# Quadratic Equations : Previous Year's Questions

One Mark Questions

Q.1       Is$x =- 2$a solution of the equation ${{\bf{x}}^{\bf{2}}}-{\rm{ }}{\bf{2x}}{\rm{ }} + {\rm{ }}{\bf{8}}{\rm{ }} = {\rm{ }}{\bf{0}}$

[AI 2008]

Sol.

We have  ${x^2} - 2x + 8 = 0$
When$x{\rm{ }} = {\rm{ }} - 2$
$LHS = \left( { - {2^2}} \right) - 2\left( { - 2} \right) + 8$
$\Rightarrow LHS = 4 + 4 + 8 = 16$
$RHS = 0$
$LHS \ne RHS$
Therefore,$x{\rm{ }} = {\rm{ }} - {\rm{ }}2$ is not a solution of the given equation.

Q.2      Is ${\bf{x}}{\rm{ }} = {\rm{ }}-{\bf{4}}$ a solution of the equation ${\bf{2}}{{\bf{x}}^{\bf{2}}} + {\rm{ }}{\bf{5x}}-{\bf{12}} = {\rm{ }}{\bf{0}}$

[AI 2008]

Sol.

We have  $2{x^2} + 5x - 12 = 0$
When$x{\rm{ }} = {\rm{ }} - 4$
$LHS = 2{\left( { - 4} \right)^2} - 5\left( { - 4} \right) - 12$
$\Rightarrow LHS = 32 - 20 - 12$
$\Rightarrow LHS = 32 - 32 = 0$
$LHS{\rm{ }} = {\rm{ }}RHS{\rm{ }} = {\rm{ }}0$
Therefore,$x{\rm{ }} = {\rm{ }} - 4$ is a solution of the given equation.

Q.3 Show that${\bf{x}}{\rm{ }} = {\rm{ }} - {\bf{3}}$ is a solution of equation${{\bf{x}}^{\bf{2}}} + {\rm{ }}{\bf{6x}}{\rm{ }} + {\rm{ }}{\bf{9}}{\rm{ }} = {\rm{ }}{\bf{0}}$

[Foreign 2008]

Sol. We have  ${x^2} + 6x + 9 = 0$
When $x{\rm{ }} = {\rm{ }} - 3$
$LHS = {( - 3)^2} + 6x( - 3) + 9$
$= {\rm{ }}9{\rm{ }} + {\rm{ }}9{\rm{ }}-{\rm{ }}18$
$= {\rm{ }}0$
$LHS = RHS = 0$
Therefore ,$x{\rm{ }} = {\rm{- }}3$ is a solution of the given equation.

Q.4 For what value of k are the roots of the quadratic equation $3{x^2} + 2kx + 27 = 0$ real and equal.

[Delhi 2008]

Sol. We have  $3{x^2} + 2kx + 27 = 0$
Here $a = 3,{\rm{ }}b{\rm{ }} = {\rm{ }}2k,{\rm{ }}c{\rm{ }} = {\rm{ }}27$
Since $D{\rm{ }} = {\rm{ }}{b^2}-{\rm{ }}4ac$
$\Rightarrow D = {\left( {2k} \right)^2} - 4\left( 3 \right)\left( {27} \right)$
$\Rightarrow D = 4{k^2} - 324$
For real and equal roots, D = 0
$\Rightarrow 4{k^2} - 324 = 0$
$\Rightarrow 4{k^2} = 324$
$\Rightarrow {k^2} = {{324} \over 4} = 81$
$\Rightarrow k =\pm 9$
So, at $k =\pm 9$ the equation has real and equal roots.

Q.5 For what value of k are roots of the quadratic equation${\bf{k}}{{\bf{x}}^{\bf{2}}} + {\rm{ }}{\bf{4x}}{\rm{ }} + {\rm{ }}{\bf{1}}{\rm{ }} = {\rm{ }}{\bf{0}}$ equals and reals.

[AI 2008C]

Sol. We have $k{x^2} + {\rm{ }}4x{\rm{ }} + {\rm{ }}1{\rm{ }} = {\rm{ }}0$
Here$a{\rm{ }} = {\rm{ }}k{\rm{ }},{\rm{ }}b{\rm{ }} = {\rm{ }}4{\rm{ }},{\rm{ }}c{\rm{ }} = {\rm{ }}1$
Since $D = {b^2} - 4ac$
$\Rightarrow D = {(4)^2} - 4 \times \left( k \right)\left( 1 \right)$
$\Rightarrow D = 16 - 4k$
For real and equal roots, D = 0
$\Rightarrow 16 - 4k = 0$
$\Rightarrow 4k = 16$
$\Rightarrow k = 4$
So, at$k{\rm{ }} = {\rm{ }}4$ equation has real and equal roots.

Q.6 For what value of k does$\left( {k - 12} \right){x^2} + 2\left( {k - 12} \right)x + 2 = 0$ have equal roots.
[AI 2008C]

Sol. Here  $a{\rm{ }} = {\rm{ }}k{\rm{ }}-{\rm{ }}12,{\rm{ }}b{\rm{ }} = {\rm{ }}2{\rm{ }}\left( {k - 12} \right),{\rm{ }}c{\rm{ }} = {\rm{ }}2$
Since $D{\rm{ }} = {\rm{ }}{b^2}-{\rm{ }}4ac$
$\Rightarrow D = {\left[ {2\left( {k - 12} \right)} \right]^2} - 4 \times 2 \times \left( {k - 12} \right)$
$\Rightarrow D = 4{\left( {k - 12} \right)^2} - 8\left( {k - 12} \right)$
$\Rightarrow D = 4\left( {k - 12} \right)[k - 12 - 2]$
$\Rightarrow D = 4\left( {k - 12} \right)\left( {k - 14} \right)$
Now for equal and real roots D = 0
Then $D = 4\left( {k - 12} \right)\left( {k - 14} \right) = 0$
$\Rightarrow k = 12\,\,or\,\,k = 14$
Because  for $k = 12$ the given equation $\left( {k - 12} \right){x^2} + 2\left( {k + 2} \right)x + 2 = 0$ becomes a linear equation. So,$k{\rm{ }} = {\rm{ }}12$ is rejected.
Therefore, only$k{\rm{ }} = {\rm{ }}14$is solution for the equation having real and equal roots.

Q.7 Find the discriminant of the quadratic equation $13\sqrt3{x^2}+10x+\sqrt 3= 0$

[AI 2009]

Sol. We have  $13\sqrt 3 {x^2}+ 10x +\sqrt 3=0$
Here  $a{\rm{ = 13}}\sqrt {\rm{3}} ,b = 10,{\rm{c = }}\sqrt {\rm{3}}$
Since$D{\rm{ }} = {\rm{ }}{b^2}-{\rm{ }}4{\rm{ }}ac$
$\Rightarrow D = \left( {10} \right)2 - 4 \times \left( {13\sqrt 3 } \right) \times \left( {\sqrt 3 } \right)$
$\Rightarrow D = 100 - 4 \times 39 \times 8$
$\Rightarrow D = 100 - 156 = - 56$
So, the discriminant of quadratic equation is$-{\rm{ }}56$

Q.8. Write the nature of roots of quadratic equation $4{x^2} + 4\sqrt 3 x + 3 = 0$

[Foreign 2009]

Sol. We have $4{x^2} + 4\sqrt 3 x + 3 = 0$
Here $a{\rm{ }} = {\rm{ }}4$, b =$4\sqrt 3$, ${\rm{ }}c{\rm{ }} = {\rm{ }}3$
Since$D{\rm{ }} = {\rm{ }}{b^2}-{\rm{ }}4ac$
$\Rightarrow D = {\left( {4\sqrt 3 } \right)^2} - 4 \times 4 \times 3$
$\Rightarrow D = 16 \times 3 - 16 \times 3$
$\Rightarrow D = 48 - 48 = 0$
As $D{\rm{ }} = {\rm{ }}0$, the equation has real and equal roots.

