}x$ as $3{x}^{2}×2=6{x}^{2}=\left(-\sqrt{6}x\right)×\left(-\sqrt{6}x\right)$

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

Hence, $\frac{\sqrt{6}}{3}$ is the repreated root of the given equation.

#### Question 27:

Solve each of the following quadratic equations:

$\sqrt{3}{x}^{2}-2\sqrt{2}x-2\sqrt{3}=0$                  [CBSE 2011]

#### Answer:

We write, $-2\sqrt{2}x=-3\sqrt{2}x+\sqrt{2}x$ as $\sqrt{3}{x}^{2}×\left(-2\sqrt{3}\right)=-6{x}^{2}=\left(-3\sqrt{2}x\right)×\left(\sqrt{2}x\right)$

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

Hence, the roots of the given equation are $\sqrt{6}$ and $-\frac{\sqrt{6}}{3}$.

#### Question 28:

Solve each of the following quadratic equations:

${x}^{2}-3\sqrt{5}x+10=0$                  [CBSE 2011]

#### Answer:

We write, $-3\sqrt{5}x=-2\sqrt{5}x-\sqrt{5}x$ as ${x}^{2}×10=10{x}^{2}=\left(-2\sqrt{5}x\right)×\left(-\sqrt{5}x\right)$

$\therefore {x}^{2}-3\sqrt{5}x+10=0\phantom{\rule{0ex}{0ex}}⇒{x}^{2}-2\sqrt{5}x-\sqrt{5}x+10=0\phantom{\rule{0ex}{0ex}}⇒x\left(x-2\sqrt{5}\right)-\sqrt{5}\left(x-2\sqrt{5}\right)=0\phantom{\rule{0ex}{0ex}}⇒\left(x-2\sqrt{5}\right)\left(x-\sqrt{5}\right)=0$

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

#### Question 29:

Solve each of the following quadratic equations:

${x}^{2}-\left(\sqrt{3}+1\right)x+\sqrt{3}=0$        [CBSE 2015]

#### Answer:

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

Hence, 1 and $\sqrt{3}$ are the roots of the given equation.

#### Question 30:

Solve each of the following quadratic equations:

${x}^{2}+3\sqrt{3}x-30=0$        [CBSE 2015]

#### Answer:

We write, $3\sqrt{3}x=5\sqrt{3}x-2\sqrt{3}x$ as ${x}^{2}×\left(-30\right)=-30{x}^{2}=5\sqrt{3}x×\left(-2\sqrt{3}x\right)$

$\therefore {x}^{2}+3\sqrt{3}x-30=0\phantom{\rule{0ex}{0ex}}⇒{x}^{2}+5\sqrt{3}x-2\sqrt{3}x-30=0\phantom{\rule{0ex}{0ex}}⇒x\left(x+5\sqrt{3}\right)-2\sqrt{3}\left(x+5\sqrt{3}\right)=0\phantom{\rule{0ex}{0ex}}⇒\left(x+5\sqrt{3}\right)\left(x-2\sqrt{3}\right)=0$

Hence, the roots of the given equation are $-5\sqrt{3}$ and $2\sqrt{3}$.

#### Question 31:

Solve each of the following quadratic equations:

$\sqrt{2}{x}^{2}+7x+5\sqrt{2}=0$       [CBSE 2013]

#### Answer:

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

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

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

#### Question 32:

Solve each of the following quadratic equations:

$5{x}^{2}+13x+8=0$                  [CBSE 2013C]

#### Answer:

We write, 13x = 5x + 8x as $5{x}^{2}×8=40{x}^{2}=5x×8x$

Hence, $-1$ and $-\frac{8}{5}$ are the roots of the given equation.

#### Question 33:

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

#### Question 34:

$9{x}^{2}+6x+1=0$

#### Question 35:

Solve each of the following quadratic equations:

$100{x}^{2}-20x+1=0$

#### Answer:

We write, $-20x=-10x-10x$ as $100{x}^{2}×1=100{x}^{2}=\left(-10x\right)×\left(-10x\right)$

$⇒{\left(10x-1\right)}^{2}=0\phantom{\rule{0ex}{0ex}}⇒10x-1=0\phantom{\rule{0ex}{0ex}}⇒x=\frac{1}{10}$
Hence, $\frac{1}{10}$ is the repreated root of the given equation.

#### Question 36:

Solve each of the following quadratic equations:

$2{x}^{2}-x+\frac{1}{8}=0$

#### Answer:

We write, $-x=-\frac{x}{2}-\frac{x}{2}$ as $2{x}^{2}×\frac{1}{8}=\frac{{x}^{2}}{4}=\left(-\frac{x}{2}\right)×\left(-\frac{x}{2}\right)$

Hence, $\frac{1}{4}$ is the repeated root of the given equation.

#### Question 37:

$10x-\frac{1}{3}=3$

#### Question 38:

$\frac{2}{{x}^{2}}-\frac{5}{x}+2=0$

#### Question 39:

Solve each of the following quadratic equations:

$2{x}^{2}+ax-{a}^{2}=0$             [CBSE 2015]

#### Answer:

We write, $ax=2ax-ax$ as $2{x}^{2}×\left(-{a}^{2}\right)=-2{a}^{2}{x}^{2}=2ax×\left(-ax\right)$

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

#### Question 40:

Solve each of the following quadratic equations:

$4{x}^{2}+4bx-\left({a}^{2}-{b}^{2}\right)=0$                [CBSE 2015]

#### Answer:

We write, $4bx=2\left(a+b\right)x-2\left(a-b\right)x$ as $4{x}^{2}×\left[-\left({a}^{2}-{b}^{2}\right)\right]=-4\left({a}^{2}-{b}^{2}\right){x}^{2}=2\left(a+b\right)x×\left[-2\left(a-b\right)x\right]$

Hence, $-\frac{a+b}{2}$ and $\frac{a-b}{2}$ are the roots of the given equation.

#### Question 41:

Solve each of the following quadratic equations:

$4{x}^{2}-4{a}^{2}x+\left({a}^{4}-{b}^{4}\right)=0$             [CBSE 2015]

#### Answer:

We write, $-4{a}^{2}x=-2\left({a}^{2}+{b}^{2}\right)x-2\left({a}^{2}-{b}^{2}\right)x$ as $4{x}^{2}×\left({a}^{4}-{b}^{4}\right)=4\left({a}^{4}-{b}^{4}\right){x}^{2}=\left[-2\left({a}^{2}+{b}^{2}\right)\right]x×\left[-2\left({a}^{2}-{b}^{2}\right)\right]x$

Hence, $\frac{{a}^{2}+{b}^{2}}{2}$ and $\frac{{a}^{2}-{b}^{2}}{2}$ are the roots of the given equation.

#### Question 42:

Solve each of the following quadratic equations:

${x}^{2}+5x-\left({a}^{2}+a-6\right)=0$                           [CBSE 2015]

#### Answer:

We write, $5x=\left(a+3\right)x-\left(a-2\right)x$ as ${x}^{2}×\left[-\left({a}^{2}+a-6\right)\right]=-\left({a}^{2}+a-6\right){x}^{2}=\left(a+3\right)x×\left[-\left(a-2\right)x\right]$

Hence, $-\left(a+3\right)$ and $\left(a-2\right)$ are the roots of the given equation.

#### Question 43:

Solve each of the following quadratic equations:

${x}^{2}-2ax-\left(4{b}^{2}-{a}^{2}\right)=0$                         [CBSE 2015]

#### Answer:

We write, $-2ax=\left(2b-a\right)x-\left(2b+a\right)x$ as ${x}^{2}×\left[-\left(4{b}^{2}-{a}^{2}\right)\right]=-\left(4{b}^{2}-{a}^{2}\right){x}^{2}=\left(2b-a\right)x×\left[-\left(2b+a\right)x\right]$

Hence, $a-2b$ and $a+2b$ are the roots of the given equation.

#### Question 44:

Solve each of the following quadratic equations:

${x}^{2}-\left(2b-1\right)x+\left({b}^{2}-b-20\right)=0$                     [CBSE 2015]

#### Answer:

We write, $-\left(2b-1\right)x=-\left(b-5\right)x-\left(b+4\right)x$ as ${x}^{2}×\left({b}^{2}-b-20\right)=\left({b}^{2}-b-20\right){x}^{2}=\left[-\left(b-5\right)x\right]×\left[-\left(b+4\right)x\right]$

Hence, $b-5$ and $b+4$ are the roots of the given equation.

#### Question 45:

Solve each of the following quadratic equations:

${x}^{2}+6x-\left({a}^{2}+2a-8\right)=0$                          [CBSE 2015]

#### Answer:

We write, $6x=\left(a+4\right)x-\left(a-2\right)x$ as ${x}^{2}×\left[-\left({a}^{2}+2a-8\right)\right]=-\left({a}^{2}+2a-8\right){x}^{2}=\left(a+4\right)x×\left[-\left(a-2\right)x\right]$

Hence, $-\left(a+4\right)$ and $\left(a-2\right)$ are the roots of the given equation.

#### Question 46:

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

#### Question 47:

Solve each of the following quadratic equations:

${x}^{2}-4ax-{b}^{2}+4{a}^{2}=0$                           [CBSE 2012]

#### Answer:

We write, $-4ax=-\left(b+2a\right)x+\left(b-2a\right)x$ as ${x}^{2}×\left(-{b}^{2}+4{a}^{2}\right)=\left(-{b}^{2}+4{a}^{2}\right){x}^{2}=-\left(b+2a\right)x×\left(b-2a\right)x$

Hence, $\left(2a+b\right)$ and $\left(2a-b\right)$ are the roots of the given equation.

#### Question 48:

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

#### Question 49:

$12ab{x}^{2}-\left(9{a}^{2}-8{b}^{2}\right)x-6ab=0$

#### Question 50:

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

#### Question 51:

Solve each of the following quadratic equations:

$9{x}^{2}-9\left(a+b\right)x+\left(2{a}^{2}+5ab+2{b}^{2}\right)=0$                 [CBSE 2009]

#### Answer:

We write, $-9\left(a+b\right)x=-3\left(2a+b\right)x-3\left(a+2b\right)x$ as $9{x}^{2}×\left(2{a}^{2}+5ab+2{b}^{2}\right)=9\left(2{a}^{2}+5ab+2{b}^{2}\right){x}^{2}=\left[-3\left(2a+b\right)x\right]×\left[-3\left(a+2b\right)x\right]$
$\therefore 9{x}^{2}-9\left(a+b\right)x+\left(2{a}^{2}+5ab+2{b}^{2}\right)=0\phantom{\rule{0ex}{0ex}}⇒9{x}^{2}-3\left(2a+b\right)x-3\left(a+2b\right)x+\left(2a+b\right)\left(a+2b\right)=0\phantom{\rule{0ex}{0ex}}⇒3x\left[3x-\left(2a+b\right)\right]-\left(a+2b\right)\left[3x-\left(2a+b\right)\right]=0\phantom{\rule{0ex}{0ex}}⇒\left[3x-\left(2a+b\right)\right]\left[3x-\left(a+2b\right)\right]=0$

Hence, $\frac{2a+b}{3}$ and $\frac{a+2b}{3}$ are the roots of the given equation.

#### Question 52:

Solve each of the following quadratic equations:

[CBSE 2014]

#### Answer:

Hence, −4 and 4 are the roots of the given equation.

#### Question 53:

Solve each of the following quadratic equations:

[CBSE 2014]

#### Answer:

Hence, −2 and 1 are the roots of the given equation.

#### Question 54:

Solve each of the following quadratic equations:

[CBSE 2014]

#### Answer:

Hence, 1 and 3 are the roots of the given equation.

#### Question 55:

Solve each of the following quadratic equations:

[CBSE 2010]

#### Answer:

$⇒{x}^{2}+4x-12=0\phantom{\rule{0ex}{0ex}}⇒{x}^{2}+6x-2x-12=0\phantom{\rule{0ex}{0ex}}⇒x\left(x+6\right)-2\left(x+6\right)=0\phantom{\rule{0ex}{0ex}}⇒\left(x+6\right)\left(x-2\right)=0$

Hence, −6 and 2 are the roots of the given equation.

#### Question 56:

Solve each of the following quadratic equations:

$\frac{1}{2a+b+2x}=\frac{1}{2a}+\frac{1}{b}+\frac{1}{2x}$               [CBSE 2013]

#### Answer:

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

Hence, $-a$ and $-\frac{b}{2}$ are the roots of the given equation.

#### Question 58:

Solve each of the following quadratic equations:

[CBSE 2010]

#### Answer:

$⇒18{x}^{2}-48x+130=105x-140\phantom{\rule{0ex}{0ex}}⇒18{x}^{2}-153x+270=0\phantom{\rule{0ex}{0ex}}⇒2{x}^{2}-17x+30=0\phantom{\rule{0ex}{0ex}}⇒2{x}^{2}-12x-5x+30=0$

Hence, 6 and $\frac{5}{2}$ are the roots of the given equation.

#### Question 59:

Solve each of the following quadratic equations:

[CBSE 2011]

#### Answer:

$⇒8{x}^{2}-8x+4=17{x}^{2}-17x\phantom{\rule{0ex}{0ex}}⇒9{x}^{2}-9x-4=0\phantom{\rule{0ex}{0ex}}⇒9{x}^{2}-12x+3x-4=0\phantom{\rule{0ex}{0ex}}⇒3x\left(3x-4\right)+1\left(3x-4\right)=0$

Hence, $\frac{4}{3}$ and $-\frac{1}{3}$ are the roots of the given equation.

#### Question 60:

Solve each of the following quadratic equations:

[CBSE 2014]

#### Answer:

$⇒30{x}^{2}+30x+15=34{x}^{2}+34x\phantom{\rule{0ex}{0ex}}⇒4{x}^{2}+4x-15=0\phantom{\rule{0ex}{0ex}}⇒4{x}^{2}+10x-6x-15=0\phantom{\rule{0ex}{0ex}}⇒2x\left(2x+5\right)-3\left(2x+5\right)=0$

Hence, $-\frac{5}{2}$ and $\frac{3}{2}$ are the roots of the given equation.

#### Question 61:

Solve each of the following quadratic equations:

[CBSE 2014]

#### Answer:

$⇒\frac{{x}^{2}-11x+29}{{x}^{2}-12x+35}=\frac{5}{3}\phantom{\rule{0ex}{0ex}}⇒3{x}^{2}-33x+87=5{x}^{2}-60x+175\phantom{\rule{0ex}{0ex}}⇒2{x}^{2}-27x+88=0$
$⇒2{x}^{2}-16x-11x+88=0\phantom{\rule{0ex}{0ex}}⇒2x\left(x-8\right)-11\left(x-8\right)=0\phantom{\rule{0ex}{0ex}}⇒\left(x-8\right)\left(2x-11\right)=0$

Hence, 8 and $\frac{11}{2}$ are the roots of the given equation.

#### Question 62:

Solve each of the following quadratic equations:

[CBSE 2010]

#### Answer:

$⇒\frac{{x}^{2}-5x+5}{{x}^{2}-6x+8}=\frac{5}{3}\phantom{\rule{0ex}{0ex}}⇒3{x}^{2}-15x+15=5{x}^{2}-30x+40\phantom{\rule{0ex}{0ex}}⇒2{x}^{2}-15x+25=0\phantom{\rule{0ex}{0ex}}⇒2{x}^{2}-10x-5x+25=0$

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

#### Question 64:

Solve each of the following quadratic equations:

[CBSE 2013C]

#### Answer:

$⇒3{x}^{2}+16x+16=5{x}^{2}+15x+10\phantom{\rule{0ex}{0ex}}⇒2{x}^{2}-x-6=0\phantom{\rule{0ex}{0ex}}⇒2{x}^{2}-4x+3x-6=0\phantom{\rule{0ex}{0ex}}⇒2x\left(x-2\right)+3\left(x-2\right)=0$

Hence, 2 and $-\frac{3}{2}$ are the roots of the given equation.

