Polynomials : Exercise - 2.2 (Mathematics NCERT Class 10th)

Q.1       Find the zeroes of the following quadratic polynomials and verify the relationship between the zeroes and the coefficients :
(i) ${x^2} - 2x - 8$
(ii) $4{s^2} - 4s + 1$
(iii) $6{x^2} - 3 - 7x$
(iv) $4{u^2} + 8u$
(v) ${t^2} - 15$
(vi) $3{x^2} - x - 4$

Sol.       (i) We have, ${x^2} - 2x - 8$ $= {x^2} + 2x - 4x - 8$
$= x\left( {x + 2} \right) - 4\left( {x + 2} \right)$
$= \left( {x + 2} \right)\left( {x - 4} \right)$

The value of ${x^2} - 2x - 8$ is zero when the value of (x + 2) (x – 4) is zero, i.e.,
when x + 2 = 0 or x – 4 = 0 , i.e., when x = – 2 or x = 4.

So, The zeroes of ${x^2} - 2x - 8$ are – 2 and 4.
Therefore , sum of the zeroes = (– 2) + 4 = 2
$= {{ - Coefficient\,of\,x} \over {Coefficient\,of\,{x^2}}}$
and product of zeroes = (– 2) (4) = – 8 $= {{ - 8} \over 1}$
$={{Cons\tan t\,term} \over {Coefficient\,of\,{x^2}}}$

(ii) We have, $4{s^2} - 4s + 1$ $= 4{s^2} - 2s - 2s + 1$
$= 2s\left( {2s - 1} \right) - 1\left( {2s - 1} \right)$
$= \left( {2s - 1} \right)\left( {2s - 1} \right)$
The value of $4{s^2} - 4s + 1$ is zero when the value of
(2s – 1) (2s – 1) is zero, i.e., when 2s – 1 = 0 or 2s – 1 = 0,
i.e., when $s = {1 \over 2}\,or\,s = {1 \over 2}$.
So, The zeroes of $4{s^2} - 4s + 1\,are\,{1 \over 2}\,and\,{1 \over 2}$
Therefore, sum of the zeroes $= {1 \over 2} + {1 \over 2} = 1$
$= {{-Coefficient\,of\,s} \over {Coefficient\,of\,{s^2}}}$
and product of zeroes $= \left( {{1 \over 2}} \right)\left( {{1 \over 2}} \right) = {1 \over 4}$
$= {{Constant\,term} \over {Coefficient \,of\,{s^2}}}$
(iii) We have, $6{x^2} - 3 - 7x$$6{x^2} - 7x - 3$
$= 6{x^2} - 9x + 2x - 3$
$= 3x\left( {2x - 3} \right) + 1\left( {2x - 3} \right)$
$= \left( {3x + 1} \right)\left( {2x - 3} \right)$
The value of $6{x^2} - 3 - 7x$ is zero when the value of (3x + 1) (2x – 3) is zero, i.e., when 3x + 1 = 0 or 2x – 3 = 0, i.e, when $x = - {1 \over 3} or x = {3 \over 2}$
So, The zeroes of $6{x^2} - 3 - 7x\,\,\,are\,\, - {1 \over 3}\,and\,\,{3 \over 2}$
Therefore, sum of the zeroes $= - {1 \over 3} + {3 \over 2} = {7 \over 6}$
$= {{ - Coefficient\,of\,x} \over {Coefficient\,of\,{x^2}}}$
and product of zeroes $= \left( { - {1 \over 3}} \right)\left( {{3 \over 2}} \right) = {{ - 1} \over 2}$
$= {{Constant\,term} \over {Coefficient \,of\,{x^2}}}$
(iv)  We have, $4{u^2} + 8u$ = 4u (u + 2)
The value of $4{u^2} + 8u$ is zero when the value of 4u(u + 2) is zero, i.e., when u = 0 or u + 2 = 0, i.e., when u = 0 or u = – 2.
So, The zeroes of $4{u^2} + 8u$ and 0 and – 2
Therefore, sum of the zeroes = 0 + (– 2) = – 2
$= {{Coefficent\,of\,u} \over {Coefficient\,of\,{u^2}}}$
and , product of zeroes = (0) (–2) = 0
$= {{Cons\tan t\,term} \over {Coefficient\,of\,{u^2}}}$
(v) We have ${t^2} - 15$ $= \left( {t - \sqrt {15} } \right)\left( {t + \sqrt {15} } \right)$
The value of ${t^2} - 15$ is zero when the value of $\left( {t - \sqrt {15} } \right)\left( {t + \sqrt {15} } \right)$ is zero,
i.e., when
$t - \sqrt {15} = 0\,\,or\,\,t + \sqrt {15}$ = 0 i.e., when $t = \sqrt {15} \,\,or\,\,t = - \sqrt {15}$
So, The zeroes of ${t^2} - 15\,are\,\sqrt {15} \,\,and\,\, - \sqrt {15}$
Therefore , sum of the zeroes = $\sqrt {15} + \left( { - \sqrt {15} } \right) = 0$
$= {{ - Coefficient\,of\,t} \over {Coefficient\,of\,{t^2}}}$
and, product of the zeroes = $\left( {\sqrt {15} } \right)\left( { - \sqrt {15} } \right) = - 15$
$= {{Cons\tan t\,term} \over {Coefficient\,of\,{t^2}}}$
(vi) We have, $3{x^2} - x - 4$ =  $3{x^2} + 3x - 4x - 4$
$= 3x\left( {x + 1} \right) - 4\left( {x + 1} \right)$
$= \left( {x + 1} \right)\left( {3x - 4} \right)$
The value of $3{x^2} - x - 4$ is zero when the value of (x + 1) (3x – 4) is zero, i.e., when x + 1 = 0 or 3x – 4 = 0, i.e., when x = – 1 or  $x = {4 \over 3}$.
So, The zeroes of $3{x^2}-x-4\,are\,-1\,and\,{4 \over 3}$
Therefore , sum of the zeroes $= - 1 + {4 \over 3} = {{ - 3 + 4} \over 3}$
$= {1 \over 3} = {{Coefficient\,of\,x} \over {Coefficient\,of\,{x^2}}}$
and, product of the zeroes $= \left( { - 1} \right)\left( {{4 \over 3}} \right) = - {4 \over 3}$
$= {{Constant\,term} \over {Coefficient\,of\,{x^2}}}$