Two Marks Questions

Q.1 Solve for $x : {x^2} - 2\left( {{a^2} + {b^2}} \right)x + {\left( {{a^2} - {b^2}} \right)^2} = 0$

[Foreign 2005; Delhi 2006 C]

Sol. We have  ${x^2} - 2\left( {{a^2} + {b^2}} \right)x + {\left( {{a^2} - {b^2}} \right)^2} = 0$
Here$A{\rm{ }} = {\rm{ }}1$, $B = -2\left({{a^2} + {b^2}} \right)$, $C = {\left( {{a^2} - {b^2}} \right)^2}$
Since $D = B{}^2 - 4AC$
$\Rightarrow D = {\left[ { - 2\left( {{a^2} + {b^2}} \right)} \right]^2} - 4 \times 1 \times {\left[ {\left( {{a^2} - {b^2}} \right)} \right]^2}$
$\Rightarrow D = 4\left[ {{{\left( {{a^2} + {b^2}} \right)}^2} - {{\left( {{a^2} - {b^2}} \right)}^2}} \right]$
$\Rightarrow D = 4\left[ {\left( {{a^2} + {b^2} + {a^2} - {b^2}} \right)\left( {{a^2} + {b^2} - a + {b^2}} \right)} \right]$
[Because ${a^2} - {b^2} = \left( {a + b} \right)\left( {a - b} \right)$]
$\Rightarrow D = 4\left( {2{b^2}} \right)\left( {2{a^2}} \right)$
$\Rightarrow D = 16{a^2}{b^2}$
Now $x = {{ - B \pm \sqrt D } \over {2A}} = {{2\left( { - 2} \right)\left( {{a^2} + {b^2}} \right) \pm \sqrt {16{a^2}{b^2}} } \over 2}$
$\Rightarrow x = {{2\left( {{a^2} + {b^2}} \right) \pm 4ab} \over 2}$
$\Rightarrow x = {a^2} + {b^2} \pm 2ab = {\left( {a + b} \right)^2},{\left( {a - b} \right)^2}$
Then values of x are ${\left( {a + b} \right)^2}$ or ${\left( {a - b} \right)^2}$

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

[Delhi 2006]

Sol. We have  $12ab{x^2} - \left( {9{a^2} - 8{b^2}} \right)x - 6ab = 0$
$\Rightarrow 12ab{x^2} - 9{a^2}x + 8{b^2}x - 6ab = 0$
$\Rightarrow 3ax\left( {4bx - 3a} \right) + 2b\left( {4bx - 39} \right) = 0$
$\Rightarrow \left( {4bx - 3a} \right)\left( {3ax + 2b} \right) = 0$

Q.3 A two digit number is such that the product of its digits is 35. When 18 is added to the number, the digits interchange their places. Find the number.

[Delhi 2006]

Sol.Let digit at ten’s place $= {\rm{ }}x$
Digit at unit’s place $= {\rm{ }}y$
Therefore, Number$= {\rm{ }}10x{\rm{ }} + {\rm{ }}y$
Now  $xy{\rm{ }} = {\rm{ }}35$,$y = {{35} \over x}$………..(1)
A.T.Q.
$10x + y + 18 = 10y + x$
$\Rightarrow 10x + y + 18 - 10y - x = 0$
$\Rightarrow ax - 9y + 18 = 0$
$\Rightarrow x - y + 2 = 0$
$\Rightarrow x - {{35} \over x} + 2 = 0$ [using (1)]
$\Rightarrow {{{x^2} - 35 + 2x} \over 2} = 0$
$\Rightarrow {x^2} + 2x - 35 = 0$
$\Rightarrow {x^2} + 7x - 5x - 35 = 0$
$\Rightarrow x\left( {x + 7} \right) - 5\left( {x + 7} \right) = 0$
$\Rightarrow \left( {x - 5} \right)\left( {x + 7} \right) = 0$
$\Rightarrow x = - 7,x = 5$
Because$x{\rm{ }} = {\rm{ }}-{\rm{ }}7$, so it can be rejected
Putting $x = {\rm{ }}5$ in eq(1) , we get
$y = {{35} \over 5} = 7$
Therefore, number$= {\rm{ }}50{\rm{ }} + {\rm{ }}7{\rm{ }} = {\rm{ }}57$

Q.4 Using the quadratic formula; solve the equation ${a^2}{b^2}{x^2} - \left( {4{b^4} - 3{a^4}} \right)x - 12{a^2}{b^2} = 0$

[AI 2006]

Sol. We have  ${a^2}{b^2}{x^2} - \left( {4{b^4} - 3{a^4}} \right)x - 12{a^2}{b^2} = 0$
Here$A{\rm{ }} = {\rm{ }}{a^2}{b^2},{\rm{ }}B{\rm{ }} = {\rm{ }}-(4{b^4} - 3{a^4}),{\rm{ }}C{\rm{ }} = {\rm{ }}-12{a^2}{b^2}$
Since $D = {B^2} - 4AC$
$\Rightarrow D = \left[ { - \left( {4{b^4} - 3{a^4}} \right) - 4\left( {{a^2}{b^2}} \right) \times \left( { - 12{a^2}{b^2}} \right)} \right]$
$\Rightarrow D = {\left( {4{b^4} - 3{b^4}} \right)^2} + 48{a^2}{b^2}$
$\Rightarrow D = 16{b^8} + 9{a^8} - 24{a^4}{b^4} + 48{a^4}{b^4}$
$\Rightarrow D = 16{b^8} + 9{a^8} + 24{a^4}{b^4}$
$\Rightarrow D = {\left( {4{b^4} + 3{a^4}} \right)^2}$
Therefore, $x = {{ - B \pm \sqrt D } \over {2A}} = {{ - \left[ { - \left( {4{b^4} - 3{a^4}} \right)} \right] \pm \sqrt {{{\left( {4{b^4} + 3{a^4}} \right)}^2}} } \over {2{a^2}{b^2}}}$
$\Rightarrow x = {{4{b^4} - 3{a^4} \pm \left( {4{b^4} + 3{a^4}} \right)} \over {2{a^2}{b^2}}}$
$\Rightarrow x = {{8{b^2}} \over {2{a^2}{b^2}}},{{ - 6{a^2}} \over {2{a^2}{b^2}}}$
$\Rightarrow x = {{4{b^2}} \over {{a^2}}},{{ - 3{a^2}} \over {{b^2}}}$

Q.5 The sum of two natural numbers is 8. Determine the numbers if sum of their reciprocals is 8/15

[AI 2006]