#### Question 65:

Solve each of the following quadratic equations:

[CBSE 2014]

#### Answer:

$⇒\frac{19{x}^{2}-42x-15}{6{x}^{2}+7x-3}=5\phantom{\rule{0ex}{0ex}}⇒19{x}^{2}-42x-15=30{x}^{2}+35x-15\phantom{\rule{0ex}{0ex}}⇒11{x}^{2}+77x=0\phantom{\rule{0ex}{0ex}}⇒11x\left(x+7\right)=0$

Hence, 0 and −7 are the roots of the given equation.

#### Question 66:

Solve each of the following quadratic equations:

[CBSE 2014]

#### Answer:

$⇒\frac{47{x}^{2}+162x-33}{35{x}^{2}-16x-3}=11\phantom{\rule{0ex}{0ex}}⇒47{x}^{2}+162x-33=385{x}^{2}-176x-33\phantom{\rule{0ex}{0ex}}⇒338{x}^{2}-338x=0\phantom{\rule{0ex}{0ex}}⇒338x\left(x-1\right)=0\phantom{\rule{0ex}{0ex}}$

Hence, 0 and 1 are the roots of the given equation.

#### Question 71:

${3}^{\left(x+2\right)}+{3}^{-x}=10$

#### Question 72:

${4}^{\left(\mathrm{x}+1\right)}+{4}^{\left(1-x\right)}=10$

#### Question 73:

${2}^{2x}-3.{2}^{\left(x+2\right)}+32=0$

#### Question 1:

Solve each of the following equations by using the method of completing the square:

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

#### Answer:

Hence, $3+\sqrt{6}$ and $3-\sqrt{6}$ are the roots of the given equation.

#### Question 2:

Solve each of the following equations by using the method of completing the square:

${x}^{2}-4x+1=0$

#### Answer:

Hence, $2+\sqrt{3}$ and $2-\sqrt{3}$ are the roots of the given equation.

#### Question 3:

Solve each of the following equations by using the method of completing the square:

${x}^{2}+8x-2=0$

#### Answer:

Hence, $\left(-4+3\sqrt{2}\right)$ and $\left(-4-3\sqrt{2}\right)$ are the roots of the given equation.

#### Question 4:

Solve each of the following equations by using the method of completing the square:

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

#### Answer:

$⇒2x+\sqrt{3}=0\phantom{\rule{0ex}{0ex}}⇒x=-\frac{\sqrt{3}}{2}$
Hence, $-\frac{\sqrt{3}}{2}$ is the repeated root of the given equation.

#### Question 5:

Solve each of the following equations by using the method of completing the square:

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

#### Answer:

Hence, $\frac{1}{2}$ and $-3$ are the roots of the given equation.

#### Question 6:

Solve each of the following equations by using the method of completing the square:

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

#### Answer:

Hence, 1 and $-\frac{2}{3}$ are the roots of the given equation.

#### Question 7:

Solve each of the following equations by using the method of completing the square:

$8{x}^{2}-14x-15=0$

#### Answer:

Hence, $\frac{5}{2}$ and $-\frac{3}{4}$ are the roots of the given equation.

#### Question 8:

Solve each of the following equations by using the method of completing the square:

$7{x}^{2}+3x-4=0$

#### Answer:

Hence, $\frac{4}{7}$ and −1 are the roots of the given equation.

#### Question 9:

Solve each of the following equations by using the method of completing the square:

$3{x}^{2}-2x-1=0$

#### Answer:

Hence, 1 and $-\frac{1}{3}$ are the roots of the given equation.

#### Question 10:

Solve each of the following equations by using the method of completing the square:

$5{x}^{2}-6x-2=0$

#### Answer:

Hence, $\frac{3+\sqrt{19}}{5}$ and $\frac{3-\sqrt{19}}{5}$ are the roots of the given equation.

#### Question 11:

Solve each of the following equations by using the method of completing the square:

$\frac{2}{{x}^{2}}-\frac{5}{x}+2=0$

#### Answer:

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

#### Question 12:

Solve each of the following equations by using the method of completing the square:

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

#### Answer:

Hence, $\frac{a-b}{2}$ and $-\frac{a+b}{2}$ are the roots of the given equation.

#### Question 13:

Solve each of the following equations by using the method of completing the square:

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

#### Answer:

Hence, $\sqrt{2}$ and 1 are the roots of the given equation.

#### Question 14:

Solve each of the following equations by using the method of completing the square:

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

#### Answer:

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

#### Question 15:

Solve each of the following equations by using the method of completing the square:

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

#### Answer:

Hence, $-\sqrt{3}$ and $-\frac{7\sqrt{3}}{3}$ are the roots of the given equation.

#### Question 16:

By using the method of completing the square, show that the equation $2{x}^{2}+x+4=0$ has no real roots.

#### Answer:

$⇒{\left(2x+\frac{1}{2}\right)}^{2}=-8+\frac{1}{4}=-\frac{31}{4}<0$
But, ${\left(2x+\frac{1}{2}\right)}^{2}$ cannot be negative for any real value of x.
So, there is no real value of x satisfying the given equation.

Hence, the given equation has no real roots.

#### Question 1:

Find the discriminant of each of the following equations:

(i) $2{x}^{2}-7x+6=0$
(ii) $3{x}^{2}-2x+8=0$
(iii) $2{x}^{2}-5\sqrt{2}x+4=0$
(iv) $\sqrt{3}{x}^{2}+2\sqrt{2}x-2\sqrt{3}=0$
(v) $\left(x-1\right)\left(2x-1\right)=0$
(vi) $1-x=2{x}^{2}$2x27x+6=02x27x+6=0

#### Answer:

(v) $\left(x-1\right)\left(2x-1\right)=0$

$⇒2{x}^{2}-3x+1=0$

Comparing it with $a{x}^{2}+bx+c=0$, we get

a = 2, b = −3 and c = 1

∴ Discriminant, D = ${b}^{2}-4ac={\left(-3\right)}^{2}-4×2×1=9-8=1$

#### Question 2:

Find the roots of each of the following equations, if they exist, by applying the quadratic formula:

${x}^{2}-4x-1=0$

#### Question 3:

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

#### Question 4:

Find the roots of each of the following equations, if they exist, by applying the quadratic formula:

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

#### Answer:

The given equation is $2{x}^{2}+x-4=0$.

Comparing it with $a{x}^{2}+bx+c=0$, we get

a = 2, b = 1 and c = −4

∴ Discriminant, D = ${b}^{2}-4ac={\left(1\right)}^{2}-4×2×\left(-4\right)=1+32=33>0$

So, the given equation has real roots.

Now, $\sqrt{D}=\sqrt{33}$

Hence, $\frac{-1+\sqrt{33}}{4}$ and $\frac{-1-\sqrt{33}}{4}$ are the roots of the given equation.

#### Question 5:

$25{x}^{2}+30x+7=0$

#### Question 6:

$16{x}^{2}=24x+1$

#### Question 7:

$15{x}^{2}-28=x$

#### Question 8:

Find the roots of each of the following equations, if they exist, by applying the quadratic formula:

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

#### Answer:

The given equation is $2{x}^{2}-2\sqrt{2}x+1=0$.

Comparing it with $a{x}^{2}+bx+c=0$, we get

a = 2, b$-2\sqrt{2}$ and c = 1

∴ Discriminant, D = ${b}^{2}-4ac={\left(-2\sqrt{2}\right)}^{2}-4×2×1=8-8=0$

So, the given equation has real roots.

Now, $\sqrt{D}=0$

Hence, $\frac{\sqrt{2}}{2}$ is the repeated root of the given equation.

#### Question 9:

Find the roots of each of the following equations, if they exist, by applying the quadratic formula:

$\sqrt{2}{x}^{2}+7x+5\sqrt{2}=0$                      [CBSE 2013]

#### Answer:

The given equation is $\sqrt{2}{x}^{2}+7x+5\sqrt{2}=0$.

Comparing it with $a{x}^{2}+bx+c=0$, we get

a = $\sqrt{2}$, b = 7 and c = $5\sqrt{2}$

∴ Discriminant, D = ${b}^{2}-4ac={\left(7\right)}^{2}-4×\sqrt{2}×5\sqrt{2}=49-40=9>0$

So, the given equation has real roots.

Now, $\sqrt{D}=\sqrt{9}=3$

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

#### Question 10:

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

#### Question 11:

Find the roots of each of the following equations, if they exist, by applying the quadratic formula:

$\sqrt{3}{x}^{2}-2\sqrt{2}x-2\sqrt{3}=0$                        [CBSE 2015]

#### Answer:

The given equation is $\sqrt{3}{x}^{2}-2\sqrt{2}x-2\sqrt{3}=0$.

Comparing it with $a{x}^{2}+bx+c=0$, we get

a = $\sqrt{3}$, b$-2\sqrt{2}$ and c = $-2\sqrt{3}$

∴ Discriminant, D = ${b}^{2}-4ac={\left(-2\sqrt{2}\right)}^{2}-4×\sqrt{3}×\left(-2\sqrt{3}\right)=8+24=32>0$

So, the given equation has real roots.

Now, $\sqrt{D}=\sqrt{32}=4\sqrt{2}$

Hence, $\sqrt{6}$ and $-\frac{\sqrt{6}}{3}$ are the roots of the given equation.

#### Question 12:

Find the roots of each of the following equations, if they exist, by applying the quadratic formula:

$2{x}^{2}+6\sqrt{3}x-60=0$                                           [CBSE 2011, 15]

#### Answer:

The given equation is $2{x}^{2}+6\sqrt{3}x-60=0$.

Comparing it with $a{x}^{2}+bx+c=0$, we get

a = 2, b$6\sqrt{3}$ and c = $-60$

∴ Discriminant, D = ${b}^{2}-4ac={\left(6\sqrt{3}\right)}^{2}-4×2×\left(-60\right)=108+480=588>0$

So, the given equation has real roots.

Now, $\sqrt{D}=\sqrt{588}=14\sqrt{3}$

Hence, $2\sqrt{3}$ and $-5\sqrt{3}$ are the roots of the given equation.

#### Question 13:

Find the roots of each of the following equations, if they exist, by applying the quadratic formula:

$4\sqrt{3}{x}^{2}+5x-2\sqrt{3}=0$                                           [CBSE 2013]

#### Answer:

The given equation is $4\sqrt{3}{x}^{2}+5x-2\sqrt{3}=0$.

Comparing it with $a{x}^{2}+bx+c=0$, we get

a = $4\sqrt{3}$, b = 5 and c = $-2\sqrt{3}$

∴ Discriminant, D = ${b}^{2}-4ac={5}^{2}-4×4\sqrt{3}×\left(-2\sqrt{3}\right)=25+96=121>0$

So, the given equation has real roots.

Now, $\sqrt{D}=\sqrt{121}=11$

Hence, $\frac{\sqrt{3}}{4}$ and $-\frac{2\sqrt{3}}{3}$ are the roots of the given equation.

#### Question 14:

Find the roots of each of the following equations, if they exist, by applying the quadratic formula:

$3{x}^{2}-2\sqrt{6}x+2=0$                                           [CBSE 2012]

#### Answer:

The given equation is $3{x}^{2}-2\sqrt{6}x+2=0$.

Comparing it with $a{x}^{2}+bx+c=0$, we get

a = 3, b$-2\sqrt{6}$ and c = 2

∴ Discriminant, D = ${b}^{2}-4ac={\left(-2\sqrt{6}\right)}^{2}-4×3×2=24-24=0$

So, the given equation has real roots.

Now, $\sqrt{D}=0$

Hence, $\frac{\sqrt{6}}{3}$ is the repeated root of the given equation.

#### Question 15:

Find the roots of each of the following equations, if they exist, by applying the quadratic formula:

$2\sqrt{3}{x}^{2}-5x+\sqrt{3}=0$                                           [CBSE 2011]

#### Answer:

The given equation is $2\sqrt{3}{x}^{2}-5x+\sqrt{3}=0$.

Comparing it with $a{x}^{2}+bx+c=0$, we get

a = $2\sqrt{3}$, b$-5$ and c = $\sqrt{3}$

∴ Discriminant, D = ${b}^{2}-4ac={\left(-5\right)}^{2}-4×2\sqrt{3}×\sqrt{3}=25-24=1>0$

So, the given equation has real roots.

Now, $\sqrt{D}=\sqrt{1}=1$

Hence, $\frac{\sqrt{3}}{2}$ and $\frac{\sqrt{3}}{3}$ are the roots of the given equation.

#### Question 16:

Find the roots of each of the following equations, if they exist, by applying the quadratic formula:

${x}^{2}+x+2=0$

#### Answer:

The given equation is ${x}^{2}+x+2=0$.

Comparing it with $a{x}^{2}+bx+c=0$, we get

a = 1, b = 1 and c = 2

∴ Discriminant, D = ${b}^{2}-4ac={1}^{2}-4×1×2=1-8=-7<0$

Hence, the given equation has no real roots (or real roots does not exist).

#### Question 17:

Find the roots of each of the following equations, if they exist, by applying the quadratic formula:

$2{x}^{2}+ax-{a}^{2}=0$                                           [CBSE 2015]

#### Answer:

The given equation is $2{x}^{2}+ax-{a}^{2}=0$.

Comparing it with $A{x}^{2}+Bx+C=0$, we get

A = 2, B = a and C = $-{a}^{2}$

∴ Discriminant, D = ${B}^{2}-4AC={a}^{2}-4×2×-{a}^{2}={a}^{2}+8{a}^{2}=9{a}^{2}\ge 0$

So, the given equation has real roots.

Now, $\sqrt{D}=\sqrt{9{a}^{2}}=3a$

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

#### Question 18:

Find the roots of each of the following equations, if they exist, by applying the quadratic formula:

${x}^{2}-\left(\sqrt{3}+1\right)x+\sqrt{3}=0$                                           [CBSE 2015]

#### Answer:

The given equation is ${x}^{2}-\left(\sqrt{3}+1\right)x+\sqrt{3}=0$.

Comparing it with $a{x}^{2}+bx+c=0$, we get

a = 1, b$-\left(\sqrt{3}+1\right)$ and c = $\sqrt{3}$

∴ Discriminant, D = ${b}^{2}-4ac={\left[-\left(\sqrt{3}+1\right)\right]}^{2}-4×1×\sqrt{3}=3+1+2\sqrt{3}-4\sqrt{3}=3-2\sqrt{3}+1={\left(\sqrt{3}-1\right)}^{2}>0$

So, the given equation has real roots.

Now, $\sqrt{D}=\sqrt{{\left(\sqrt{3}-1\right)}^{2}}=\sqrt{3}-1$

Hence, $\sqrt{3}$ and 1 are the roots of the given equation.

#### Question 19:

Find the roots of each of the following equations, if they exist, by applying the quadratic formula:

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

#### Answer:

The given equation is $2{x}^{2}+5\sqrt{3}x+6=0$.