Q.2       Find a quadratic polynomial each with the given numbers as the sum and product of its zeroes respectively.
(i) ${1 \over 4}, - 1$
(ii) $\sqrt 2 ,{1 \over 3}$
(iii) $0,\sqrt 5$
(iv) 1, 1
(v) $- {1 \over 4},{1 \over 4}$

(vi) 4, 1
Sol.      (i) Let the polynomial be $a{x^2} + bx + c$, and its zeroes be $\alpha \,\,and\,\,\beta$. Then ,
$\alpha \, + \,\,\beta = {1 \over 4} = {{ - b} \over a}$
and,                            $\alpha \,\beta = - 1 = {{ - 4} \over 4} = {c \over a}$
If a = 4,  then  b = – 1 and c = – 4.
Therefore, one quadratic polynomial which fits the given conditions is $4{x^2} - x - 4$.

(ii) Let the polynomial be $a{x^2} + bx + c$, and its zeroes be $\alpha \,and\,\beta$. Then,
$\alpha \, + \,\beta = \sqrt 2 = {{3\sqrt 2 } \over 3} = {{ - b} \over a}$
and                            $\alpha \,\beta = {1 \over 3} = {c \over a}$
If a = 3, then b $= 3\sqrt 2 \,and\,c = 1$
So, One quadratic polynomial which fits the given conditions is $3{x^2} - 3\sqrt 2 \,x + 1$.

(iii) Let the polynomial  be $a{x^2} + bx + c$, and its zeroes be $\alpha \,and\,\beta$. Then,

$\alpha + \beta = 0 = {0 \over 1} = {{ - b} \over a}$
and                           $\alpha \beta = \sqrt 5 = {{\sqrt 5 } \over 1} = {c \over a}$
If  a = 1, then b = 0 and c = ${\sqrt 5 }$
So, one quadratic polynomial which fits the given conditions is ${x^2} - 0.x + \sqrt 5 ,\,i.e.,\,{x^2} + \sqrt 5$.

(iv) Let the polynomial be $a{x^2} + bx + c$ and its zeroes be $\alpha \,and\,\beta$. Then,
$\alpha \, + \,\beta = 1$
$= {{ - \left( { - 1} \right)} \over 1} = {{ - b} \over a}$
a
nd                               $\alpha \beta = 1$ $= {1 \over 1} = {c \over a}$
If  a = 1, then b = – 1 and c = 1.
So, one quadratic polynomial which fits the given conditions is ${x^2} - x + 1$.

(v) Let the polynomial be $a{x^2} + bx + c$ and its zeroes be $\alpha \,and\,\beta$. Then
$\alpha + \beta = - {1 \over 4} = {{ - b} \over a}$
and                               $\alpha \beta = {1 \over 4} = {c \over a}$
If  a = 4 then b = – 1 and c = 1.
So, one quadratic polynomial which fits the given conditions is $4{x^2} - x + 1$.

(vi) Let the polynomial be $a{x^2} + bx + c$ and its zeroes be $\alpha \,and\,\beta$. Then,
$\alpha \, + \,\beta = 4 = {{ - b} \over a}$

$\alpha \,\,\beta = 1 = {1 \over 1} = {c \over a}$
If a = 1, then b = – 4 and c = 1
Therefore, one quadratic polynomial which fits the given conditions is ${x^2} - 4x + 1$.