Sol.Let one number$= {\rm{ }}x$
Therefore, other natural number$= {\rm{ }}y$
A.T.Q.
$x{\rm{ }} + {\rm{ }}y{\rm{ }} = {\rm{ }}8$ …………..(1)
So  $y{\rm{ }} = {\rm{ }}8{\rm{ }}-x$ (By (1))
A.T.Q.
${1 \over x} + {1 \over {8 - x}} = {8 \over {15}}$
$\Rightarrow {{8 - x + x} \over {x\left( {8 - x} \right)}} = {8 \over {15}}$
$\Rightarrow {8 \over {x\left( {8 - x} \right)}} = {8 \over {15}}$
$\Rightarrow x\left( {8 - x} \right) = 15$
$\Rightarrow 8x - {x^2} = 15$
$\Rightarrow {x^2} - 8x + 15 = 0$
$\Rightarrow x = 3,x = 5$
When $x{\rm{ }} = {\rm{ }}3$, other number$= {\rm{ }}8{\rm{ }}-{\rm{ }}3{\rm{ }} = {\rm{ }}5$
When$x = 5$, other number$= {\rm{ }}8{\rm{ }}-{\rm{ }}5{\rm{ }} = {\rm{ }}3$
Therefore, Numbers are 3 and 5

Q.6 Solve for $x :{\left( {a + b} \right)^2}{x^2} + 8\left( {{a^2} - {b^2}} \right)x + 16{\left( {a - b} \right)^2} = 0$

[Foreign 2006]

Sol. ${\left( {a + b} \right)^2}{x^2} + 8\left( {{a^2} - {b^2}} \right)x + 16{\left( {a - b} \right)^2} = 0$
A = ${\left( {a + b} \right)^2}$, B = $8\left( {{a^2} - {b^2}} \right)$, C =$16{\left( {a - b} \right)^2}$
Since  $D = {B^2} - 4AC$
$\Rightarrow D = {\left[ {8\left( {{a^2} - {b^2}} \right)} \right]^2} - 4 \times {\left( {a + b} \right)^2}16{\left( {a - b} \right)^2}$
$\Rightarrow D = 64{\left( {{a^2} - {b^2}} \right)^2} - 16{\left( {{a^2} - {b^2}} \right)^2} = 0$
Therefore, $x = {{ - B} \over {2A}} = {{ - 8\left( {{a^2} - {b^2}} \right)} \over {2{{\left( {a + b} \right)}^2}}} = {{ - 4\left[ {\left( {a - b} \right)} \right]} \over {a + b}}$
$\Rightarrow x = {{ - 4\left( {a - b} \right)} \over {a + b}}$

Q.7 Two numbers differ by 3 and their product is 504. Find the numbers.

[Foreign 2006]

Sol. Let one number = x
Therefore, other number = x + 3
A.T.Q.  $x\left( {x + 3} \right) = 504$
$\Rightarrow {x^2} + 3x - 504 = 0$
$\Rightarrow {x^2} + 24x - 21x - 504 = 0$
$\Rightarrow x\left( {x + 24} \right) - 21\left( {x + 24} \right) = 0$
$\Rightarrow \left( {x + 24} \right)\left( {x - 21} \right) = 0$
$\Rightarrow x =- 24,x = 21$
When$x{\rm{ }} ={\rm{ }}21$, other number$= {\rm{ }}21{\rm{ }} + {\rm{ }}3{\rm{ }} = {\rm{ }}24$
When$x{\rm{ }} =-{\rm{ }}24$, other number$=- 24{\rm{ }} + {\rm{ }}3{\rm{ }} = {\rm{ }} - {\rm{ }}21$
Therefore, Numbers are 21, 24 or – 24 , – 21

Q.8 Rewrite the following as a quadratic equation in x and then solve for x. ${4 \over x} - 3 = {5 \over {2x + 3}},x \ne 0,{3 \over 2}$

[AI 2006 C]

Sol. We have  ${4 \over x} - 3 = {5 \over {2x + 3}}$
$\Rightarrow {{4 - 3x} \over x} = {5 \over {2x + 3}}$
$\Rightarrow \left( {4 - 3x} \right)\left( {2x + 3} \right) = 5x$
$\Rightarrow 8x + 12 - 6{x^2} - 9x = 5x$
$\Rightarrow 6{x^2} + 6x - 12 = 0$
$\Rightarrow {x^2} + x - 2 = 0$
$\Rightarrow \left( {x + 2} \right)\left( {x - 1} \right) = 0$
$\Rightarrow x = - 2\,,\,x = 1$

Q.9 Find the value of p so that the quadratic equation $px\left( {x - 3} \right) + 9 = 0$ has two equal roots.

[Delhi 2011]

Sol. We have  $px\left( {x - 3} \right) + 9 = 0$
$\Rightarrow p{x^2} - 3px + 9 = 0$
Here ${\rm{ }}a{\rm{ }} = {\rm{ }}p,{\rm{ }}b{\rm{ }} = {\rm{ }} - 3p,{\rm{ }}c = {\rm{ }}9$
For equal roots$D{\rm{ }} = {\rm{ }}0$
Since $D{\rm{ }} = {\rm{ }}{b^2} - {\rm{ }}4ac$
$\Rightarrow D = {\left( { - 3p} \right)^2} - 4 \times p \times 9$
$\Rightarrow D = 9{p^2} - 36p = 0$
$\Rightarrow D = 9p\left( {p - 4} \right) = 0$
$\Rightarrow 9p = 0\,\,\,,\,\,\,p - 4 = 0$
$\Rightarrow p = 0\,\,,\,\,p = 4$
But $p \ne 0$ [Because for p=0 quadratic equation becomes  linear equation]
So,$p{\rm{ }} = {\rm{ }}4$ is the root of the equation.

Q.10 Find the value of m so that the quadratic equation $mx\left( {x - 7} \right) + 49 = 0$ has two equal roots.

[AI 2011]

Sol. We have  $mx\left( {x - 7} \right) + 49 = 0$
$\Rightarrow m{x^2} - 7mx + 49 = 0$
Here $a{\rm{ }} = {\rm{ }}m,{\rm{ }}b{\rm{ }} = {\rm{ }}7m,{\rm{ }}c{\rm{ }} = {\rm{ }}49$
For equal roots D = 0
Since $D{\rm{ }} = {\rm{ }}{b^2}-{\rm{ }}4ac$
$\Rightarrow {\left( { - 7m} \right)^2} - 4 \times m \times 49 = 0$
$\Rightarrow 49{m^2} - 196m = 0$
$\Rightarrow 49m = 0\,\,,\,\,m - 4 = 0$
$\Rightarrow m = 0\,\,or\,\,\,m = 4$
But $m \ne 0$ [Because for m = 0  quadratic equation becomes  linear equation]
So,$m{\rm{ }} = {\rm{ }}4$ is the root of the equation.

Q.11 For what value of k does the quadratic equation $\left( {k - 5} \right){x^2} + 2\left( {k - 5} \right)x + 2 = 0$ has two equal roots.

[AI 2011]

Sol. We have  $\left( {k - 5} \right){x^2} + 2\left( {k - 5} \right)x + 2 = 0$
Here$a{\rm{ }} = {\rm{ }}k{\rm{ }}-{\rm{ }}5,{\rm{ }}b{\rm{ }} = {\rm{ }}2{\rm{ }}\left( {k{\rm{ }}-{\rm{ }}5} \right){\rm{ }},{\rm{ }}c{\rm{ }} = {\rm{ }}2$
For equal roots $D{\rm{ }} = {\rm{ }}0$
$\Rightarrow {b^2} - 4ac = 0$
$\Rightarrow {\left( 2 \right)^2}{\left[ {\left( {k - 5} \right)} \right]^2} - 4 \times \left( {k - 5} \right) \times 2 = 0$
$\Rightarrow 4{k^2} + 100 - 40k - 8k + 40 = 0$
$\Rightarrow {k^2} - 12k + 35 = 0$
$\Rightarrow {k^2} - 7k - 5k + 35 = 0$
$\Rightarrow k\left( {k - 7} \right) - 5\left( {k - 7} \right) = 0$
$\Rightarrow \left( {k - 5} \right)\left( {k - 7} \right) = 0$
$\Rightarrow k = 5\,\,,\,\,7$
But $k \ne 5$ [Because for  k = 5  quadratic equation  becomes linear equation]
So$k{\rm{ }} = {\rm{ }}7$ is the root of the equation.