Comparing it with $a{x}^{2}+bx+c=0$, we get

a = 2, b$5\sqrt{3}$ and c = 6

∴ Discriminant, D = ${b}^{2}-4ac={\left(5\sqrt{3}\right)}^{2}-4×2×6=75-48=27>0$

So, the given equation has real roots.

Now, $\sqrt{D}=\sqrt{27}=3\sqrt{3}$

Hence, $-\frac{\sqrt{3}}{2}$ and $-2\sqrt{3}$ are the roots of the given equation.

#### Question 20:

Find the roots of each of the following equations, if they exist, by applying the quadratic formula:

$3{x}^{2}-2x+2=0$

#### Answer:

The given equation is $3{x}^{2}-2x+2=0$.

Comparing it with $a{x}^{2}+bx+c=0$, we get

a = 3, b = −2 and c = 2

∴ Discriminant, D = ${b}^{2}-4ac={\left(-2\right)}^{2}-4×3×2=4-24=-20<0$

Hence, the given equation has no real roots (or real roots does not exist).

#### Question 21:

Find the roots of each of the following equations, if they exist, by applying the quadratic formula:

#### Answer:

The given equation is

This equation is of the form $a{x}^{2}+bx+c=0$, where a = 1, b = −3 and c = 1.

∴ Discriminant, D = ${b}^{2}-4ac={\left(-3\right)}^{2}-4×1×1=9-4=5>0$

So, the given equation has real roots.

Now, $\sqrt{D}=\sqrt{5}$

Hence, $\frac{3+\sqrt{5}}{2}$ and $\frac{3-\sqrt{5}}{2}$ are the roots of the given equation.

#### Question 22:

Find the roots of each of the following equations, if they exist, by applying the quadratic formula:

[CBSE 2010]

#### Answer:

The given equation is

$⇒3{x}^{2}-6x+2=0$

This equation is of the form $a{x}^{2}+bx+c=0$, where a = 3, b = −6 and c = 2.

∴ Discriminant, D = ${b}^{2}-4ac={\left(-6\right)}^{2}-4×3×2=36-24=12>0$

So, the given equation has real roots.

Now, $\sqrt{D}=\sqrt{12}=2\sqrt{3}$

Hence, $\frac{3+\sqrt{3}}{3}$ and $\frac{3-\sqrt{3}}{3}$ are the roots of the given equation.

#### Question 23:

Find the roots of each of the following equations, if they exist, by applying the quadratic formula:

[CBSE 2010]

#### Answer:

The given equation is

This equation is of the form $a{x}^{2}+bx+c=0$, where a = 1, b = −3 and c = −1.

∴ Discriminant, D = ${b}^{2}-4ac={\left(-3\right)}^{2}-4×1×\left(-1\right)=9+4=13>0$

So, the given equation has real roots.

Now, $\sqrt{D}=\sqrt{13}$

Hence, $\frac{3+\sqrt{13}}{2}$ and $\frac{3-\sqrt{13}}{2}$ are the roots of the given equation.

#### Question 24:

Find the roots of each of the following equations, if they exist, by applying the quadratic formula:

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

#### Answer:

The given equation is

$\frac{m}{n}{x}^{2}+\frac{n}{m}=1-2x\phantom{\rule{0ex}{0ex}}⇒\frac{{m}^{2}{x}^{2}+{n}^{2}}{mn}=1-2x\phantom{\rule{0ex}{0ex}}⇒{m}^{2}{x}^{2}+{n}^{2}=mn-2mnx\phantom{\rule{0ex}{0ex}}⇒{m}^{2}{x}^{2}+2mnx+{n}^{2}-mn=0$

This equation is of the form $a{x}^{2}+bx+c=0$, where a = ${m}^{2}$, b = 2mn and c = ${n}^{2}-mn$.

∴ Discriminant, D = ${b}^{2}-4ac={\left(2mn\right)}^{2}-4×{m}^{2}×\left({n}^{2}-mn\right)=4{m}^{2}{n}^{2}-4{m}^{2}{n}^{2}+4{m}^{3}n=4{m}^{3}n>0$

So, the given equation has real roots.

Now, $\sqrt{D}=\sqrt{4{m}^{3}n}=2m\sqrt{mn}$

Hence, $\frac{-n+\sqrt{mn}}{m}$ and $\frac{-n-\sqrt{mn}}{m}$ are the roots of the given equation.

#### Question 25:

Find the roots of each of the following equations, if they exist, by applying the quadratic formula:

$36{x}^{2}-12ax+\left({a}^{2}-{b}^{2}\right)=0$

#### Answer:

The given equation is $36{x}^{2}-12ax+\left({a}^{2}-{b}^{2}\right)=0$.

Comparing it with $A{x}^{2}+Bx+C=0$, we get

A = 36, B$-12a$ and C = ${a}^{2}-{b}^{2}$

∴ Discriminant, D = ${B}^{2}-4AC={\left(-12a\right)}^{2}-4×36×\left({a}^{2}-{b}^{2}\right)=144{a}^{2}-144{a}^{2}+144{b}^{2}=144{b}^{2}>0$

So, the given equation has real roots.

Now, $\sqrt{D}=\sqrt{144{b}^{2}}=12b$

Hence, $\frac{a+b}{6}$ and $\frac{a-b}{6}$ are the roots of the given equation.

#### Question 26:

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

#### Question 27:

Find the roots of each of the following equations, if they exist, by applying the quadratic formula:

${x}^{2}-2ax-\left(4{b}^{2}-{a}^{2}\right)=0$               [CBSE 2015]

#### Answer:

The given equation is ${x}^{2}-2ax-\left(4{b}^{2}-{a}^{2}\right)=0$.

Comparing it with $A{x}^{2}+Bx+C=0$, we get

A = 1, B$-2a$ and C = $-\left(4{b}^{2}-{a}^{2}\right)$

∴ Discriminant, D = ${B}^{2}-4AC={\left(-2a\right)}^{2}-4×1×\left[-\left(4{b}^{2}-{a}^{2}\right)\right]=4{a}^{2}+16{b}^{2}-4{a}^{2}=16{b}^{2}>0$

So, the given equation has real roots.

Now, $\sqrt{D}=\sqrt{16{b}^{2}}=4b$

Hence, $a+2b$ and $a-2b$ are the roots of the given equation.

#### Question 28:

Find the roots of each of the following equations, if they exist, by applying the quadratic formula:

${x}^{2}+6x-\left({a}^{2}+2a-8\right)=0$                  [CBSE 2015]

#### Answer:

The given equation is ${x}^{2}+6x-\left({a}^{2}+2a-8\right)=0$.

Comparing it with $A{x}^{2}+Bx+C=0$, we get

A = 1, B = 6 and C = $-\left({a}^{2}+2a-8\right)$

∴ Discriminant, D = ${B}^{2}-4AC={6}^{2}-4×1×\left[-\left({a}^{2}+2a-8\right)\right]=36+4{a}^{2}+8a-32=4{a}^{2}+8a+4=4\left({a}^{2}+2a+1\right)=4{\left(a+1\right)}^{2}>0$

So, the given equation has real roots.

Now, $\sqrt{D}=\sqrt{4{\left(a+1\right)}^{2}}=2\left(a+1\right)$

Hence, $\left(a-2\right)$ and $-\left(a+4\right)$ are the roots of the given equation.

#### Question 29:

Find the roots of each of the following equations, if they exist, by applying the quadratic formula:

${x}^{2}+5x-\left({a}^{2}+a-6\right)=0$                  [CBSE 2015]

#### Answer:

The given equation is ${x}^{2}+5x-\left({a}^{2}+a-6\right)=0$.

Comparing it with $A{x}^{2}+Bx+C=0$, we get

A = 1, B = 5 and C = $-\left({a}^{2}+a-6\right)$

∴ Discriminant, D = ${B}^{2}-4AC={5}^{2}-4×1×\left[-\left({a}^{2}+a-6\right)\right]=25+4{a}^{2}+4a-24=4{a}^{2}+4a+1={\left(2a+1\right)}^{2}>0$

So, the given equation has real roots.

Now, $\sqrt{D}=\sqrt{{\left(2a+1\right)}^{2}}=2a+1$

Hence, $\left(a-2\right)$ and $-\left(a+3\right)$ are the roots of the given equation.

#### Question 30:

Find the roots of each of the following equations, if they exist, by applying the quadratic formula:

${x}^{2}-4ax-{b}^{2}+4{a}^{2}=0$                  [CBSE 2012]

#### Answer:

The given equation is ${x}^{2}-4ax-{b}^{2}+4{a}^{2}=0$.

Comparing it with $A{x}^{2}+Bx+C=0$, we get

A = 1, B = −4a and C = $-{b}^{2}+4{a}^{2}$

∴ Discriminant, D = ${B}^{2}-4AC={\left(-4a\right)}^{2}-4×1×\left(-{b}^{2}+4{a}^{2}\right)=16{a}^{2}+4{b}^{2}-16{a}^{2}=4{b}^{2}>0$

So, the given equation has real roots.

Now, $\sqrt{D}=\sqrt{4{b}^{2}}=2b$

Hence, $\left(2a+b\right)$ and $\left(2a-b\right)$ are the roots of the given equation.

#### Question 31:

Find the roots of each of the following equations, if they exist, by applying the quadratic formula:

$4{x}^{2}-4{a}^{2}x+\left({a}^{4}-{b}^{4}\right)=0$           [CBSE 2015]

#### Answer:

The given equation is $4{x}^{2}-4{a}^{2}x+\left({a}^{4}-{b}^{4}\right)=0$.

Comparing it with $A{x}^{2}+Bx+C=0$, we get

A = 4, B = −4a2 and C = ${a}^{4}-{b}^{4}$

∴ Discriminant, D = ${B}^{2}-4AC={\left(-4{a}^{2}\right)}^{2}-4×4×\left({a}^{4}-{b}^{4}\right)=16{a}^{4}-16{a}^{4}+16{b}^{4}=16{b}^{4}>0$

So, the given equation has real roots.

Now, $\sqrt{D}=\sqrt{16{b}^{4}}=4{b}^{2}$

Hence, $\frac{1}{2}\left({a}^{2}+{b}^{2}\right)$ and $\frac{1}{2}\left({a}^{2}-{b}^{2}\right)$ are the roots of the given equation.

#### Question 32:

Find the roots of each of the following equations, if they exist, by applying the quadratic formula:

$4{x}^{2}+4bx-\left({a}^{2}-{b}^{2}\right)=0$              [CBSE 2015]

#### Answer:

The given equation is $4{x}^{2}+4bx-\left({a}^{2}-{b}^{2}\right)=0$.

Comparing it with $A{x}^{2}+Bx+C=0$, we get

A = 4, B = 4b and C = $-\left({a}^{2}-{b}^{2}\right)$

∴ Discriminant, D = ${B}^{2}-4AC={\left(4b\right)}^{2}-4×4×\left[-\left({a}^{2}-{b}^{2}\right)\right]=16{b}^{2}+16{a}^{2}-16{b}^{2}=16{a}^{2}>0$

So, the given equation has real roots.

Now, $\sqrt{D}=\sqrt{16{a}^{2}}=4a$

Hence, $\frac{1}{2}\left(a-b\right)$ and $-\frac{1}{2}\left(a+b\right)$ are the roots of the given equation.

#### Question 33:

Find the roots of each of the following equations, if they exist, by applying the quadratic formula:

${x}^{2}-\left(2b-1\right)x+\left({b}^{2}-b-20\right)=0$               [CBSE 2015]

#### Answer:

The given equation is ${x}^{2}-\left(2b-1\right)x+\left({b}^{2}-b-20\right)=0$.

Comparing it with $A{x}^{2}+Bx+C=0$, we get

A = 1, B = $-\left(2b-1\right)$ and C = ${b}^{2}-b-20$

∴ Discriminant, D = ${B}^{2}-4AC={\left[-\left(2b-1\right)\right]}^{2}-4×1×\left({b}^{2}-b-20\right)=4{b}^{2}-4b+1-4{b}^{2}+4b+80=81>0$

So, the given equation has real roots.

Now, $\sqrt{D}=\sqrt{81}=9$

Hence, $\left(b+4\right)$ and $\left(b-5\right)$ are the roots of the given equation.

#### Question 35:

Find the roots of each of the following equations, if they exist, by applying the quadratic formula:

${a}^{2}{b}^{2}{x}^{2}-\left(4{b}^{4}-3{a}^{4}\right)x-12{a}^{2}{b}^{2}=0$, $a\ne 0$ and $b\ne 0$                  [CBSE 2006]

#### Answer:

The given equation is ${a}^{2}{b}^{2}{x}^{2}-\left(4{b}^{4}-3{a}^{4}\right)x-12{a}^{2}{b}^{2}=0$.

Comparing it with $A{x}^{2}+Bx+C=0$, we get

A = ${a}^{2}{b}^{2}$, B = $-\left(4{b}^{4}-3{a}^{4}\right)$ and C = $-12{a}^{2}{b}^{2}$

∴ Discriminant, D = ${B}^{2}-4AC={\left[-\left(4{b}^{4}-3{a}^{4}\right)\right]}^{2}-4×{a}^{2}{b}^{2}×\left(-12{a}^{2}{b}^{2}\right)=16{b}^{8}-24{a}^{4}{b}^{4}+9{a}^{8}+48{a}^{4}{b}^{4}=16{b}^{8}+24{a}^{4}{b}^{4}+9{a}^{8}={\left(4{b}^{4}+3{a}^{4}\right)}^{2}>0$

So, the given equation has real roots.

Now, $\sqrt{D}=\sqrt{{\left(4{b}^{4}+3{a}^{4}\right)}^{2}}=4{b}^{4}+3{a}^{4}$

Hence, $\frac{4{b}^{2}}{{a}^{2}}$ and $-\frac{3{a}^{2}}{{b}^{2}}$ are the roots of the given equation.

#### Question 36:

$12ab{x}^{2}-\left(9{a}^{2}-8{b}^{2}\right)x-6ab=0$, where a$\ne$0 and b$\ne$0

#### Question 1:

Find the nature of the roots of the following quadratic equations:
(i) $2{x}^{2}-8x+5=0$
(ii) $3{x}^{2}-2\sqrt{6}x+2=0$
(iii) $5{x}^{2}-4x+1=0$
(iv) $5x\left(x-2\right)+6=0$
(v) $12{x}^{2}-4\sqrt{15}x+5=0$
(vi) ${x}^{2}-x+2=0$

#### Answer:

(i) The given equation is $2{x}^{2}-8x+5=0$.

This is of the form $a{x}^{2}+bx+c=0$, where a = 2, b = −8 and c = 5.

∴ Discriminant, D = ${b}^{2}-4ac={\left(-8\right)}^{2}-4×2×5=64-40=24>0$

Hence, the given equation has real and unequal roots.

(ii) The given equation is $3{x}^{2}-2\sqrt{6}x+2=0$.

This is of the form $a{x}^{2}+bx+c=0$, where a = 3, b$-2\sqrt{6}$ and c = 2.

∴ Discriminant, D = ${b}^{2}-4ac={\left(-2\sqrt{6}\right)}^{2}-4×3×2=24-24=0$

Hence, the given equation has real and equal roots.

(iii) The given equation is $5{x}^{2}-4x+1=0$.

This is of the form $a{x}^{2}+bx+c=0$, where a = 5, b = −4 and c = 1.

∴ Discriminant, D = ${b}^{2}-4ac={\left(-4\right)}^{2}-4×5×1=16-20=-4<0$

Hence, the given equation has no real roots.