Q.12 Find the value of k for which the roots of the quadratic equation $\left( {k - 4} \right){x^2} + 2\left( {k - 4} \right)x + 2 = 0$ are equal.

[Delhi 2013]

Sol. We have   $\left( {k - 4} \right){x^2} + 2\left( {k - 4} \right)x + 2 = 0$
Here  $a{\rm{ }} = {\rm{ }}k{\rm{ }}-{\rm{ }}4{\rm{ }},{\rm{ }}b{\rm{ }} = {\rm{ }}2\left( {k{\rm{ }}-{\rm{ }}4} \right),{\rm{ }}c{\rm{ }} = {\rm{ }}2$
$D = {\rm{ }}0$ for real and equal roots
$\Rightarrow {b^2} - 4ac = 0$
$\Rightarrow {\left[ {2\left( {k - 4} \right)} \right]^2} - 4x\left( {k - 4} \right) \times 2 = 0$
$\Rightarrow 4{\left( {k - 4} \right)^2} - 8\left( {k - 4} \right) = 0$
$\Rightarrow k = 4\,\,\,or\,\,\,k = 6$
But $k \ne 4$ [Because for  k = 4  quadratic equation becomes linear equation]
So, ${\rm{ }}k{\rm{ }} = {\rm{ }}6$ is the root of equation.

Q.13 Solve for $x\,:\,4\sqrt 3 {x^2} + 5x - 2\sqrt 3 = 0$

[Delhi 2013]

SolWe have  $4\sqrt 3 {x^2} + 5x - 2\sqrt 3 = 0$
$\Rightarrow 4\sqrt 3 {x^2} + 8x - 3x - 253 = 0$
$\Rightarrow 4x\left( {\sqrt 3 x + 2} \right) - \sqrt 3 \left( {\sqrt 3 x + 2} \right) = 0$
$\Rightarrow \left( {\sqrt 3 x + 2} \right)\left( {4x - \sqrt 3 } \right) = 0$
$\Rightarrow \sqrt 3 x + 2 = 0$, $4x - \sqrt 3 = 0$
$\Rightarrow x = {2 \over {\sqrt 3 }}$  or $x = {{\sqrt 3 } \over 4}$

Q.14 Solve for $x:\sqrt 2 {x^2} + 7x + 5\sqrt 2 = 0$

[AI 2013]

Sol. We have  $\sqrt 2 {x^2} + 7x + 5\sqrt 2 = 0$
$\Rightarrow \sqrt 2 {x^2} + 5x + 2x + 5\sqrt 2 = 0$
$\Rightarrow x\left( {\sqrt 2 x + 5} \right) + \sqrt 2 \left( {\sqrt 2 x + 5} \right) = 0$
$\Rightarrow \left( {x + \sqrt 2 } \right)\left( {\sqrt 2 x + 5} \right) = 0$
$\Rightarrow \left( {x + \sqrt 2 } \right) = 0\,\,\,;\,\,\,\sqrt 2 x + 5 = 0$
$\Rightarrow x = - \sqrt 2$ or $x = - 5/\sqrt 2$

Q.15 Solve for $x:{x^2} - \left( {\sqrt 2 + 1}\right)x + \sqrt 2 = 0$

[Foreign 2013]

Sol. We have  ${x^2} -\left( {\sqrt 2 + 1}\right)x + \sqrt 2 = 0$
$\Rightarrow {x^2} - \sqrt 2 x - x + \sqrt 2 = 0$
$\Rightarrow \left({x - 1} \right)\left( {x - \sqrt 2 }\right) = 0$
$\Rightarrow x - 1 = 0\,\,or\,\,x - \sqrt 2 = 0$
$\Rightarrow x =\sqrt 2$  or ${\rm{ }}x{\rm{ }} = {\rm{ }}1$

Three Marks Questions

Q.1 A passenger train takes 2 hours less for a journey of 300 km. It its speed is increased by 5 km/hr from its usual speed. Find the usual speed of the train.

[Delhi 2005 C, 2006]

Sol. Let usual speed of the train = x km / hr
Distance = 300 km
Time taken = ${{300} \over x}$  [Because $t = {D \over S}$]
It speed = (x + 5) km/hr
Then time taken = ${{300} \over {x + 5}}hr$
A.T.Q.
${{300} \over x} - {{300} \over {x + 5}} = 2$
$\Rightarrow {{300\left( {x + 5} \right) - 300x} \over {x\left( {x + 5} \right)}} = 2$
$\Rightarrow {{300x + 300 \times 5 - 300x} \over {{x^2} + 5x}} = 2$
$\Rightarrow 1500 = 2\left( {{x^2} + 5x} \right)$
$\Rightarrow {x^2} + 5x - 750 = 0$
$\Rightarrow {x^2} + 30x - 25x - 750 = 0$
$\Rightarrow x\left( {x + 30} \right) - 25\left( {x + 30} \right) = 0$
$\Rightarrow \left( {x - 25} \right)\left( {x + 30} \right) = 0$
$\Rightarrow x = 25, - 30$ [reject x=-30 because speed cannot negative)
Therefore, usual speed = 25 km / hr

Q.2 A speed of a boat in still water is 11 km / hr. It can go 12 km upstream and return downstream to the original point in 2 hr 45 minutes. Find the speed of the stream.

[AI 2006]

Sol. Let speed of the stream = x km / hr
Speed of the boat in still water = 11 km / hr
Therefore, up stream speed = (1 – x) km / hr
and downstream speed =(11 + x) km / hr
Distance = 12 km
Time taken for downstream direction = ${{12} \over {11 + x}}hr$
Time taken for upstream direction = ${{12} \over {11 - x}}hr$
A.T.Q.
${{12} \over {11 + x}} + {{12} \over {11 - x}} =$ 2 hr 45 min
$\Rightarrow {{12} \over {11 + x}} + {{12} \over {11 - x}} = 2{{45} \over {60}}hr$
$\Rightarrow {{12\left( {11 - x} \right) + 12\left( {11 + x} \right)} \over {\left( {11 + x} \right)\left( {11 - x} \right)}} = {{11} \over 4}hr$
$\Rightarrow {{132 - 12x + 132 + 12x} \over {121 - {x^2}}} = {{11} \over 4}$
$\Rightarrow {{132 + 132} \over {121 - {x^2}}} = {{11} \over 4}$
$\Rightarrow 4 \times 264 = 11\left( {121 - {x^2}} \right)$
$\Rightarrow {{4 \times 264} \over {11}} = 121 - {x^2}$
$\Rightarrow 96 = 121 - {x^2}$
$\Rightarrow {x^2} = 25$
$\Rightarrow x =\pm 5$[Reject x=-5 because speed cannot be negative]
So x = 5 km/hr is the speed of the stream.

Q.3 A fast train takes 3 hours less than a slow train for a journey of 600 km. If the speed of the slow train is 10 km / hr less than that of the fast train. Find the speeds of the two trains..

[Foreign 2006]

Sol. Let speed of fast train = x km / hr
$\Rightarrow$ Speed of slow train = $\left( {x - 10} \right)km/hr$
A.T.Q.
${{600} \over {x - 10}} - {{600} \over x} = 3$
${{600x - 600\left( {x - 10} \right)} \over {x\left( {x - 10} \right)}} = 3$
$\Rightarrow {{600x - 600x + 6000} \over {{x^2} - 10x}} = 3$
$\Rightarrow {{6000} \over {{x^2} - 10x}} = 3$
$\Rightarrow 2000 = {x^2} - 10x$
$\Rightarrow {x^2} - 10x - 2000 = 0$
$\Rightarrow {x^2} - 50x + 40x - 2000 = 0$
$\Rightarrow \left( {x + 40} \right)\left( {x - 50} \right) = 0$
$\Rightarrow x = 50, - 40$  [rejected]
So, speed so of the that train = 50 km / hr
And speed of the slow train = 40 km / hr.

Q.4 Seven years ago Varun’s age was five times the square of Swati’s age. three years hence Swati’s age will be two-fifth of Varun’s age. Find their present ages.

[Delhi 2006 C]

Sol. Let Varun’s present age = x years.
And Swati’s age = y years
Case 1 : Seven years ago
Varun’s age was (x – 7) years and
A.T.Q.
$\left( {x - 7} \right) = 5{\left( {y - 7} \right)^2}$
$\Rightarrow x = 5{\left( {y - 7} \right)^2} + 7$ ………..(1)
Case II : Three years hence
Varun’s age will be (x + 3) years and
Swati’s age will be (y + 3) years
A.T.Q.
$y + 3 = {2 \over 5}\left( {x + 3} \right)$
$\Rightarrow y + 3 = {2 \over 5}\left[ {5{{\left( {y - 7} \right)}^2} + 7 + 3} \right]$
$\Rightarrow y + 3 = {2 \over 5} \times 5{\left( {y - 7} \right)^2} + {2 \over 5} \times 10$
$\Rightarrow y + 3 = 2{\left( {y - 7} \right)^2} + 4$
$\Rightarrow y + 3 = 2\left( {{y^2} + 49 - 14y} \right) + 1$
$\Rightarrow y = 2{y^2} + 98 - 28y + 1$
$\Rightarrow 2{y^2} - 2ay + 99 = 0$
$\Rightarrow 2{y^2} - 18y - 11y + 99 = 0$
$\Rightarrow 2y\left( {y - 9} \right) - 11\left( {y - 9} \right) = 0$
$\Rightarrow \left( {2y - 11} \right)\left( {y - 9} \right) = 0$
$\Rightarrow y = 9$, $y = {{11} \over 2}$
y = 9   [Rejecting  y=11/2 because age cannot be in fraction]
Putting y=9 in (1),We get
$x = 5{\left( {9 - 7} \right)^2} + 7$
$\Rightarrow x{\rm{ }} = {\rm{ }}27$
Therefore, Present age of Swati = 9 years and Varun = 27 years.

Q.5 A 2-digit number is such that product of its digits is 18. When 63 is subtracted from the number, the digits interchange their places. Find the number.

[AI 2006 C]

Sol. Let digit at unit’s place = x
And digit at ten’s place = y
Therefore, number = 10y + x.
A.T.Q.
$xy = 18 \Rightarrow y = {{18} \over x}.........(1)$
And $10y + x - 63 = 10x + y$
$\Rightarrow 10y + x - 63 - 10x - y = 0$
$\Rightarrow 9y - 9x - 63 = 0$
$\Rightarrow y - x - 7 = 0$
$\Rightarrow {{18} \over x} - x - 7 = 0$  [at using (1)]
$\Rightarrow {{18 - {x^2} - 7x} \over x} = 0$
$\Rightarrow {x^2} + 7x - 18 = 0$
$\Rightarrow {x^2} + 9x - 2x - 18 = 0$
$\Rightarrow x\left( {x + 9} \right) - \left( {x + 9} \right) = 0$
$\Rightarrow \left( {x - 2} \right)\left( {x + 9} \right) = 0$
$\Rightarrow x = 2,\, - 9$[Reject x=-9]
When x = 2, $y = {{18} \over 2} = 9$
Therefore, number =92

Q.6 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.

[AI 2006 C]

Sol.Let uniform speed of the train = x km/hr
Distance = 90 km
Therefore, Time taken = ${{90} \over x}$  hr
If speed was (x + 15) km / hr
Then time taken = ${{90} \over {x + 15}}hr$
A.T.Q.
${{90} \over x} - {{90} \over {x + 15}} = {1 \over 2}$
$\Rightarrow {{90\left( {x + 15} \right) - 90x} \over {x\left( {x + 15} \right)}} = {1 \over 2}$
$\Rightarrow {{90x + 90 \times 15 - 90x} \over {{x^2} + 15x}} = {1 \over 2}$
$\Rightarrow 90 \times 15 \times 2 = {x^2} + 15x$
$\Rightarrow {x^2} + 15x - 2700 = 0$
$\Rightarrow {x^2} + 60x - 45x - 2700 = 0$
$\Rightarrow x\left( {x + 60} \right) - 45\left( {x + 60} \right) = 0$
$\Rightarrow \left( {x - 45} \right)\left( {x + 60} \right)$
$\Rightarrow x = 45,\,or\,x =- 60$[Rejecting x=-60 because speed cannot be negative]
Therefore, x = 45 km / hr
Hence uniform speed of the train = 45 km/hr

Q.7 The difference of two numbers is 5 and the difference of their reciprocals is ${1 \over {10}}$ Find the numbers.

[Delhi 2007]

Sol. Let one number is x
Therefore, other number = x + 5
A.T.Q.
${1 \over x} - {1 \over {x + 5}} = {1 \over {10}}$
$\Rightarrow {{x + 5 - x} \over {x\left( {x + 5} \right)}} = {1 \over {10}}$
$\Rightarrow 50 = {x^2} + 5x$
$\Rightarrow {x^2} + 5x - 50 = 0$
$\Rightarrow {x^2} + 10x - 5x - 50 = 0$
$\Rightarrow x\left( {x + 10} \right) - 5\left( {x + 10} \right) = 0$
$\Rightarrow \left( {x - 5} \right)\left( {x + 10} \right) = 0$
$\Rightarrow x = - 10,x = 5$
Hence, numbers are -10, -5 or 5, 10

Q.8 By increasing the list price of a book by Rs. 10. A person can buy 10 less books for Rs. 1200. Find the original list price of the book.

[Delhi 2007]

Sol. Let list price of the book = Rs. X
Total cost = Rs. 1200
Therefore, number of books = ${{1200} \over x}$
If list price of the book = Rs. (x+10)
Then number of books = ${{1200} \over {x + 10}}$
A.T.Q.
${{1200} \over x} - {{1200} \over {x + 10}} = 10$
$\Rightarrow {{1200\left( {x + 10} \right) - 1200x} \over {x\left( {x + 10} \right)}} = 10$
$\Rightarrow {{1200x + 12000 - 1200x} \over {x\left( {x + 10} \right)}} = 10$
$\Rightarrow {{12000} \over {x\left( {x + 10} \right)}} = 10$
$\Rightarrow 1200 = {x^2} + 10x$
$\Rightarrow {x^2} + 10x - 1200 = 0$
$\Rightarrow {x^2} + 40x - 30x - 1200 = 0$
$\Rightarrow x\left( {x + 40} \right) - 30\left( {x + 40} \right) = 0$
$\Rightarrow \left( {x - 30} \right)\left( {x + 40} \right) = 0$
$\Rightarrow x = 30,or\,x =- 40$  [Because price cannot be negative so rejected]
Therefore, list price of the book = Rs. 30

Q.9 The numerator of a fraction is one less than its denominator. If three is added to each of the numerator and denominator, the fraction is increased by ${3 \over {28}}$.  Find the fraction.