(iv) The given equation is

$5x\left(x-2\right)+6=0\phantom{\rule{0ex}{0ex}}⇒5{x}^{2}-10x+6=0$

This is of the form $a{x}^{2}+bx+c=0$, where a = 5, b = −10 and c = 6.

∴ Discriminant, D = ${b}^{2}-4ac={\left(-10\right)}^{2}-4×5×6=100-120=-20<0$

Hence, the given equation has no real roots.

(v) The given equation is $12{x}^{2}-4\sqrt{15}x+5=0$.

This is of the form $a{x}^{2}+bx+c=0$, where a = 12, b$-4\sqrt{15}$ and c = 5.

∴ Discriminant, D = ${b}^{2}-4ac={\left(-4\sqrt{15}\right)}^{2}-4×12×5=240-240=0$

Hence, the given equation has real and equal roots.

(vi) The given equation is ${x}^{2}-x+2=0$.

This is of the form $a{x}^{2}+bx+c=0$, where a = 1, b = −1 and c = 2.

∴ Discriminant, D = ${b}^{2}-4ac={\left(-1\right)}^{2}-4×1×2=1-8=-7<0$

Hence, the given equation has no real roots.

#### Question 2:

If a and b are distinct real numbers, show that the quadratic equation $2\left({a}^{2}+{b}^{2}\right){x}^{2}+2\left(a+b\right)x+1=0$ has no real roots.

#### Answer:

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

Hence, the given equation has no real roots.

#### Question 3:

Show that the roots of the equation ${x}^{2}+px-{q}^{2}=0$ are real for all real value of p and q.

#### Question 4:

For what values of k are the roots of the quadratic equation $3{x}^{2}+2kx+27=0$ real and equal?

#### Question 5:

For what value of k are the roots of the quadratic equation $kx\left(x-2\sqrt{5}\right)+10=0$ real and equal?                                         [CBSE 2013]

#### Answer:

The given equation is

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

This is of the form $a{x}^{2}+bx+c=0$, where a = k, b = $-2\sqrt{5}k$ and c = 10.

$\therefore D={b}^{2}-4ac={\left(-2\sqrt{5}k\right)}^{2}-4×k×10=20{k}^{2}-40k$

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

But, for k = 0, we get 10 = 0, which is not true.

Hence, 2 is the required value of k.

#### Question 6:

For what values of p are the roots of the equation $4{x}^{2}+px+3=0$ real and equal?                                                             [CBSE 2014]

#### Answer:

The given equation is $4{x}^{2}+px+3=0$.

This is of the form $a{x}^{2}+bx+c=0$, where a = 4, b = p and c = 3.

$\therefore D={b}^{2}-4ac={p}^{2}-4×4×3={p}^{2}-48$

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

$\therefore {p}^{2}-48=0\phantom{\rule{0ex}{0ex}}⇒{p}^{2}=48\phantom{\rule{0ex}{0ex}}⇒p=±\sqrt{48}=±4\sqrt{3}$

Hence, $4\sqrt{3}$ and $-4\sqrt{3}$ are the required values of p.

#### Question 7:

Find the non-zero value of k for which the roots of the quadratic equation $9{x}^{2}-3kx+k=0$ are real and equal.                             [CBSE 2014]

#### Answer:

The given equation is $9{x}^{2}-3kx+k=0$.

This is of the form $a{x}^{2}+bx+c=0$, where a = 9, b = −3k and c = k.

$\therefore D={b}^{2}-4ac={\left(-3k\right)}^{2}-4×9×k=9{k}^{2}-36k$

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

But, k ≠ 0        (Given)

Hence, the required value of k is 4.

#### Question 8:

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 real and equal roots.                                  [CBSE 2014]

#### Answer:

The given equation is $\left(3k+1\right){x}^{2}+2\left(k+1\right)x+1=0$.

This is of the form $a{x}^{2}+bx+c=0$, where a = 3k +1, b = 2(k + 1) and c = 1.

$=4{k}^{2}-4k$

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

Hence, 0 and 1 are the required values of k.

#### Question 9:

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 real and equal roots.                           [CBSE 2014]

#### Answer:

The given equation is $\left(2p+1\right){x}^{2}-\left(7p+2\right)x+\left(7p-3\right)=0$.

This is of the form $a{x}^{2}+bx+c=0$, where a = 2p +1, b = (7p + 2) and c = 7p − 3.

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

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

$\therefore -7{p}^{2}+24p+16=0\phantom{\rule{0ex}{0ex}}⇒7{p}^{2}-24p-16=0\phantom{\rule{0ex}{0ex}}⇒7{p}^{2}-28p+4p-16=0\phantom{\rule{0ex}{0ex}}⇒7p\left(p-4\right)+4\left(p-4\right)=0$

Hence, 4 and $-\frac{4}{7}$ are the required values of p.

#### Question 10:

Find the values of p for which the quadratic equation has equal roots. Hence, find the roots of the equation.                                                                                                                                                                       [CBSE 2015]

#### Answer:

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

This is of the form $a{x}^{2}+bx+c=0$, where a = p +1, b = 6(p + 1) and c = 3(p + 9).

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

But, $p\ne -1$           (Given)

Thus, the value of p is 3.

Putting p = 3, the given equation becomes $4{x}^{2}-24x+36=0$.

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

Hence, 3 is the repeated root of this equation.

#### Question 11:

If −5 is a root of the quadratic equation $2{x}^{2}+px-15=0$ and the quadratic equation $p\left({x}^{2}+x\right)+k=0$ has equal roots, find the value of k.
[CBSE 2014]

#### Answer:

It is given that −5 is a root of the quadratic equation $2{x}^{2}+px-15=0$.

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

The roots of the equation $p{x}^{2}+px+k=0$ = 0 are equal.

$\therefore D=0\phantom{\rule{0ex}{0ex}}⇒{p}^{2}-4pk=0\phantom{\rule{0ex}{0ex}}⇒{\left(7\right)}^{2}-4×7×k=0\phantom{\rule{0ex}{0ex}}⇒49-28k=0$
$⇒k=\frac{49}{28}=\frac{7}{4}$

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

#### Question 12:

If 3 is a root of the quadratic equation ${x}^{2}-x+k=0$, find the value of p so that the roots of the equation ${x}^{2}+k\left(2x+k+2\right)+p=0$ are equal.
[CBSE 2015]

#### Answer:

It is given that 3 is a root of the quadratic equation ${x}^{2}-x+k=0$.

$\therefore {\left(3\right)}^{2}-3+k=0\phantom{\rule{0ex}{0ex}}⇒k+6=0\phantom{\rule{0ex}{0ex}}⇒k=-6$

The roots of the equation ${x}^{2}+2kx+\left({k}^{2}+2k+p\right)=0$ are equal.

$\therefore D=0\phantom{\rule{0ex}{0ex}}⇒{\left(2k\right)}^{2}-4×1×\left({k}^{2}+2k+p\right)=0\phantom{\rule{0ex}{0ex}}⇒4{k}^{2}-4{k}^{2}-8k-4p=0\phantom{\rule{0ex}{0ex}}⇒-8k-4p=0$
$⇒p=\frac{8k}{-4}=-2k\phantom{\rule{0ex}{0ex}}⇒p=-2×\left(-6\right)=12$

Hence, the value of p is 12.

#### Question 13:

If −4 is a root of the quadratic equation ${x}^{2}+2x+4p=0$, find the value of k for which the quadratic equation ${x}^{2}+px\left(1+3k\right)+7\left(3+2k\right)=0$ has equal roots.                                                                                                                                                                             [CBSE 2015]

#### Answer:

It is given that −4 is a root of the quadratic equation ${x}^{2}+2x+4p=0$.

$\therefore {\left(-4\right)}^{2}+2×\left(-4\right)+4p=0\phantom{\rule{0ex}{0ex}}⇒16-8+4p=0\phantom{\rule{0ex}{0ex}}⇒4p+8=0\phantom{\rule{0ex}{0ex}}⇒p=-2$

The equation ${x}^{2}+px\left(1+3k\right)+7\left(3+2k\right)=0$ has equal roots.

$⇒4\left(1+6k+9{k}^{2}-21-14k\right)=0\phantom{\rule{0ex}{0ex}}⇒9{k}^{2}-8k-20=0\phantom{\rule{0ex}{0ex}}⇒9{k}^{2}-18k+10k-20=0\phantom{\rule{0ex}{0ex}}⇒9k\left(k-2\right)+10\left(k-2\right)=0$

Hence, the required value of k is 2 or $-\frac{10}{9}$.

#### Question 14:

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

#### Question 15:

If the roots of the equation $\left({c}^{2}-ab\right){x}^{2}-2\left({a}^{2}-bc\right)x+\left({b}^{2}-ac\right)=0$ are real and equal, show that either

#### Question 16:

Determine the values of p for which the quadratic equation $2{x}^{2}+px+8=0$ has real roots.

#### Question 17:

Find the value of α for which the equation $\left(\alpha -12\right){x}^{2}+2\left(\alpha -12\right)x+2=0$ has equal roots.

#### Question 18:

Find the values k for which of roots of $9{x}^{2}+8kx+16=0$ are real and equal

#### Question 19:

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

(i) $k{x}^{2}+6x+1=0$
(ii) ${x}^{2}-kx+9=0$
(iii) $9{x}^{2}+3kx+4=0$
(iv) $5{x}^{2}-kx+1=0$

#### Answer:

(i) The given equation is $k{x}^{2}+6x+1=0$.

$\therefore D={6}^{2}-4×k×1=36-4k$

The given equation has real and distinct roots if D > 0.

$\therefore 36-4k>0\phantom{\rule{0ex}{0ex}}⇒4k<36\phantom{\rule{0ex}{0ex}}⇒k<9$

(ii) The given equation is ${x}^{2}-kx+9=0$.

$\therefore D={\left(-k\right)}^{2}-4×1×9={k}^{2}-36$

The given equation has real and distinct roots if D > 0.

(iii) The given equation is $9{x}^{2}+3kx+4=0$.

$\therefore D={\left(3k\right)}^{2}-4×9×4=9{k}^{2}-144$

The given equation has real and distinct roots if D > 0.

(iv) The given equation is $5{x}^{2}-kx+1=0$.

$\therefore D={\left(-k\right)}^{2}-4×5×1={k}^{2}-20$

The given equation has real and distinct roots if D > 0.

#### Question 20:

If a and b are real and ab then show that the roots of the equation $\left(a-b\right){x}^{2}+5\left(a+b\right)x-2\left(a-b\right)=0$ are real and unequal.

#### Answer:

The given equation is $\left(a-b\right){x}^{2}+5\left(a+b\right)x-2\left(a-b\right)=0$.

Since a and b are real and ab, so ${\left(a-b\right)}^{2}>0$ and ${\left(a+b\right)}^{2}>0$.

$8{\left(a-b\right)}^{2}>0$    .....(1)               (Product of two positive numbers is always positive)

Also, $25{\left(a+b\right)}^{2}>0$             .....(2)              (Product of two positive numbers is always positive)

Adding (1) and (2), we get

$25{\left(a+b\right)}^{2}+8{\left(a-b\right)}^{2}>0$                 (Sum of two positive numbers is always positive)

$⇒D>0$

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

#### Question 21:

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

#### Answer:

It is given that the roots of the equation $\left({a}^{2}+{b}^{2}\right){x}^{2}-2\left(ac+bd\right)x+\left({c}^{2}+{d}^{2}\right)=0$ are equal.
$\therefore D=0\phantom{\rule{0ex}{0ex}}⇒{\left[-2\left(ac+bd\right)\right]}^{2}-4\left({a}^{2}+{b}^{2}\right)\left({c}^{2}+{d}^{2}\right)=0\phantom{\rule{0ex}{0ex}}⇒4\left({a}^{2}{c}^{2}+{b}^{2}{d}^{2}+2abcd\right)-4\left({a}^{2}{c}^{2}+{a}^{2}{d}^{2}+{b}^{2}{c}^{2}+{b}^{2}{d}^{2}\right)=0\phantom{\rule{0ex}{0ex}}⇒4\left({a}^{2}{c}^{2}+{b}^{2}{d}^{2}+2abcd-{a}^{2}{c}^{2}-{a}^{2}{d}^{2}-{b}^{2}{c}^{2}-{b}^{2}{d}^{2}\right)=0$
$⇒\left(-{a}^{2}{d}^{2}+2abcd-{b}^{2}{c}^{2}\right)=0\phantom{\rule{0ex}{0ex}}⇒-\left({a}^{2}{d}^{2}-2abcd+{b}^{2}{c}^{2}\right)=0\phantom{\rule{0ex}{0ex}}⇒{\left(ad-bc\right)}^{2}=0\phantom{\rule{0ex}{0ex}}$
$⇒ad-bc=0\phantom{\rule{0ex}{0ex}}⇒ad=bc\phantom{\rule{0ex}{0ex}}⇒\frac{a}{b}=\frac{c}{d}$
Hence Proved.

#### Question 22:

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

#### Answer:

It is given that the roots of the equation $a{x}^{2}+2bx+c=0$ are real.

Also, the roots of the equation $b{x}^{2}-2\sqrt{ac}x+b=0$ are real.
$\therefore {D}_{2}={\left(-2\sqrt{ac}\right)}^{2}-4×b×b\ge 0\phantom{\rule{0ex}{0ex}}⇒4\left(ac-{b}^{2}\right)\ge 0\phantom{\rule{0ex}{0ex}}⇒-4\left({b}^{2}-ac\right)\ge 0$

The roots of the given equations are simultaneously real if (1) and (2) holds true together. This is possible if
${b}^{2}-ac=0\phantom{\rule{0ex}{0ex}}⇒{b}^{2}=ac$

#### Question 1:

The sum of a natural number and its square is 156. Find the number.

#### Answer:

Let the required natural number be x.

According to the given condition,

$x+{x}^{2}=156\phantom{\rule{0ex}{0ex}}⇒{x}^{2}+x-156=0\phantom{\rule{0ex}{0ex}}⇒{x}^{2}+13x-12x-156=0\phantom{\rule{0ex}{0ex}}⇒x\left(x+13\right)-12\left(x+13\right)=0$

x = 12     (x cannot be negative)

Hence, the required natural number is 12.

#### Question 2:

The sum of a natural number and its positive square root is 132. Find the number.

#### Answer:

Let the required natural number be x.

According to the given condition,

$x+\sqrt{x}=132$

Putting $\sqrt{x}=y$ or x = y2, we get

${y}^{2}+y=132\phantom{\rule{0ex}{0ex}}⇒{y}^{2}+y-132=0\phantom{\rule{0ex}{0ex}}⇒{y}^{2}+12y-11y-132=0\phantom{\rule{0ex}{0ex}}⇒y\left(y+12\right)-11\left(y+12\right)=0$

y = 11         (y cannot be negative)

Now,

$\sqrt{x}=11\phantom{\rule{0ex}{0ex}}⇒x={\left(11\right)}^{2}=121$

Hence, the required natural number is 121.

#### Question 3:

The sum of two natural numbers is 28 and their product is 192. Find the numbers.

#### Answer:

Let the required numbers be x and (28 − x).

According to the given condition,

$x\left(28-x\right)=192\phantom{\rule{0ex}{0ex}}⇒28x-{x}^{2}=192\phantom{\rule{0ex}{0ex}}⇒{x}^{2}-28x+192=0\phantom{\rule{0ex}{0ex}}⇒{x}^{2}-16x-12x+192=0$

When x = 12,

28 − x = 28 − 12 = 16

When x = 16,

28 − x = 28 − 16 = 12

Hence, the required numbers are 12 and 16.

#### Question 4:

The sum of the squares of two consecutive positive integers is 365. Find the integers.

#### Answer:

Let the required two consecutive positive integers be x and (x + 1).

According to the given condition,
${x}^{2}+{\left(x+1\right)}^{2}=365\phantom{\rule{0ex}{0ex}}⇒{x}^{2}+{x}^{2}+2x+1=365\phantom{\rule{0ex}{0ex}}⇒2{x}^{2}+2x-364=0\phantom{\rule{0ex}{0ex}}⇒{x}^{2}+x-182=0$

x = 13                (x is a positive integer)

When x = 13,
x + 1 = 13 + 1 = 14

Hence, the required positive integers are 13 and 14.

#### Question 5:

The sum of the squares of two consecutive positive odd numbers is 514. Find the numbers.

#### Answer:

Let the two consecutive positive odd numbers be x and (x + 2).

According to the given condition,
${x}^{2}+{\left(x+2\right)}^{2}=514\phantom{\rule{0ex}{0ex}}⇒{x}^{2}+{x}^{2}+4x+4=514\phantom{\rule{0ex}{0ex}}⇒2{x}^{2}+4x-510=0\phantom{\rule{0ex}{0ex}}⇒{x}^{2}+2x-255=0$

x = 15             (x is a positive odd number)

When x = 15,
x + 2 = 15 + 2 = 17

Hence, the required numbers are 15 and 17.

#### Question 6:

The sum of the squares of two consecutive positive even numbers is 452. Find the numbers.

#### Answer:

Let the two consecutive positive even numbers be x and (x + 2).

According to the given condition,
${x}^{2}+{\left(x+2\right)}^{2}=452\phantom{\rule{0ex}{0ex}}⇒{x}^{2}+{x}^{2}+4x+4=452\phantom{\rule{0ex}{0ex}}⇒2{x}^{2}+4x-448=0\phantom{\rule{0ex}{0ex}}⇒{x}^{2}+2x-224=0$

x = 14             (x is a positive even number)

When x = 14,
x + 2 = 14 + 2 = 16

Hence, the required numbers are 14 and 16.

#### Question 7:

The product of two consecutive positive integers is 306. Find the integers.

#### Answer:

Let the two consecutive positive integers be x and (x + 1).

According to the given condition,
$x\left(x+1\right)=306\phantom{\rule{0ex}{0ex}}⇒{x}^{2}+x-306=0\phantom{\rule{0ex}{0ex}}⇒{x}^{2}+18x-17x-306=0\phantom{\rule{0ex}{0ex}}⇒x\left(x+18\right)-17\left(x+18\right)=0$

x = 17           (x is a positive integer)

When x = 17,
x + 1 = 17 + 1 = 18

Hence, the required integers are 17 and 18.

#### Question 8:

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

#### Question 9:

Find two consecutive multiples of 3 whose product is 648.

#### Answer:

Let the required consecutive multiples of 3 be 3x and 3(x + 1).

According to the given condition,
$3x×3\left(x+1\right)=648\phantom{\rule{0ex}{0ex}}⇒9\left({x}^{2}+x\right)=648\phantom{\rule{0ex}{0ex}}⇒{x}^{2}+x=72\phantom{\rule{0ex}{0ex}}⇒{x}^{2}+x-72=0$

x = 8           (Neglecting the negative value)

When x = 8,
3x = 3 × 8 = 24
3(x + 1) = 3 × (8 + 1) = 3 × 9 = 27

Hence, the required multiples are 24 and 27.

#### Question 10:

Find two consecutive positive odd integers whose product is 483.

#### Answer:

Let the two consecutive positive odd integers be x and (x + 2).

According to the given condition,
$x\left(x+2\right)=483\phantom{\rule{0ex}{0ex}}⇒{x}^{2}+2x-483=0\phantom{\rule{0ex}{0ex}}⇒{x}^{2}+23x-21x-483=0\phantom{\rule{0ex}{0ex}}⇒x\left(x+23\right)-21\left(x+23\right)=0$

x = 21           (x is a positive odd integer)

When x = 21,
x + 2 = 21 + 2 = 23

Hence, the required integers are 21 and 23.

#### Question 11:

Find two consecutive positive even integers whose product is 288.

#### Answer:

Let the two consecutive positive even integers be x and (x + 2).

According to the given condition,
$x\left(x+2\right)=288\phantom{\rule{0ex}{0ex}}⇒{x}^{2}+2x-288=0\phantom{\rule{0ex}{0ex}}⇒{x}^{2}+18x-16x-288=0\phantom{\rule{0ex}{0ex}}⇒x\left(x+18\right)-16\left(x+18\right)=0$

x = 16           (x is a positive even integer)

When x = 16,
x + 2 = 16 + 2 = 18

Hence, the required integers are 16 and 18.

#### Question 12:

The sum of two natural numbers is 9 and the sum of their reciprocals is $\frac{1}{2}$. Find the numbers.                              [CBSE 2012]

#### Answer:

Let the required natural numbers be x and (9 − x).

According to the given condition,
$\frac{1}{x}+\frac{1}{9-x}=\frac{1}{2}\phantom{\rule{0ex}{0ex}}⇒\frac{9-x+x}{x\left(9-x\right)}=\frac{1}{2}\phantom{\rule{0ex}{0ex}}⇒\frac{9}{9x-{x}^{2}}=\frac{1}{2}\phantom{\rule{0ex}{0ex}}⇒9x-{x}^{2}=18$
$⇒{x}^{2}-9x+18=0\phantom{\rule{0ex}{0ex}}⇒{x}^{2}-6x-3x+18=0\phantom{\rule{0ex}{0ex}}⇒x\left(x-6\right)-3\left(x-6\right)=0\phantom{\rule{0ex}{0ex}}⇒\left(x-3\right)\left(x-6\right)=0$

When x = 3,
9 − x = 9 − 3 = 6

When x = 6,
9 − x = 9 − 6 = 3

Hence, the required natural numbers are 3 and 6.

#### Question 13:

The sum of two natural numbers is 15 and the sum of their reciprocals is $\frac{3}{10}$. Find the numbers.                              [CBSE 2005]

#### Answer:

Let the required natural numbers be x and (15 − x).

According to the given condition,
$\frac{1}{x}+\frac{1}{15-x}=\frac{3}{10}\phantom{\rule{0ex}{0ex}}⇒\frac{15-x+x}{x\left(15-x\right)}=\frac{3}{10}\phantom{\rule{0ex}{0ex}}⇒\frac{15}{15x-{x}^{2}}=\frac{3}{10}\phantom{\rule{0ex}{0ex}}⇒15x-{x}^{2}=50$
$⇒{x}^{2}-15x+50=0\phantom{\rule{0ex}{0ex}}⇒{x}^{2}-10x-5x+50=0\phantom{\rule{0ex}{0ex}}⇒x\left(x-10\right)-5\left(x-10\right)=0\phantom{\rule{0ex}{0ex}}⇒\left(x-5\right)\left(x-10\right)=0$

When x = 5,
15 − x = 15 − 5 = 10

When x = 10,
15 − x = 15 − 10 = 5

Hence, the required natural numbers are 5 and 10.

#### Question 14:

The difference of two natural numbers is 3 and the difference of their reciprocals is $\frac{3}{28}$. Find the numbers.                            [CBSE 2014]

#### Answer:

Let the required natural numbers be x and (x + 3).

Now, x < x + 3
$\therefore \frac{1}{x}>\frac{1}{x+3}$

According to the given condition,
$\frac{1}{x}-\frac{1}{x+3}=\frac{3}{28}\phantom{\rule{0ex}{0ex}}⇒\frac{x+3-x}{x\left(x+3\right)}=\frac{3}{28}\phantom{\rule{0ex}{0ex}}⇒\frac{3}{{x}^{2}+3x}=\frac{3}{28}\phantom{\rule{0ex}{0ex}}⇒{x}^{2}+3x=28$
$⇒{x}^{2}+3x-28=0\phantom{\rule{0ex}{0ex}}⇒{x}^{2}+7x-4x-28=0\phantom{\rule{0ex}{0ex}}⇒x\left(x+7\right)-4\left(x+7\right)=0\phantom{\rule{0ex}{0ex}}⇒\left(x+7\right)\left(x-4\right)=0$

x = 4             (−7 is not a natural number)

When x = 4,
x + 3 = 4 + 3 = 7

Hence, the required natural numbers are 4 and 7.

#### Question 15:

The difference of two natural numbers is 5 and the difference of their reciprocals is $\frac{5}{14}$. Find the numbers.                            [CBSE 2014]

#### Answer:

Let the required natural numbers be x and (x + 5).

Now, x < x + 5
$\therefore \frac{1}{x}>\frac{1}{x+5}$

According to the given condition,
$\frac{1}{x}-\frac{1}{x+5}=\frac{5}{14}\phantom{\rule{0ex}{0ex}}⇒\frac{x+5-x}{x\left(x+5\right)}=\frac{5}{14}\phantom{\rule{0ex}{0ex}}⇒\frac{5}{{x}^{2}+5x}=\frac{5}{14}\phantom{\rule{0ex}{0ex}}⇒{x}^{2}+5x=14$
$⇒{x}^{2}+5x-14=0\phantom{\rule{0ex}{0ex}}⇒{x}^{2}+7x-2x-14=0\phantom{\rule{0ex}{0ex}}⇒x\left(x+7\right)-2\left(x+7\right)=0\phantom{\rule{0ex}{0ex}}⇒\left(x+7\right)\left(x-2\right)=0$

x = 2             (−7 is not a natural number)

When x = 2,
x + 5 = 2 + 5 = 7

Hence, the required natural numbers are 2 and 7.

#### Question 16:

The sum of the squares of two consecutive multiples of 7 is 1225. Find the multiples.

#### Answer:

Let the required consecutive multiples of 7 be 7x and 7(x + 1).

According to the given condition,
${\left(7x\right)}^{2}+{\left[7\left(x+1\right)\right]}^{2}=1225\phantom{\rule{0ex}{0ex}}⇒49{x}^{2}+49\left({x}^{2}+2x+1\right)=1225\phantom{\rule{0ex}{0ex}}⇒49{x}^{2}+49{x}^{2}+98x+49=1225\phantom{\rule{0ex}{0ex}}⇒98{x}^{2}+98x-1176=0$
$⇒{x}^{2}+x-12=0\phantom{\rule{0ex}{0ex}}⇒{x}^{2}+4x-3x-12=0\phantom{\rule{0ex}{0ex}}⇒x\left(x+4\right)-3\left(x+4\right)=0\phantom{\rule{0ex}{0ex}}⇒\left(x+4\right)\left(x-3\right)=0$

x = 3            (Neglecting the negative value)

When x = 3,
7x = 7 × 3 = 21
7(x + 1) = 7(3 + 1) = 7 × 4 = 28

Hence, the required multiples are 21 and 28.

#### Question 17:

The sum of a natural number and its reciprocal is $\frac{65}{8}$. Find the number.

#### Answer:

Let the natural number be x.

According to the given condition,
$x+\frac{1}{x}=\frac{65}{8}\phantom{\rule{0ex}{0ex}}⇒\frac{{x}^{2}+1}{x}=\frac{65}{8}\phantom{\rule{0ex}{0ex}}⇒8{x}^{2}+8=65x\phantom{\rule{0ex}{0ex}}⇒8{x}^{2}-65x+8=0$

x = 8         (x is a natural number)

Hence, the required number is 8.

#### Question 18:

Divide 57 into two parts whose product is 680.

#### Answer:

Let the two parts be x and (57 − x).

According to the given condition,
$x\left(57-x\right)=680\phantom{\rule{0ex}{0ex}}⇒57x-{x}^{2}=680\phantom{\rule{0ex}{0ex}}⇒{x}^{2}-57x+680=0\phantom{\rule{0ex}{0ex}}⇒{x}^{2}-40x-17x+680=0$

When x = 40,
57 − x = 57 − 40 = 17

When x = 17,
57 − x = 57 − 17 = 40

Hence, the required parts are 17 and 40.

#### Question 19:

Divide 27 into two parts such that the sum of their reciprocals is $\frac{3}{20}$.

#### Answer:

Let the two parts be x and (27 − x).

According to the given condition,
$\frac{1}{x}+\frac{1}{27-x}=\frac{3}{20}\phantom{\rule{0ex}{0ex}}⇒\frac{27-x+x}{x\left(27-x\right)}=\frac{3}{20}\phantom{\rule{0ex}{0ex}}⇒\frac{27}{27x-{x}^{2}}=\frac{3}{20}\phantom{\rule{0ex}{0ex}}⇒27x-{x}^{2}=180$
$⇒{x}^{2}-27x+180=0\phantom{\rule{0ex}{0ex}}⇒{x}^{2}-15x-12x+180=0\phantom{\rule{0ex}{0ex}}⇒x\left(x-15\right)-12\left(x-15\right)=0\phantom{\rule{0ex}{0ex}}⇒\left(x-12\right)\left(x-15\right)=0$

When x = 12,
27 − x = 27 − 12 = 15

When x = 15,
27 − x = 27 − 15 = 12

Hence, the required parts are 12 and 15.

#### Question 20:

Divide 16 into two parts such that twice the squares of the larger part exceeds the square of the smaller part by 164.

#### Question 21:

Find two natural numbers, the sum of whose squares is 25 times their sum and also equal to 50 times their difference.

#### Question 22:

The difference of the squares of two natural numbers is 45. The squares of the smaller number is four times the largest number. Find the numbers.

#### Question 23:

Three consecutive positive integers are such that the sum of the square of the first and the product of the other two is 46. Find the integers.                                                                                                                                                                                       [CBSE 2010]

#### Answer:

Let the three consecutive positive integers be x, x + 1 and x + 2.

According to the given condition,
${x}^{2}+\left(x+1\right)\left(x+2\right)=46\phantom{\rule{0ex}{0ex}}⇒{x}^{2}+{x}^{2}+3x+2=46\phantom{\rule{0ex}{0ex}}⇒2{x}^{2}+3x-44=0\phantom{\rule{0ex}{0ex}}⇒2{x}^{2}+11x-8x-44=0$

x = 4           (x is a positive integer)

When x = 4,
x + 1 = 4 + 1 = 5
x + 2 = 4 + 2 = 6

Hence, the required integers are 4, 5 and 6.

#### Question 24:

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

#### Question 25:

A two-digit number is such that the product of its digits is 14. If 45 is added to the number, the digits interchange their places. Find the number.

​​

#### Question 26:

The denominator of a fraction is 3 more than its numerator. The sum of the fraction and its reciprocal is $2\frac{9}{10}$. Find the fraction.

#### Question 27:

The numerator of a fraction is 3 less than its denominator. If 1 is added to the denominator, the fraction is decreased by $\frac{1}{15}$. Find the fraction.                                                                                                                                                                                          [CBSE 2012]

#### Answer:

Let the denominator of the required fraction be x.

Numerator of the required fraction = x − 3

∴ Original fraction = $\frac{x-3}{x}$
If 1 is added to the denominator, then the new fraction obtained is $\frac{x-3}{x+1}$.