[AI 2007]

Sol. Let Denominator = x
Therefore, numerator = x – 1
Fraction = ${{x - 1} \over x}$
A.T.Q.
${{x - 1 + 3} \over {x + 3}} = {{x - 1} \over x} + {3 \over {28}}$
$\Rightarrow {{x + 2} \over {x + 3}} - {{x - 1} \over x} = {3 \over {28}}$
$\Rightarrow {{x\left( {x + 2} \right) - \left( {x - 1} \right)\left( {x + 3} \right)} \over {\left( {x + 3} \right)x}} = {3 \over {28}}$
$\Rightarrow {{{x^2} + 2x - \left( {{x^2} + 3x - x - 3} \right)} \over {{x^2} + 3x}} = {3 \over {28}}$
$\Rightarrow {{{x^2} + 2x - {x^2} - 2x + 3} \over {{x^2} + 3x}} = {3 \over {28}}$
$\Rightarrow {x^2} + 3x - 28 = 0$
$\Rightarrow {x^2} + 7x - 4x - 28 = 0$
$\Rightarrow x\left( {x + 7} \right) - 4\left( {x + 7} \right) = 0$
$\Rightarrow x = - 7$ or $x{\rm{ }} = {\rm{ }}4$  [Rejecting x = -7]
$\Rightarrow x{\rm{ }} = {\rm{ }}4$
Fraction is ${{4 - 1} \over 4}$ $= {3 \over 4}$

Q.10 The difference of squares of two natural number is 45. The square of the smaller number is four times the larger number. Find the numbers.

[AI 2007]

Sol.Let larger number be = x
Smaller number be = y
A.T.Q.
${x^2} - {y^2} = 45$
${y^2} = {x^2} - 45$ ………….(1)
Also ${y^2} = 4x$
$\Rightarrow {x^2} - 45 = 4x$  [using (1)]
$\Rightarrow {x^2} - 4x - 45 = 0$
$\Rightarrow {x^2} - 9x + 5x - 45 = 0$
$\Rightarrow x\left( {x - 9} \right) + 5\left( {x - 9} \right) = 0$
$\Rightarrow x = 9,x= - 5$
So larger numbers is 9.
Putting x=9 in (1),We get
${y^2} = {9^2} - 45$
$\Rightarrow y = 81 - 45 = \sqrt {36} = \pm 6$ [Rejecting y=-6]
$\Rightarrow y = 6$
So Numbers are: 9 and 6

Q.11 Find the roots of the following equation

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

[Delhi 2008]

Sol. We have ${1 \over {x + 4}} - {1 \over {x - 7}} = {{11} \over {30}}$
$\Rightarrow {{x - 7 - \left( {x + 4} \right)} \over {\left( {x + 4} \right)\left( {x - 7} \right)}} = {{11} \over {30}}$
$\Rightarrow {{ - 11} \over {x{}^2 - 3x - 28}} = {{11} \over {30}}$
$\Rightarrow {x^2} - 3x - 28 = - 30$
$\Rightarrow {x^2} - 3x + 2 = 0$
$\Rightarrow \left( {x - 2} \right)\left( {x - 1} \right) = 0$
$\Rightarrow x = 2,1$
Therefore, required roots are 2 , 1

Q.12 Find the roots of the following equation $2\sqrt 3 \,{x^2} - 5x +\sqrt 3 = 0$

[Delhi 2011]

Sol. We have   $2\sqrt 3 \,{x^2} - 5x + \sqrt 3 = 0$
Here a =$2\sqrt 3$, b = -5 , c = $\sqrt 3$
Since $D{\rm{ }} = {\rm{ }}{b^2}-{\rm{ }}4ac$
$\Rightarrow D = {\left( { - 5} \right)^2} - 4 \times 2\sqrt 3 {\mkern 1mu} {\mkern 1mu} \times\sqrt 3$
$\Rightarrow D = 25 - 24 = 1$
$x = {{ - b \pm \sqrt D } \over {2a}}$
$\Rightarrow x = {{5 \pm 1} \over {2 \times 2\sqrt 3 }}$
$\Rightarrow x = {6 \over {4\sqrt 3 }} \times {{\sqrt 3 } \over {\sqrt 3 }} \Rightarrow {{\sqrt 3 } \over 2}$
Or $x = {{5 - 1} \over {4\sqrt 3 }} = {4 \over {4\sqrt 3 }}$
$\Rightarrow x = {1 \over {\sqrt 3 }}$
So required roots are ${{\sqrt 3 } \over 2}$  or $x = {1 \over {\sqrt 3 }}$

Q.13 Solve the following quadratic equation for $x:{x^2} - 4ax - {b^2} + 4{a^2} = 0$

[AI 2012]

Sol. We have ${x^2} - 4ax - {b^2} + 4{a^2} = 0$
Here A = 1, B = - 4a , C =$- {b^2} + 4{a^2}$
Since D = ${B^2} - 4AC$
$\Rightarrow D = {\left( { - 4a} \right)^2} - 4 \times 1 \times \left( { - {b^2} + 4{a^2}} \right)$
$\Rightarrow D = 16{a^2} + 4{b^2} - 16{a^2}$
$\Rightarrow D = 4{b^2}$
Therefore, $x = {{ - B + \sqrt D } \over {2A}},{{ - B + \sqrt D } \over {2A}}$
$\Rightarrow x = {{4a + 2b} \over 2},{{4a - 2b} \over 2}$
$\Rightarrow x = 2a + b\,,\,2a - b$
Therefore, required roots are $2a + b,\,\,2a - b$

Q.14 If the sum of two natural number is 8 and their product is 15. Find the numbers

[AI 2012]

Sol. Let the first natural number = x
A.T.Q.
x + y = 8 ……………..(1)
Where, y is another natural number
So,First natural number = x
Second natural number =8 – x [By equation (1)]
A.T.Q.
$x\left( {8 - x} \right) = 15$
$\Rightarrow 8x - {x^2} = 15$
$\Rightarrow {x^2} - 8x + 15 = 0$
$\Rightarrow {x^2} - 5x - 3x + 15 = 0$
$\Rightarrow x\left( {x - 5} \right) - 3\left( {x - 5} \right) = 0$
$\Rightarrow \left( {x - 3} \right)\left( {x - 5} \right) = 0$
$\Rightarrow x{\rm{ }} = {\rm{ }}3$  or  $x{\rm{ }} = {\rm{ 5}}$
Now if x = 3 then another number is 5
If x = 5 then another number is 3

Q.15 for what value of k, are the roots of the quadratic equation $\left( {k - 4} \right){x^2} + \left( {k - 4} \right)x + 4 = 0$ equal ?

[Foreign 2013]

Sol. We have $\left( {k - 4} \right){x^2} + \left( {k - 4} \right)x + 4 = 0$
Here  $a{\rm{ }} = {\rm{ }}k{\rm{ }}-{\rm{ }}4,{\rm{ }}b{\rm{ }} = {\rm{ }}k{\rm{ }}-{\rm{ }}4,{\rm{ }}c{\rm{ }} = {\rm{ }}4$
For equal roots D = 0
Since  $D{\rm{ }} = {\rm{ }}{b^2}-{\rm{ }}4ac$
$\Rightarrow {b^2} - 4ac = 0$
$\Rightarrow {\left( {k - 4} \right)^2} - 4 \times 4 \times \left( {k - 4} \right) = 0$
$\Rightarrow \left( {k - 4} \right)\left[ {k - 4 - 16} \right] = 0$
$\Rightarrow k = 4$ or $k = 20$
$\Rightarrow k \ne 4$ [ If k=4 then quadratic equation becomes linear Equation ]
$\Rightarrow k = 20$ is the required root.