According to the given condition,
$\frac{x-3}{x+1}=\frac{x-3}{x}-\frac{1}{15}\phantom{\rule{0ex}{0ex}}⇒\frac{x-3}{x}-\frac{x-3}{x+1}=\frac{1}{15}\phantom{\rule{0ex}{0ex}}⇒\frac{\left(x-3\right)\left(x+1\right)-x\left(x-3\right)}{x\left(x+1\right)}=\frac{1}{15}\phantom{\rule{0ex}{0ex}}⇒\frac{{x}^{2}-2x-3-{x}^{2}+3x}{{x}^{2}+x}=\frac{1}{15}$
$⇒\frac{x-3}{{x}^{2}+x}=\frac{1}{15}\phantom{\rule{0ex}{0ex}}⇒{x}^{2}+x=15x-45\phantom{\rule{0ex}{0ex}}⇒{x}^{2}-14x+45=0\phantom{\rule{0ex}{0ex}}⇒{x}^{2}-9x-5x+45=0$

When x = 5,
$\frac{x-3}{x}=\frac{5-3}{5}=\frac{2}{5}$

When x = 9,
$\frac{x-3}{x}=\frac{9-3}{9}=\frac{6}{9}=\frac{2}{3}$             (This fraction is neglected because this does not satisfies the given condition.)

Hence, the required fraction is $\frac{2}{5}$.

#### Question 28:

The sum of a number and its reciprocal is $2\frac{1}{30}$. Find the number.

#### Answer:

Let the required number be x.

According to the given condition,
$x+\frac{1}{x}=2\frac{1}{30}\phantom{\rule{0ex}{0ex}}⇒\frac{{x}^{2}+1}{x}=\frac{61}{30}\phantom{\rule{0ex}{0ex}}⇒30{x}^{2}+30=61x\phantom{\rule{0ex}{0ex}}⇒30{x}^{2}-61x+30=0$

Hence, the required number is $\frac{5}{6}$ or $\frac{6}{5}$.

#### Question 29:

A teacher on attempting to arrange the students for mass drill in the form of a solid square found that 24 students were left. When he increased the size of the square by one student, he found that he was short of 25 students. Find the number of students.

#### Question 30:

300 apples are distributed equally among a certain number of students. Had there been 10 more students, each would have received one apple less. Find the number of student.

#### Question 31:

In a class test, the sum of Kamal's marks in Mathematics and English is 40. Had he got 3 marks more in Mathematics and 4 marks less in English, the product of the marks would have been 360. Find his marks in two subjects separately.

#### Question 32:

Some students planned a picnic. The total budget for food was ₹2000. But, 5 students failed to attend the picnic and thus the cost for food for each member increased by ₹20. How many students attended the picnic and how much did each student pay for the food?                                                                                                                                                                                                    [CBSE 2010]

#### Answer:

Let x be the number of students who planned a picnic.

∴ Original cost of food for each member = ₹$\frac{2000}{x}$

Five students failed to attend the picnic. So, (x − 5) students attended the picnic.

∴ New cost of food for each member = ₹$\frac{2000}{\left(x-5\right)}$

According to the given condition,
$\frac{2000}{x-5}$ − ₹$\frac{2000}{x}$ = ₹20
$⇒\frac{2000x-2000x+10000}{x\left(x-5\right)}=20\phantom{\rule{0ex}{0ex}}⇒\frac{10000}{{x}^{2}-5x}=20\phantom{\rule{0ex}{0ex}}⇒{x}^{2}-5x=500\phantom{\rule{0ex}{0ex}}⇒{x}^{2}-5x-500=0$

x = 25                (Number of students cannot be negative)

Number of students who attended the picnic = x − 5 = 25 − 5 = 20

Amount paid by each student for the food = ₹$\frac{2000}{\left(25-5\right)}$ = ₹$\frac{2000}{20}$ = ₹100

#### Question 33:

If the price of a book is reduced by ₹5, a person can buy 4 more books for ₹600. Find the original price of the book.

#### Answer:

Let the original price of the book be ₹x.

∴ Number of books bought at original price for ₹600 = $\frac{600}{x}$

If the price of a book is reduced by ₹5, then the new price of the book is ₹(x − 5).

∴ Number of books bought at reduced price for ₹600 = $\frac{600}{x-5}$

According to the given condition,
$\frac{600}{x-5}-\frac{600}{x}=4\phantom{\rule{0ex}{0ex}}⇒\frac{600x-600x+3000}{x\left(x-5\right)}=4\phantom{\rule{0ex}{0ex}}⇒\frac{3000}{{x}^{2}-5x}=4\phantom{\rule{0ex}{0ex}}⇒{x}^{2}-5x=750$
$⇒{x}^{2}-5x-750=0\phantom{\rule{0ex}{0ex}}⇒{x}^{2}-30x+25x-750=0\phantom{\rule{0ex}{0ex}}⇒x\left(x-30\right)+25\left(x-30\right)=0\phantom{\rule{0ex}{0ex}}⇒\left(x-30\right)\left(x+25\right)=0$

x = 30               (Price cannot be negative)

Hence, the original price of the book is ₹30.

#### Question 34:

A person on tour has ₹10800 for his expenses. If he extends his tour by 4 days, he has to cut down his daily expenses by ₹90. Find the original duration of the tour.

#### Answer:

Let the original duration of the tour be x days.

∴ Original daily expenses = ₹$\frac{10,800}{x}$

If he extends his tour by 4 days, then his new daily expenses = ₹$\frac{10,800}{x+4}$

According to the given condition,

$\frac{10,800}{x}$ − ₹$\frac{10,800}{x+4}$ = ₹90

$⇒\frac{10800x+43200-10800x}{x\left(x+4\right)}=90\phantom{\rule{0ex}{0ex}}⇒\frac{43200}{{x}^{2}+4x}=90\phantom{\rule{0ex}{0ex}}⇒{x}^{2}+4x=480\phantom{\rule{0ex}{0ex}}⇒{x}^{2}+4x-480=0$

x = 20              (Number of days cannot be negative)

Hence, the original duration of the tour is 20 days.

#### Question 35:

In a class test, the sum of the marks obtained by P in mathematics and science is 28. Had he got 3 more marks in mathematics and 4 marks less in science, the product of marks obtained in the two subjects would have been 180. Find the marks obtained him in the two subjects separately.                                                                                                                                                                                           [CBSE 2008]

#### Answer:

Let the marks obtained by P in mathematics and science be x and (28 − x), respectively.

According to the given condition,
$\left(x+3\right)\left(28-x-4\right)=180\phantom{\rule{0ex}{0ex}}⇒\left(x+3\right)\left(24-x\right)=180\phantom{\rule{0ex}{0ex}}⇒-{x}^{2}+21x+72=180\phantom{\rule{0ex}{0ex}}⇒{x}^{2}-21x+108=0$

When x = 12,
28 − x = 28 − 12 = 16

When x = 9,
28 − x = 28 − 9 = 19

Hence, he obtained 12 marks in mathematics and 16 marks in science or 9 marks in mathematics and 19 marks in science.

#### Question 36:

A man busy a number of pens for Rs 80. If he had bought 4 more pens for the same amount, each pen would have cost him Rs 1 less. How many pens did he buy?

#### Question 37:

A dealer sells an article for ₹75 and gains as much percent as the cost price of the article. Find the cost price of the article.
[CBSE 2011]

#### Answer:

Let the cost price of the article be ₹x.

∴ Gain percent = x%

According to the given condition,
x + ₹$\left(\frac{x}{100}×x\right)$ = ₹75                            (Cost price + Gain = Selling price)
$⇒\frac{100x+{x}^{2}}{100}=75\phantom{\rule{0ex}{0ex}}⇒{x}^{2}+100x=7500\phantom{\rule{0ex}{0ex}}⇒{x}^{2}+100x-7500=0\phantom{\rule{0ex}{0ex}}⇒{x}^{2}+150x-50x-7500=0$

x = 50                   (Cost price cannot be negative)

Hence, the cost price of the article is ₹50.

#### Question 38:

One year ago, a man was 8 times as old as his son. Now, his age is equal to the square of his son's age. Find their present ages.

#### Answer:

Let the present age of the son be x years.

∴ Present age of the man = x2 years

One year ago,

Age of the son = (x − 1) years

Age of the man = (x2 − 1) years

According to the given condition,

Age of the man = 8 × Age of the son

$\therefore {x}^{2}-1=8\left(x-1\right)\phantom{\rule{0ex}{0ex}}⇒{x}^{2}-1=8x-8\phantom{\rule{0ex}{0ex}}⇒{x}^{2}-8x+7=0\phantom{\rule{0ex}{0ex}}⇒{x}^{2}-7x-x+7=0$

x = 7                (Man's age cannot be 1 year)

Present age of the son = 7 years

Present age of the man = 72 years = 49 years

#### Question 39:

The sum of the reciprocals of Meena's ages (in years) 3 years ago and 5 years hence is $\frac{1}{3}$. Find her present age.

#### Answer:

Let the present age of Meena be x years.

Meena's age 3 years ago = (x − 3) years

Meena's age 5 years hence = (x + 5) years

According to the given condition,
$\frac{1}{x-3}+\frac{1}{x+5}=\frac{1}{3}\phantom{\rule{0ex}{0ex}}⇒\frac{x+5+x-3}{\left(x-3\right)\left(x+5\right)}=\frac{1}{3}\phantom{\rule{0ex}{0ex}}⇒\frac{2x+2}{{x}^{2}+2x-15}=\frac{1}{3}\phantom{\rule{0ex}{0ex}}⇒{x}^{2}+2x-15=6x+6$
$⇒{x}^{2}-4x-21=0\phantom{\rule{0ex}{0ex}}⇒{x}^{2}-7x+3x-21=0\phantom{\rule{0ex}{0ex}}⇒x\left(x-7\right)+3\left(x-7\right)=0\phantom{\rule{0ex}{0ex}}⇒\left(x-7\right)\left(x+3\right)=0$

x = 7                 (Age cannot be negative)

Hence, the present age of Meena is 7 years.

#### Question 40:

The sum of the ages of a boy and his brother is 25 years, and the product of their ages in years is 126. Find their ages.

#### Question 41:

The product of Tanvy's age (in years) 5 years ago and her age 8 years later is 30. Find her present age.

#### Question 42:

Two years ago, a man's age was three times the square of his son's age. In three years time, his age will be four times his son's age. Find their present ages.

#### Answer:

Let son's age 2 years ago be x years. Then,

Man's age 2 years ago = 3x2 years

∴ Son's present age = (x + 2) years

Man's present age = (3x2 + 2) years

In three years time,

Son's age = (x + 2 + 3) years = (x + 5) years

Man's age = (3x2 + 2 + 3) years = (3x2 + 5) years

According to the given condition,

Man's age = 4 × Son's age

∴ 3x2 + 5 = 4(x + 5)

$⇒3{x}^{2}+5=4x+20\phantom{\rule{0ex}{0ex}}⇒3{x}^{2}-4x-15=0\phantom{\rule{0ex}{0ex}}⇒3{x}^{2}-9x+5x-15=0\phantom{\rule{0ex}{0ex}}⇒3x\left(x-3\right)+5\left(x-3\right)=0$

x = 3              (Age cannot be negative)

Son's present age = (x + 2) years = (3 + 2) years = 5 years

Man's present age = (3x2 + 2) years = (3 × 9 + 2) years = 29 years

#### Question 43:

A truck covers a distance of 150 km at a certain average speed and then covers another 200 km at an average speed which is 20 km per hour more than the first speed. If the truck covers the total distance in 5 hours, find the first speed of the truck.                                 [CBSE 2015]

#### Answer:

Let the first speed of the truck be x km/h.

∴ Time taken to cover 150 km = $\frac{150}{x}$ h

New speed of the truck = (x + 20) km/h

∴ Time taken to cover 200 km = $\frac{200}{x+20}$ h

According to the given condition,

Time taken to cover 150 km + Time taken to cover 200 km = 5 h
$\therefore \frac{150}{x}+\frac{200}{x+20}=5\phantom{\rule{0ex}{0ex}}⇒\frac{150x+3000+200x}{x\left(x+20\right)}=5\phantom{\rule{0ex}{0ex}}⇒350x+3000=5\left({x}^{2}+20x\right)\phantom{\rule{0ex}{0ex}}⇒350x+3000=5{x}^{2}+100x$
$⇒5{x}^{2}-250x-3000=0\phantom{\rule{0ex}{0ex}}⇒{x}^{2}-50x-600=0\phantom{\rule{0ex}{0ex}}⇒{x}^{2}-60x+10x-600=0\phantom{\rule{0ex}{0ex}}⇒x\left(x-60\right)+10\left(x-60\right)=0$

x = 60                  (Speed cannot be negative)

Hence, the first speed of the truck is 60 km/h.

#### Question 44:

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/hour.
Find the original speed of the plane.
Do you appreciate the values shown by the pilot, namely promptness in providing help to the injured and his efforts to reach in time?
[CBSE 2013]

#### Answer:

Let the original speed of the plane be x km/h.

∴ Actual speed of the plane = (x + 100) km/h

Distance of the journey = 1500 km

Time taken to reach the destination at original speed = $\frac{1500}{x}$ h                             $\left(\mathrm{Time}=\frac{\mathrm{Distance}}{\mathrm{Speed}}\right)$

Time taken to reach the destination at actual speed = $\frac{1500}{x+100}$ h

According to the given condition,

Time taken to reach the destination at original speed = Time taken to reach the destination at actual speed + 30 min

$⇒{x}^{2}+100x=300000\phantom{\rule{0ex}{0ex}}⇒{x}^{2}+100x-300000=0\phantom{\rule{0ex}{0ex}}⇒{x}^{2}+600x-500x-300000=0\phantom{\rule{0ex}{0ex}}⇒x\left(x+600\right)-500\left(x+600\right)=0$

x = 500                 (Speed cannot be negative)

Hence, the original speed of the plane is 500 km/h.

Yes, we appreciate the values shown by the pilot, namely promptness in providing help to the injured and his efforts to reach in time. This reflects the caring nature of the pilot and his dedication to the work.

#### Question 45:

A train covers a distance of 480 km at a uniform speed. If the speed had been 8 km/h less then it would have taken 3 hours more to cover the same distance. Find the usual speed of the train.                                                                                                                             [CBSE 2013C]

#### Answer:

Let the usual speed of the train be x km/h.

∴ Reduced speed of the train = (x − 8) km/h

Total distance to be covered = 480 km

Time taken by the train to cover the distance at usual speed = $\frac{480}{x}$ h                  $\left(\mathrm{Time}=\frac{\mathrm{Distance}}{\mathrm{Speed}}\right)$

Time taken by the train to cover the distance at reduced speed = $\frac{480}{x-8}$ h

According to the given condition,

Time taken by the train to cover the distance at reduced speed = Time taken by the train to cover the distance at usual speed + 3 h

$\therefore \frac{480}{x-8}=\frac{480}{x}+3\phantom{\rule{0ex}{0ex}}⇒\frac{480}{x-8}-\frac{480}{x}=3\phantom{\rule{0ex}{0ex}}⇒\frac{480x-480x+3840}{x\left(x-8\right)}=3\phantom{\rule{0ex}{0ex}}⇒\frac{3840}{{x}^{2}-8x}=3$
$⇒{x}^{2}-8x=1280\phantom{\rule{0ex}{0ex}}⇒{x}^{2}-8x-1280=0\phantom{\rule{0ex}{0ex}}⇒{x}^{2}-40x+32x-1280=0\phantom{\rule{0ex}{0ex}}⇒x\left(x-40\right)+32\left(x-40\right)=0$

x = 40              (Speed cannot be negative)

Hence, the usual speed of the train is 40 km/h.

#### Question 46:

A train travels at a certain average speed for a distance of 54 km and then travels a distance of 63 km at an average of 6 km/hr more than the first speed. If it takes 3 hours to complete the total journey, what is its first speed?                                                                        [CBSE 2015]

#### Answer:

Let the first speed of the train be x km/h.

∴ Time taken to cover 54 km = $\frac{54}{x}$ h

New speed of the train = (x + 6) km/h

∴ Time taken to cover 63 km = $\frac{63}{x+6}$ h

According to the given condition,

Time taken to cover 54 km + Time taken to cover 63 km = 3 h
$\therefore \frac{54}{x}+\frac{63}{x+6}=3\phantom{\rule{0ex}{0ex}}⇒\frac{54x+324+63x}{x\left(x+6\right)}=3\phantom{\rule{0ex}{0ex}}⇒117x+324=3\left({x}^{2}+6x\right)\phantom{\rule{0ex}{0ex}}⇒117x+324=3{x}^{2}+18x$
$⇒3{x}^{2}-99x-324=0\phantom{\rule{0ex}{0ex}}⇒{x}^{2}-33x-108=0\phantom{\rule{0ex}{0ex}}⇒{x}^{2}-36x+3x-108=0\phantom{\rule{0ex}{0ex}}⇒x\left(x-36\right)+3\left(x-36\right)=0$

x = 36                  (Speed cannot be negative)

Hence, the first speed of the train is 36 km/h.

#### Question 47:

A train travels 180 km at a uniform speed. If the speed had been 9 km/hr more, it would have taken 1 hour less for the same journey. Find the speed of the train.       [CBSE 2013]

#### Answer:

Distance covered by the train = 180 km
We know that distance covered$\left(d\right)=\mathrm{speed}\left(s\right)×\mathrm{time}\left(t\right)$

Also, given that if the speed is increased by 9km/h, time of travel gets reduced by 1 hour.

Put the value of (i) in (ii), we get

Ignore the negative value
So, time taken = 5 hours
From (i)
$s=\frac{180}{t}=\frac{180}{5}=36$
Hence, the speed is 36 km/h.

#### Question 48:

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

#### Question 49:

A passenger train takes 2 hours less for a journey of 300 km if its speed is increased by 5 km/hr from its usual speed. Find its usual speed.

#### Question 50:

The distance between Mumbai and Pune is 192 km. Travelling by the Deccan Queen, it takes 48 minutes less than another train. Calculate the speed of the Deccan Queen if the speed of the two trains differ by 20 km/hr.

#### Question 51:

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

#### Question 52:

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.

#### Question 53:

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

#### Question 54:

A takes 10 days 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.

#### Question 55:

Two pipes running together can fill a cistern in $3\frac{1}{13}$ minutes. If one pipe takes 3 minutes more than the other to fill it, find the time in which each pipe would fill the cistern.

#### Question 56:

Two pipes running together can fill a tank in $11\frac{1}{9}$ 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.                                                                                                                [CBSE 2010]

#### Answer:

Let the time taken by one pipe to fill the tank be x minutes.

∴ Time taken by the other pipe to fill the tank = (x + 5) min

Suppose the volume of the tank be V.

Volume of the tank filled by one pipe in x minutes = V

∴ Volume of the tank filled by one pipe in 1 minute = $\frac{V}{x}$

⇒ Volume of the tank filled by one pipe in $11\frac{1}{9}$ minutes = $\frac{V}{x}×11\frac{1}{9}=\frac{V}{x}×\frac{100}{9}$

Similarly,

Volume of the tank filled by the other pipe in $11\frac{1}{9}$ minutes = $\frac{V}{\left(x+5\right)}×11\frac{1}{9}=\frac{V}{\left(x+5\right)}×\frac{100}{9}$

Now,

Volume of the tank filled by one pipe in $11\frac{1}{9}$ minutes + Volume of the tank filled by the other pipe in $11\frac{1}{9}$ minutes = V
$\therefore V\left(\frac{1}{x}+\frac{1}{x+5}\right)×\frac{100}{9}=V\phantom{\rule{0ex}{0ex}}⇒\frac{1}{x}+\frac{1}{x+5}=\frac{9}{100}\phantom{\rule{0ex}{0ex}}⇒\frac{x+5+x}{x\left(x+5\right)}=\frac{9}{100}\phantom{\rule{0ex}{0ex}}⇒\frac{2x+5}{{x}^{2}+5x}=\frac{9}{100}$
$⇒200x+500=9{x}^{2}+45x\phantom{\rule{0ex}{0ex}}⇒9{x}^{2}-155x-500=0\phantom{\rule{0ex}{0ex}}⇒9{x}^{2}-180x+25x-500=0\phantom{\rule{0ex}{0ex}}⇒9x\left(x-20\right)+25\left(x-20\right)=0$

x = 20                    (Time cannot be negative)

Time taken by one pipe to fill the tank = 20 min

Time taken by other pipe to fill the tank = (20 + 5) = 25 min

#### Question 57:

Two water taps together can fill a tank in 6 hours. The tap of larger diameter takes 9 hours less than the smaller one to fill the tank separately. Find the time in which each tap can separately fill the tank.                                                                                                             [CBSE 2011]

#### Answer:

Let the tap of smaller diameter fill the tank in x hours.

∴ Time taken by the tap of larger diameter to fill the tank = (x − 9) h

Suppose the volume of the tank be V.

Volume of the tank filled by the tap of smaller diameter in x hours = V

∴ Volume of the tank filled by the tap of smaller diameter in 1 hour = $\frac{V}{x}$

⇒ Volume of the tank filled by the tap of smaller diameter in 6 hours = $\frac{V}{x}×6$

Similarly,

Volume of the tank filled by the tap of larger diameter in 6 hours = $\frac{V}{\left(x-9\right)}×6$

Now,

Volume of the tank filled by the tap of smaller diameter in 6 hours + Volume of the tank filled by the tap of larger diameter in 6 hours = V
$\therefore V\left(\frac{1}{x}+\frac{1}{x-9}\right)×6=V\phantom{\rule{0ex}{0ex}}⇒\frac{1}{x}+\frac{1}{x-9}=\frac{1}{6}\phantom{\rule{0ex}{0ex}}⇒\frac{x-9+x}{x\left(x-9\right)}=\frac{1}{6}\phantom{\rule{0ex}{0ex}}⇒\frac{2x-9}{{x}^{2}-9x}=\frac{1}{6}$
$⇒12x-54={x}^{2}-9x\phantom{\rule{0ex}{0ex}}⇒{x}^{2}-21x+54=0\phantom{\rule{0ex}{0ex}}⇒{x}^{2}-18x-3x+54=0\phantom{\rule{0ex}{0ex}}⇒x\left(x-18\right)-3\left(x-18\right)=0$

For x = 3, time taken by the tap of larger diameter to fill the tank is negative which is not possibe.

x = 18

Time taken by the tap of smaller diameter to fill the tank = 18 h

Time taken by the tap of larger diameter to fill the tank = (18 − 9) = 9 h

Hence, the time taken by the taps of smaller and larger diameter to fill the tank is 18 hours and 9 hours, respectively.

#### Question 58:

The length of a rectangle is twice its breadth and its area is 288 cm2. Find the dimensions of the rectangle.

#### Question 59:

The length of a rectangular field is three times its breadth. If the area of the field by 147 sq metres, find the length of the field.

#### Question 60:

The length of a hall is 3 metres more than its breadth. If the area of the hall is 238 square metres, calculate its length and breadth.

#### Question 61:

The perimeter of a rectangular plot is 62 m and its area is 228 sq metres. Find the dimensions of the plot.

#### Question 62:

A rectangular field is 16 m long and 10 m wide. There is a path of uniform width all around it, having an area of 120 m2. Find the width of the path.

#### Question 63:

The sum of the areas of two squares is 640 m2. If the difference in their perimeters be 64 m, find the sides of the two squares.

#### Answer:

Let the length of the side of the first and the second square be $x$ and y, respectively.

Putting the value of $x$ in (i), we get:

#### Question 64:

The length of a rectangle is thrice as long as the side of a square. The side of the square is 4 cm more than width of a the rectangle. Their areas being equal, find their dimensions.

#### Question 65:

A farmer prepares a rectangular vegetable garden of area 180 sq metres. With 39 metres of barbed wire, he can fence the three sides of the garden, leaving one of the longer sides unfenced. Find the dimensions of the garden.

#### Question 66:

The area of a right-triangle is  600 cm2. If the base of the triangle exceeds the altitude by 10 cm, find the dimensions of the triangle.

#### Answer:

Let the altitude of the triangle be x cm.
Therefore, the base of the triangle will be (+ 10) cm.

#### Question 67:

The area of a right-angled triangle is 96 sq metres. If the base is three times the altitude, find the base.

#### Answer:

Let the altitude of the triangle be x m.
Therefore, the base will be 3x m.

Area of a triangle =

The value of cannot be negative.
Therefore, the altitude and base of the triangle are 8 m and (3 $×$ 8 = 24 m), respectively.

#### Question 68:

The area of a right-angled triangle is 165 sq metres. Determine its base and altitude if the latter exceeds the former by 7 metres.

#### Answer:

Let the base be $x$ m.
Therefore, the altitude will be m.

Area of a triangle =

The value of $x$ cannot be negative.
Therefore, the base is 15 m and the altitude is {(15 + 7) = 22 m}.

#### Question 69:

The hypotenuse of a right-angled triangle is 20 metres. If the difference between the length of the other sides be 4 metres, find the other sides.

#### Answer:

Let one side of the right-angled triangle be $x$ m and the other side be  m.
On applying Pythagoras theorem, we have:

The value of x cannot be negative.
Therefore, the base is 12 m and the other side is {(12 + 4) = 16 m}.

#### Question 70:

The length of the hypotenuse of a right-angled triangle exceeds the length of the base by 2 cm and exceeds twice the length of the altitude by 1 cm. Find the length of each side of the triangle.

#### Answer:

Let the base and altitude of the right-angled triangle be $x$ and y cm, respectively.
Therefore, the hypotenuse will be  cm.

Thus, the base, altitude and hypotenuse of the triangle are 15 cm, 8 cm and 17 cm, respectively.

#### Question 71:

The hypotenuse of a right-angled triangle is 1 metre less than twice the shortest side. If the third is 1 metre more than the shortest side, find the sides of the triangle.

#### Answer:

Let the shortest side be $x$ m.
Therefore, according to the question:
Hypotenuse =
Third side =  m
On applying Pythagoras theorem, we get:

The length of the side cannot be 0; therefore, the shortest side is 3 m.
Therefore,
Hypotenuse = 5 m
Third side = (3 + 1) = 4 m

#### Question 1:

Which of the following is a quadratic equation?
(a) ${x}^{2}-3\sqrt{x}+2=0$
(b) $x+\frac{1}{x}={x}^{2}$
(c) ${x}^{2}+\frac{1}{{x}^{2}}=5$
(d)

(d)

#### Question 2:

Which of the following is a quadratic equation?
(a) $\left({x}^{2}+1\right)=\left(2-x{\right)}^{2}+3$
(b) ${x}^{3}-{x}^{2}=\left(x-1{\right)}^{3}$
(c) $2{x}^{2}+3=\left(5+x\right)\left(2x-3\right)$
(d) None of these

(b)

#### Question 3:

Which of the following is not a quadratic equation?
(a) $3x-{x}^{2}={x}^{2}+5$
(b) $\left(x+2{\right)}^{2}=2\left({x}^{2}-5\right)$
(c) $\left(\sqrt{2}x+3{\right)}^{2}=2{x}^{2}+6$
(d) $\left(x-1{\right)}^{2}=3{x}^{2}+x-2$

(c)

#### Question 4:

If x = 3 is a solution of the equation $3{x}^{2}+\left(k-1\right)x+9=0,$ then k = ?
(a) 11
(b) −11
(c) 13
(d) −13

(b) −11

#### Question 5:

If one root of the equation $2{x}^{2}+ax+6=0$ is 2, then a = ?
(a) 7
(b) −7
(c) $\frac{7}{2}$
(d) $\frac{-7}{2}$

(b) −7

#### Question 6:

The sum of the roots of the equation ${x}^{2}-6x+2=0$ is
(a) 2
(b) −2
(c) 6
(d) −6

(c) 6

#### Question 7:

If the product of the roots of the equation ${x}^{2}-3x+k=10$ is −2, then the value of k is
(a) −2
(b) −8
(c)  8
(d) 12

(c)  8

#### Question 8:

The ratio of the sum and product of the roots of the equation $7{x}^{2}-12x+18=0$
(a) 7 : 12
(b) 7 : 18
(c) 2 : 3
(d) 3 : 2

(c) 2 : 3

#### Question 9:

If one root of the equation  then the other root is
(a) $\frac{-1}{3}$
(b) $\frac{1}{3}$
(c) −3
(d) 3

(d) 3

#### Question 10:

If one root of $5{x}^{2}+13x+k=0$ be the reciprocal of the other root, then the value of k is
(a) 0
(b) 1
(c) 2
(d) 5

(d) 5

#### Question 11:

If the sum of the roots of the equation $k{x}^{2}+2x+3x=0$ is equal to their product, then the value of k is
(a) $\frac{1}{3}$
(b) $\frac{-1}{3}$
(c) $\frac{2}{3}$
(d) $\frac{-2}{3}$

#### Answer:

(d) $\frac{-2}{3}$

#### Question 12:

The root of a quadratic equation are 5 and −2. Then, the equation is
(a) ${x}^{2}-3x+10=0$
(b) ${x}^{2}-3x-10=0$
(c) ${x}^{2}+3x-10=0$
(d) ${x}^{2}+3x+10=0$

(b)

#### Question 13:

If the sum of the roots of a quadratic equation is 6 and their product is 6, the equation is
(a) ${x}^{2}-6x+6=0$
(b) ${x}^{2}+6x-6=0$
(c) ${x}^{2}-6x-6=0$
(d) ${x}^{2}+6x+6=0$

(a)

#### Question 14:

Objective Questions (MCQ)

If α and β are the roots of the equation $3{x}^{2}+8x+2=0$ then $\left(\frac{1}{\alpha }+\frac{1}{\beta }\right)$ = ?

(a) $\frac{-3}{8}$                            (b) $\frac{2}{3}$                            (c) −4                            (d) 4

#### Answer:

It is given that α and β are the roots of the equation $3{x}^{2}+8x+2=0$.

and $\alpha \beta =\frac{2}{3}$
$\frac{1}{\alpha }+\frac{1}{\beta }=\frac{\alpha +\beta }{\alpha \beta }=\frac{-\frac{8}{3}}{\frac{2}{3}}=-4$

Hence, the correct answer is option C.