Five Marks Questions

Q.1 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 maths and 4 marks less in science, the product of marks obtained in the two subjects would have been 180. Find the marks obtained in the two subjects separately.

[Delhi 2008]

Sol. Let marks obtained in mathematics be x and marks obtained in science be y.
A.T.Q.
$x + y = 28 \Rightarrow y = 28 - x$ …………….(1)
Also $\left( {x + 3} \right)\left( {y + 4} \right) = 180$
$\Rightarrow \left( {x + 3} \right)\left( {28 - x - 4} \right) = 180$  [using (1)]
$\Rightarrow \left( {x + 3} \right)\left( {24 - x} \right) = 180$
$\Rightarrow 24x - {x^2} + 72 - 3x = 180$
$\Rightarrow {x^2} - 21x + 108 = 0$
$\Rightarrow \left( {x - 12} \right)\left( {x - 9} \right) = 0$

$\Rightarrow x = 12$ or $x = 9$
Therefore, marks obtained in mathematics = 12 or 9
If marks obtained in mathematics = 12
$\Rightarrow$ marks obtained in science = $= 28 - 12 = 16$
If marks obtained in mathematics = 9
$\Rightarrow$ marks obtained in science = $28 - 9 = 19$

Q.2 The sum of the areas of two squares is 640m2.If the difference in their perimeters be 64m. Find the sides of the two squares.

[Delhi 2008C, Delhi 2008]

Sol. Let side of bigger square  = x m
And side of smaller square = y m
A.T.Q.
${x^2} + {y^2} = 640$  …………….(1)
And $4x - 4y = 64$
$\Rightarrow 4\left( {x - y} \right) = 64$
$\Rightarrow x - y = 16$
$\Rightarrow x = 16 + y$
Substituting in equation (1), we get
${\left( {16 + y} \right)^2} + {y^2} = 640$
$\Rightarrow 256 + {y^2} + 32y + {y^2} = 640$
$\Rightarrow 2{y^2} + 32y - 384 = 0$
$\Rightarrow {y^2} + 16y - 192 = 0$
$\Rightarrow \left( {y + 24} \right)\left( {y - 8} \right) = 0$
$\Rightarrow y = 8$ or $y = 24$
If y = 8, then $x = 16 + 8 = 24$
Therefore, sides of squares are 8 m and 24 m

Q.3 A motor boat whose speed is 18 km/hr in still water takes 1 hr more to go 24 km upstream and than to return downstream to the same spot. Find the speed of stream .

[AI 2008]

Sol. Speed of boat = 18 km / hr
Let speed of stream = x km/hr
Upstream speed $= \left( {18 - x} \right)$  km/hr
Downstream speed $= \left( {18 + x} \right)$  km/hr
Time taken to cover 24 km upstream = ${{24} \over {\left( {18 - x} \right)}}$ hrs.
Time taken to cover 24 km downstream = ${{24} \over {\left( {18 + x} \right)}}$ hrs.
A.T.Q.
${{24} \over {18 - x}} - {{24} \over {18 + x}} = 1$
$\Rightarrow {{24\left( {18 + x} \right) - 24\left( {18 - x} \right)} \over {\left( {18 - x} \right)\left( {18 + x} \right)}} = 1$
$\Rightarrow 24 \times 18 + 24x - 24 \times 18 + 24x = {\left( {18} \right)^2} - {x^2}$
$\Rightarrow 48x + {x^2} = 324$
$\Rightarrow {x^2} + 48x - 324 = 0$
$\Rightarrow {x^2} + 54x - 6x - 324 = 0$
$\Rightarrow x\left( {x + 54} \right) - 6\left( {x + 54} \right) = 0$
$\Rightarrow \left( {x - 6} \right)\left( {x + 54} \right) = 0$
$\Rightarrow$ x = 6, x =- 54
$x \ne 54$[ Because speed cannot be  negative]
Therefore, speed of the stream = 6 km  / hr

Q.4 Two water taps together can fill the tank in$9{3 \over 8}$ hrs. The tap of larger diameter takes 10 hrs. less than smaller one to fill the tank. Find the time in which each tap can separately fill the tank.

[AI 2008]

Sol. Let the two water taps fill the tank separately in (x) and (x-10) hrs
Since, both the taps together fill the tank in ${{75} \over 8}$ hr
Therefore, ${1 \over x} + {1 \over {x - 10}} = {8 \over {75}}$
$\Rightarrow {{x - 10 + x} \over {x\left( {x - 10} \right)}} = {8 \over {75}}$
$\Rightarrow {{2x - 10} \over {{x^2} - 10x}} = {8 \over {75}}$
$\Rightarrow 150x - 750 = 8{x^2} - 80x$
$\Rightarrow 8{x^2} - 530x + 750 = 0$
$\Rightarrow 4{x^2} - 115x + 375 = 0$
$\Rightarrow 4{x^2} - 100x - 15x + 375 = 0$
$\Rightarrow 4x\left( {x - 25} \right) - 15\left( {x - 25} \right) = 0$
$\Rightarrow \left( {4x - 15} \right)\left( {x - 25} \right) = 0$
Therefore, $x = {{15} \over 4}$, 25
Rejecting $x = {{15} \over 4}$, we get x  = 25
Therefore, two taps separately fill the tank in (x) and (x – 10) i.e. 25 hrs and 15 hrs.

Q.5 A peacock is sitting on the top of a pillar which is 9 m high. From a point 27 m away from the bottom of the pillar a snake is coming to its hole at the base of the pillar. Seeing the snake the peacock pounces on it. If their speeds are equal at what distance from the hole is the snake caught?

Sol. Let the distance covered by peacock $\left( {AD} \right) = \,'x'\,m$
Therefore, Distance covered by snake$\left( {DC} \right) = \,'x'\,m$
In  right angle$\Delta ABD,$
$A{d^2} = A{B^2} + B{D^2}$
$\Rightarrow {x^2} = {\left( 9 \right)^2} + {\left( {27 - x} \right)^2}$
$\Rightarrow {x^2} = 81 + 729 + {x^2} - 54x$
$\Rightarrow 810 = 54x$
$\Rightarrow {\rm{ }}x{\rm{ }} = {\rm{ }}15$
And $27{\rm{ }}-{\rm{ }}x{\rm{ }} = {\rm{ }}27{\rm{ }}-{\rm{ }}15{\rm{ }} = {\rm{ }}12$
Therefore, the snake is caught at 12 m from the hole.

Q.6 A person on tour has Rs. 4200 for his expenses. If he extends his tour for 3 days  he has to cut down his daily expenses by Rs. 70. Find the duration of the tour.