#### Question 15:

The roots of the equation $a{x}^{2}+bx+c=0$ will be reciprocal of each other if
(a) a = b
(b) bc
(c) c = a
(d) none of these

(c) c  =  a

#### Question 16:

If the roots of the equation $a{x}^{2}+bx+c=0$ are equal, then then c = ?
(a) $\frac{-b}{2a}$
(b) $\frac{b}{2a}$
(c) $\frac{-{b}^{2}}{4a}$
(d) $\frac{{b}^{2}}{4a}$

#### Answer:

(d) $\frac{{b}^{2}}{4a}$

#### Question 17:

If the equation  has equal roots, then k = ?
(a) 2 or 0
(b) −2 or 0
(c) 2 or −2
(d) 0 only

(c) 2 or −2

#### Question 18:

If the equation ${x}^{2}+2\left(k+2\right)x+9k=0$ has equal rots, then k = ?
(a) 1 or 4
(b) −1 or 4
(c) 1 or −4
(d) −1 or −4

(a) 1 or 4

#### Question 19:

If the equation $4{x}^{2}-3kx+1=0$ has equal roots, then k = ?
(a) $±\frac{2}{3}$
(b) $±\frac{1}{3}$
(c) $±\frac{3}{4}$
(d) $±\frac{4}{3}$

#### Answer:

(d) $±\frac{4}{3}$

#### Question 20:

The roots of  are real and unequal, if $\left({b}^{2}-4ac\right)$
(a) > 0
(b) = 0
(c) < 0
(d) none of these

(a) >  0

#### Question 21:

Objective Questions (MCQ)

In the equation $a{x}^{2}+bx+c=0$, it is given that $D=\left({b}^{2}-4ac\right)>0$. Then, the roots of the equation are

(a) real and equal                                                        (b) real and unequal
(c) imaginary                                                              (d) none of these

#### Answer:

We know that when discriminant, $D>0$, the roots of the given quadratic equation are real and unequal.

Hence, the correct answer is option B.

#### Question 22:

The roots of the equation $2{x}^{2}-6x+7=0$ are
(a) real, unequal and rational
(b) real, unequal and irrational
(c) real and equal
(d) imaginary

(d) imaginary

#### Question 23:

The roots of the equation $2{x}^{2}-6x+3=0$ are
(a) real, unequal and rational
(b) real, unequal and irrational
(c) real and equal
(d) imaginary

#### Answer:

(b) real, unequal and irrational

#### Question 24:

If the roots of $5{x}^{2}-kx+1=0$ are real and distinct, then
(a) $-2\sqrt{5}
(b) $k>2\sqrt{5}$ only
(c) $k<-2\sqrt{5}$ only
(d) either $k>2\sqrt{5}$ or $k<-2\sqrt{5}$

(d) either or

#### Question 25:

If the equation ${x}^{2}+5kx+16=0$ has no real roots, then
(a) $k>\frac{8}{5}$
(b) $k<\frac{-8}{5}$
(c) $\frac{-8}{5}\frac{8}{5}$
(d) none of these

(c)

#### Question 26:

If the equation ${x}^{2}-kx+1=0$ has no real roots, then
(a) k < −2
(b) k > 2
(c) −2 < k < 2
(d) none of these

(c) −2  <   <  2

#### Question 27:

For what values of k, the equation $k{x}^{2}-6x-2=0$ has real roots?
(a) $k\le \frac{-9}{2}$
(b) $k\ge \frac{-9}{2}$
(c) $k\le -2$
(d) None of these

(b)

#### Question 28:

The sum of a number and its reciprocal is $2\frac{1}{20}.$ The number is
(a)
(b)
(c)
(d)

(a)

#### Question 29:

The perimeter of a rectangle is 82 m and its area is 400 m2. The breadth of the rectangle is
(a) 25 m
(b) 20 m
(c) 16 m
(d) 9 m

(c) 16 m

#### Question 30:

Objective Questions (MCQ)

The length of a rectangular field exceeds its breadth by 8 m and the area of the field is 240 m2. The breadth of the field is

(a) 20 m                                (b) 30 m                                (c) 12 m                                (d) 16 m                                           [CBSE 2014]

#### Answer:

Let the breadth of the rectangular field be x m.

∴ Length of the rectangular field = (x + 8) m

Area of the rectangular field = 240 m2                    (Given)

$\therefore \left(x+8\right)×x=240$                       (Area = Length × Breadth)
$⇒{x}^{2}+8x-240=0\phantom{\rule{0ex}{0ex}}⇒{x}^{2}+20x-12x-240=0\phantom{\rule{0ex}{0ex}}⇒x\left(x+20\right)-12\left(x+20\right)=0\phantom{\rule{0ex}{0ex}}⇒\left(x+20\right)\left(x-12\right)=0$

x = 12              (Breadth cannot be negative)

Thus, the breadth of the field is 12 m.

Hence, the correct answer is option C.

#### Question 31:

Objective Questions (MCQ)

The roots of the quadratic equation $2{x}^{2}-x-6=0$ are

(a)                                     (b)                                     (c)                                     (d)                      [CBSE 2012]

#### Answer:

The given quadratic equation is $2{x}^{2}-x-6=0$.

$2{x}^{2}-x-6=0\phantom{\rule{0ex}{0ex}}⇒2{x}^{2}-4x+3x-6=0\phantom{\rule{0ex}{0ex}}⇒2x\left(x-2\right)+3\left(x-2\right)=0\phantom{\rule{0ex}{0ex}}⇒\left(x-2\right)\left(2x+3\right)=0$

Thus, the roots of the given equation are 2 and $\frac{-3}{2}$.

Hence, the correct answer is option B.

#### Question 32:

The sum of two natural numbers is 8 and their product is 15. Find the numbers.                                          [CBSE 2012]

#### Answer:

Let the required natural numbers be x and (8 − x).

It is given that the product of the two numbers is 15.
$\therefore x\left(8-x\right)=15\phantom{\rule{0ex}{0ex}}⇒8x-{x}^{2}=15\phantom{\rule{0ex}{0ex}}⇒{x}^{2}-8x+15=0\phantom{\rule{0ex}{0ex}}⇒{x}^{2}-5x-3x+15=0$

Hence, the required numbers are 3 and 5.

#### Question 33:

Very-Short-Answer Questions

Show that x = −3 is a solution of ${x}^{2}+6x+9=0$.                         [CBSE 2008]

#### Answer:

The given equation is ${x}^{2}+6x+9=0$.

Putting x = −3 in the given equation, we get

LHS = ${\left(-3\right)}^{2}+6×\left(-3\right)+9=9-18+9=0$ = RHS

∴ x = −3 is a solution of the given equation.

#### Question 34:

Very-Short-Answer Questions

Show that = −2 is a solution of $3{x}^{2}+13x+14=0$.                          [CBSE 2008]

#### Answer:

The given equation is $3{x}^{2}+13x+14=0$.

Putting x = −2 in the given equation, we get

LHS = $3×{\left(-2\right)}^{2}+13×\left(-2\right)+14=12-26+14=0$ = RHS

∴ x = −2 is a solution of the given equation.

#### Question 35:

Very-Short-Answer Questions

If $x=\frac{-1}{2}$ is a solution of the quadratic equation $3{x}^{2}+2kx-3=0$, find the value of k.                     [CBSE 2015]

#### Answer:

It is given that $x=\frac{-1}{2}$ is a solution of the quadratic equation $3{x}^{2}+2kx-3=0$.

$\therefore 3×{\left(\frac{-1}{2}\right)}^{2}+2k×\left(\frac{-1}{2}\right)-3=0\phantom{\rule{0ex}{0ex}}⇒\frac{3}{4}-k-3=0\phantom{\rule{0ex}{0ex}}⇒k=\frac{3-12}{4}=-\frac{9}{4}$

Hence, the value of k is $-\frac{9}{4}$.

#### Question 36:

Very-Short-Answer Questions

Find the roots of the quadratic equation $2{x}^{2}-x-6=0$.                                 [CBSE 2012]

#### Answer:

The given quadratic equation is $2{x}^{2}-x-6=0$.
$2{x}^{2}-x-6=0\phantom{\rule{0ex}{0ex}}⇒2{x}^{2}-4x+3x-6=0\phantom{\rule{0ex}{0ex}}⇒2x\left(x-2\right)+3\left(x-2\right)=0\phantom{\rule{0ex}{0ex}}⇒\left(x-2\right)\left(2x+3\right)=0$

Hence, the roots of the given equation are 2 and $-\frac{3}{2}$.

#### Question 37:

Very-Short-Answer Questions

Find the solution of the quadratic equation $3\sqrt{3}{x}^{2}+10x+\sqrt{3}=0$.                    [CBSE 2009]

#### Answer:

The given quadratic equation is $3\sqrt{3}{x}^{2}+10x+\sqrt{3}=0$.
$3\sqrt{3}{x}^{2}+10x+\sqrt{3}=0\phantom{\rule{0ex}{0ex}}⇒3\sqrt{3}{x}^{2}+9x+x+\sqrt{3}=0\phantom{\rule{0ex}{0ex}}⇒3\sqrt{3}x\left(x+\sqrt{3}\right)+1\left(x+\sqrt{3}\right)=0\phantom{\rule{0ex}{0ex}}⇒\left(x+\sqrt{3}\right)\left(3\sqrt{3}x+1\right)=0$

Hence, $-\sqrt{3}$ and $-\frac{\sqrt{3}}{9}$ are the solutions of the given equation.

#### Question 38:

Very-Short-Answer Questions

If the roots of the quadratic equation $2{x}^{2}+8x+k=0$ are equal then find the value of k.                     [CBSE 2014]

#### Answer:

It is given that the roots of the quadratic equation $2{x}^{2}+8x+k=0$ are equal.
$\therefore D=0\phantom{\rule{0ex}{0ex}}⇒{8}^{2}-4×2×k=0\phantom{\rule{0ex}{0ex}}⇒64-8k=0\phantom{\rule{0ex}{0ex}}⇒k=8$
Hence, the value of k is 8.

#### Question 39:

Very-Short-Answer Questions

If the quadratic equation $p{x}^{2}-2\sqrt{5}px+15=0$ has two equal roots then find the value of p.                   [CBSE 2015]

#### Answer:

It is given that the quadratic equation $p{x}^{2}-2\sqrt{5}px+15=0$ has two equal roots.

$\therefore D=0\phantom{\rule{0ex}{0ex}}⇒{\left(-2\sqrt{5}p\right)}^{2}-4×p×15=0\phantom{\rule{0ex}{0ex}}⇒20{p}^{2}-60p=0\phantom{\rule{0ex}{0ex}}⇒20p\left(p-3\right)=0$

For p = 0, we get 15 = 0, which is not true.

p ≠ 0

Hence, the value of p is 3.

#### Question 40:

Very-Short-Answer Questions

If 1 is a root of the equation $a{y}^{2}+ay+3=0$ and ${y}^{2}+y+b=0$ then find the value of ab.                 [CBSE 2012]

#### Answer:

It is given that y = 1 is a root of the equation $a{y}^{2}+ay+3=0$.
$\therefore a×{\left(1\right)}^{2}+a×1+3=0\phantom{\rule{0ex}{0ex}}⇒a+a+3=0\phantom{\rule{0ex}{0ex}}⇒2a+3=0\phantom{\rule{0ex}{0ex}}⇒a=-\frac{3}{2}$

Also, y = 1 is a root of the equation ${y}^{2}+y+b=0$.
$\therefore {\left(1\right)}^{2}+1+b=0\phantom{\rule{0ex}{0ex}}⇒1+1+b=0\phantom{\rule{0ex}{0ex}}⇒b+2=0\phantom{\rule{0ex}{0ex}}⇒b=-2$

$\therefore ab=\left(-\frac{3}{2}\right)×\left(-2\right)=3$

Hence, the value of ab is 3.

#### Question 41:

Very-Short-Answer Questions

If one zero of the polynomial ${x}^{2}-4x+1$ is $\left(2+\sqrt{3}\right)$, write the other zero.                    [CBSE 2010]

#### Answer:

Let the other zero of the given polynomial be $\alpha$.

Now,
Sum of the zeroes of the given polynomial = $\frac{-\left(-4\right)}{1}$= 4

Hence, the other zero of the given polynomial is $\left(2-\sqrt{3}\right)$.

#### Question 42:

Very-Short-Answer Questions

If one root of the quadratic equation $3{x}^{2}-10x+k=0$ is reciprocal of the other, find the value of k.                  [CBSE 2014]

#### Answer:

Let $\alpha$ and $\beta$ be the roots of the equation $3{x}^{2}-10x+k=0$.
(Given)

Hence, the value of k is 3.

#### Question 43:

Very-Short-Answer Questions

If the roots of the quadratic equation $px\left(x-2\right)+6=0$ are equal, find the value of p.                         [CBSE 2013]

#### Answer:

It is given that the roots of the quadratic equation $p{x}^{2}-2px+6=0$ are equal.
$\therefore D=0\phantom{\rule{0ex}{0ex}}⇒{\left(-2p\right)}^{2}-4×p×6=0\phantom{\rule{0ex}{0ex}}⇒4{p}^{2}-24p=0\phantom{\rule{0ex}{0ex}}⇒4p\left(p-6\right)=0$

For p = 0, we get 6 = 0, which is not true.

p ≠ 0

Hence, the value of p is 6.

#### Question 44:

Very-Short-Answer Questions

Find the values of k so that the quadratic equation ${x}^{2}-4kx+k=0$ has equal roots.                         [CBSE 2012]

#### Answer:

It is given that the quadratic equation ${x}^{2}-4kx+k=0$ has equal roots.

Hence, 0 and $\frac{1}{4}$ are the required values of k.

#### Question 45:

Very-Short-Answer Questions

Find the values of k for which the quadratic equation $9{x}^{2}-3kx+k=0$ has equal roots.                         [CBSE 2014]

#### Answer:

It is given that the quadratic equation $9{x}^{2}-3kx+k=0$ has equal roots.

Hence, 0 and 4 are the required values of k.

#### Question 46:

Short-Answer Questions

Solve: ${x}^{2}-\left(\sqrt{3}+1\right)x+\sqrt{3}=0$                             [CBSE 2015]

#### Answer:

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

Hence, 1 and $\sqrt{3}$ are the roots of the given equation.

#### Question 47:

Short-Answer Questions

Solve: $2{x}^{2}+ax-{a}^{2}=0$                             [CBSE 2014]

#### Answer:

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

Hence, −a and $\frac{a}{2}$ are the roots of the given equation.

#### Question 48:

Short-Answer Questions

Solve: $3{x}^{2}+5\sqrt{5}x-10=0$                             [CBSE 2014]

#### Answer:

$3{x}^{2}+5\sqrt{5}x-10=0\phantom{\rule{0ex}{0ex}}⇒3{x}^{2}+6\sqrt{5}x-\sqrt{5}x-10=0\phantom{\rule{0ex}{0ex}}⇒3x\left(x+2\sqrt{5}\right)-\sqrt{5}\left(x+2\sqrt{5}\right)=0\phantom{\rule{0ex}{0ex}}⇒\left(x+2\sqrt{5}\right)\left(3x-\sqrt{5}\right)=0$

Hence, $-2\sqrt{5}$ and $\frac{\sqrt{5}}{3}$ are the roots of the given equation.

#### Question 49:

Short-Answer Questions

Solve: $\sqrt{3}{x}^{2}+10x-8\sqrt{3}=0$                             [CBSE 2014]

#### Answer:

$\sqrt{3}{x}^{2}+10x-8\sqrt{3}=0\phantom{\rule{0ex}{0ex}}⇒\sqrt{3}{x}^{2}+12x-2x-8\sqrt{3}=0\phantom{\rule{0ex}{0ex}}⇒\sqrt{3}x\left(x+4\sqrt{3}\right)-2\left(x+4\sqrt{3}\right)=0\phantom{\rule{0ex}{0ex}}⇒\left(x+4\sqrt{3}\right)\left(\sqrt{3}x-2\right)=0$