[AI 2008]

Sol. Let number of days tour = x
Total expenses = Rs. 4200
Therefore, daily expenses = Rs. ${{4200} \over x}$
If number of days of tour = x + 3
Then daily expenses  = Rs. ${{4200} \over {x + 3}}$
A.T.Q.
${{4200} \over x} - {{4200} \over {x + 3}} = 70$
$\Rightarrow {{4200\left( {x + 3} \right) - 4200x} \over {x\left( {x + 3} \right)}} = 70$
$\Rightarrow 4200x + 12600 - 4200x = 70\left( {{x^2} + 3x} \right)$
$\Rightarrow 4200x + 12600 - 4200x = 70\left( {{x^2} + 3x} \right)$
$\Rightarrow 180 = {x^2} + 3x$
$\Rightarrow {x^2} + 3x - 180 = 0$
$\Rightarrow {x^2} + 15x - 12x - 180 = 0$
$\Rightarrow x\left( {x + 15} \right) - 12\left( {x + 15} \right) = 0$
$\Rightarrow \left( {x - 12} \right)\left( {x - 15} \right) = 0$
$\Rightarrow x = 12$  or  $x = - 15$ [Rejecting x=-15 because days cannot be negative]
${\rm{ }} \Rightarrow x{\rm{ }} = {\rm{ }}12$
So, number of days of tour is 12 days.

Q.7 A trader bought a number of articles for Rs. 900,five articles were found damaged. He sold each of the remaining articles at Rs. 2 more than what he paid for it. He got a profit of Rs. 80 on the whole transaction. Find the number of articles he bought .

[Foreign 2009]

Sol. Number of articles = x
Cost of each articles = y
A.T.Q.
$xy{\rm{ }} = {\rm{ }}900$
$\Rightarrow y = {{900} \over x}$  …………….(1)
$\Rightarrow \left( {x - 5} \right)\left( {{{900} \over x} + 2} \right) = 900 + 80$
$\Rightarrow \left( {x - 5} \right)\left( {{{900 + 2x} \over x}} \right) = 980$
$\Rightarrow {{900x + 2{x^2} - 4500 - 10x} \over x} = 980$
$\Rightarrow 2{x^2} - 890x - 4500 = 980x$
$\Rightarrow 2{x^2} - 90x - 4500 = 0$
$\Rightarrow {x^2} - 45x - 2250 = 0$
$\Rightarrow {x^2} - 75x + 30x - 2250 = 0$
$\Rightarrow \left( {x + 30} \right)\left( {x - 75} \right) = 0$
$\Rightarrow x = - 30$  (rejected), x = 75
Therefore, No of articles = 75

Q.8 Two years ago a man’s age was three times the square of his son’s age. Three years hence his age will be four times his son’s age. Find their present ages.

[Foreign 2009]

Sol. Age of man = x years
Age of son = y years
A.T.Q.
$\left( {x - 2} \right) = 3{\left( {y - 2} \right)^2}$
$\Rightarrow x - 2 = 3\left( {{y^2} + 4 - 4y} \right)$
$\Rightarrow x - 2 = 3{y^2} + 12 - 12y$
$\Rightarrow x = 3{y^2} - 12y + 14$ ……………….(1)
Also $\left( {x + 3} \right) = 4\left( {y + 3} \right)$
$\Rightarrow x + 3 = 4y + 12$………………..(2)
$\Rightarrow 3{y^2} - 12y + 14 + 3 = 4y + 12$ [ Using (1)]
$\Rightarrow 3{y^2} - 16y + 5 = 0$
$\Rightarrow 3{y^2} - 15y - y + 5 = 0$
$\Rightarrow 3y\left( {y - 5} \right) - \left( {y - 5} \right) = 0$
$\Rightarrow \left( {3y - 1} \right)\left( {y - 5} \right) = 0$
$\Rightarrow y = {1 \over 3}$(rejected, age can never be in fraction form)
$\Rightarrow y{\rm{ }} = {\rm{ }}5$
Putting y=5 in (1),We get
$x = 3{\left( 5 \right)^2} - 10 \times 5 + 14$
$\Rightarrow x = 75 + 14 - 60 = 29$
Hence, Present age of Son is 5 years and Present age of Father is 29 years.

Multiple choice questions

Q.1 The roots of the equation${x^2} + x - p\left( {p + 1} \right) = 0,$where p is constant, are

[Delhi 2011]

(a) $\left( {p,\left( {p + 1} \right)} \right)$
(b) $p, - \left( {p + 1} \right)$
(c) $\left( { - p} \right),\left( {p + 1} \right)$
(d) $- p, - \left( {p + 1} \right)$

Sol. (b) We have${x^2} + x - p\left( {p + 1} \right) = 0,$
$\Rightarrow {x^2} + \left( {p + 1} \right)x - px - p\left( {p + 1} \right) = 0$
$\Rightarrow x\left( {x + p + 1} \right) - p\left( {x + p + 1} \right) = 0$
$\Rightarrow \left( {x + p + 1} \right)\left( {x - p} \right) = 0$
$\Rightarrow x = - p - 1,$ $x = p$

Q.2 The roots of the equation ${x^2} - 3x - m\left( {m + 3} \right) = 0$, where m is a constant, are -

[AI 2011]

(a) $m,m + 3$
(b) $- m,m + 3$
(c) $m, - \left( {m + 3} \right)$
(d) $- m, - m - 3$

Sol. We have ${x^2} - 3x - m\left( {m + 3} \right) = 0$
$\Rightarrow {x^2} - \left( {3 + m} \right)x + mx - m\left( {m + 3} \right) = 0$
$\Rightarrow x\left[ {x - \left( {3 + m} \right)} \right] + m\left[ {x - \left( {m + 3} \right)} \right] = 0$
$\Rightarrow \left( {x + m} \right)\left[ {x - \left( {3 + m} \right)} \right]$
$\Rightarrow x = 3 + m,x = - m$

Q.3 The roots of the quadratic equation${x^2} + 5x - \left({\alpha + 1} \right)\left( {\alpha + 6} \right) = 0$, Where $\alpha$ is a constant, are

[Foreign 2011]

(a) $x + 1,x + 6$
(b) $x + 1,\left( {x + 6} \right)$
(c) $- \left( {x + 1} \right),\left( {x + 6} \right)$
(d) $- \left( {x + 1} \right), - \left( {x + 6} \right)$

Sol. (b)${x^2} + 5x - \left( {\alpha + 1} \right)\left( {\alpha + 6} \right) = 0$
$\Rightarrow {x^2} + \left( {\alpha + 6} \right)x - \left( {\alpha + 1} \right)x - \left( {\alpha + 1} \right)\left( {\alpha + 6} \right) = 0$
$\Rightarrow x\left( {x + \alpha+ 6} \right) - \left( {x + 1} \right)\left[ {\left( {x + \alpha + 6} \right)} \right] = 0$
$\Rightarrow \left[ {x - \left( {\alpha + 1} \right)} \right]\left[ {x + \left( {\alpha + 6} \right)} \right] = 0$
$\Rightarrow x = - \left( {\alpha + 6} \right),x = \left( {\alpha + 1} \right)$

Q.4 If the quadratic equation $m{x^2} + 2x + m = 0$ has two equal roots, then the values of m are

[Foreign 2012]

(a) $\pm 1$
(b) 0, 2
(c) 0, 1
(d) -1, 0

Sol. We have $m{x^2} + 2x + m = 0$
Here  $a{\rm{ }} = {\rm{ }}m,{\rm{ }}b{\rm{ }} = {\rm{ }}2,{\rm{ }}c{\rm{ }} = {\rm{ }}m$
$D = {b^2} - 4ac = {\left( 2 \right)^2} - 4m \times m$
$\Rightarrow D = {\left( 2 \right)^2} - 4m \times m$
For equal roots  D = 0
$\Rightarrow 4 - 4{m^2} = 0$
$\Rightarrow {m^2} = 1$
$\Rightarrow m = \pm 1$

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