Exercise 2.2 (Revised) - Chapter 2 - Polynomials - Ncert Solutions class 10 - Maths

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NCERT Solutions Class 10 Maths: Chapter 2 - Polynomials | Comprehensive Answers

Ex 2.2 Question 1.

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

Answer.

(i) $x^2-2 x-8$
Comparing given polynomial with general form of quadratic polynomial $a x^2+b x+c$,
We get $\mathrm{a}=1, \mathrm{~b}=-2$ and $\mathrm{c}=-8$
We have, $x^2-2 x-8$
$=x^2-4 x+2 x-8$
$=x(x-4)+2(x-4)=(x-4)(x+2)$
Equating this equal to 0 will find values of 2 zeroes of this polynomial.
$
(x-4)(x+2)=0
$
$\Rightarrow x=4,-2$ are two zeroes.

Sum of zeroes $=4+(-2)=2=$
$
\Rightarrow \frac{-(-2)}{1}=\frac{-b}{a}=\frac{- \text { Coefficient of } x}{\text { Coefficient of } x^2}
$

Product of zeroes $=4 \times(-2)=-8$
$
=\frac{-8}{1}=\frac{c}{a}=\frac{\text { Constant term }}{\text { Coefficient of } x^2}
$
(ii) $4 s^2-4 s+1$

Here, $a=4, b=-4$ and $c=1$
We have, $4 s^2-4 s+1$
$
\begin{aligned}
& =4 s^2-2 s-2 s+1 \\
& =2 s(2 s-1)-1(2 s-1) \\
& =(2 s-1)(2 s-1)
\end{aligned}
$

Equating this equal to 0 will find values of 2 zeroes of this polynomial.
$
\begin{aligned}
& \Rightarrow(2 s-1)(2 s-1)=0 \\
& \Rightarrow s=\frac{1}{2}, \frac{1}{2}
\end{aligned}
$

Therefore, two zeroes of this polynomial are $\frac{1}{2}, \frac{1}{2}$
Sum of zeroes $=\frac{1}{2}+\frac{1}{2}=1=\frac{-(-1)}{1} \times \frac{4}{4}=\frac{-(-4)}{4}$
$
=\frac{-b}{a}=\frac{- \text { Coefficient of } x}{\text { Coefficient of } x^2}
$

$\text { Product of Zeroes }=\frac{1}{2} \times \frac{1}{2}=\frac{1}{4}$

$
=\frac{c}{a}=\frac{\text { Constant term }}{\text { Coefficient of } x^2}
$
(iii) $6 x^2-3-7 x \quad \Rightarrow \quad 6 x^2-7 x-3$

Here, $a=6, b=-7$ and $c=-3$
We have, $6 x^2-7 x-3$
$
\begin{aligned}
& =6 x^2-9 x+2 x-3 \\
& =3 x(2 x-3)+1(2 x-3)=(2 x-3)(3 x+1)
\end{aligned}
$

Equating this equal to 0 will find values of 2 zeroes of this polynomial.
$\Rightarrow(2 x-3)(3 x+1)=0$
$\Rightarrow x=\frac{3}{2}, \frac{-1}{3}$
Therefore, two zeroes of this polynomial are $\frac{3}{2}, \frac{-1}{3}$
Sum of zeroes $=\frac{3}{2}+\frac{-1}{3}=\frac{9-2}{6}=\frac{7}{6}=\frac{-(-7)}{6}=\frac{-b}{a}=\frac{- \text { Coefficient of } x}{\text { Coefficient of } x^2}$
Product of Zeroes $=\frac{3}{2} \times \frac{-1}{3}=\frac{-1}{2}=\frac{c}{a}=\frac{\text { Constant term }}{\text { Coefficient of } x^2}$
(iv) $4 u^2+8 u$

Here, $a=4, b=8$ and $c=0$
$
4 u^2+8 u=4 u(u+2)
$

Equating this equal to 0 will find values of 2 zeroes of this polynomial.

$\begin{aligned}
& \Rightarrow 4 u(u+2)=0 \\
& \Rightarrow u=0,-2
\end{aligned}$

Therefore, two zeroes of this polynomial are $0,-2$

Sum of zeroes $=0-2=-2$
$
=\frac{-2}{1} \times \frac{4}{4}=\frac{-8}{4}=\frac{-b}{a}=\frac{- \text { Coefficient of } x}{\text { Coefficient of } x^2}
$

Product of Zeroes $=0 \times-2=0$
$
=\frac{0}{4}=\frac{c}{a}=\frac{\text { Constant term }}{\text { Coefficient of } x^2}
$
(v) $t^2-15$

Here, $\mathrm{a}=1, \mathrm{~b}=0$ and $\mathrm{c}=-15$
We have, $t^2-15 \Rightarrow t^2=15 \Rightarrow t= \pm \sqrt{15}$
Therefore, two zeroes of this polynomial are $\sqrt{15},-\sqrt{15}$
Sum of zeroes $=\sqrt{15}+(-\sqrt{15})=0=\frac{0}{1}=\frac{-b}{a}=\frac{- \text { Coefficient of } x}{\text { Coefficient of } x^2}$
Product of Zeroes $=\sqrt{15} \times(-\sqrt{15})=-15$
$
=\frac{-15}{1}=\frac{c}{a}=\frac{\text { Constant term }}{\text { Coefficient of } x^2}
$
(vi) $3 x^2-x-4$

Here, $\mathrm{a}=3, \mathrm{~b}=-1$ and $\mathrm{c}=-4$
We have, $3 x^2-x-4=3 x^2-4 x+3 x-4$
$
=x(3 x-4)+1(3 x-4)=(3 x-4)(x+1)
$

Equating this equal to 0 will find values of 2 zeroes of this polynomial.
$
\Rightarrow(3 x-4)(x+1)=0
$

$
\Rightarrow x=\frac{4}{3},-1
$

Therefore, two zeroes of this polynomial are $\frac{4}{3},-1$
$
\begin{aligned}
& \text { Sum of zeroes }=\frac{4}{3}+(-1)=\frac{4-3}{3}=\frac{1}{3}=\frac{-(-1)}{3}=\frac{-b}{a}=\frac{- \text { Coefficient of } x}{\text { Coefficient of } x^2} \\
& \text { Product of Zeroes }=\frac{4}{3} \times(-1)=\frac{-4}{3}=\frac{c}{a}=\frac{\text { Constant term }}{\text { Coefficient of } x^2}
\end{aligned}
$
Ex 2.2 Question 2.

Find a quadratic polynomial each with the given numbers as the sum and product of its zeroes respectively.
(i) $\frac{1}{4},-1$
(ii) $\sqrt{2}, 13$
(iii) $0, \sqrt{5}$
(iv) 1,1
(v) $\frac{-1}{4}, \frac{1}{4}$
(vi) 4,1

Answer.

(i) $\frac{1}{4},-1$
Let quadratic polynomial be $a x^2+b x+c$
Let $\alpha$ and $\beta$ are two zeroes of above quadratic polynomial.

$\alpha+\beta=\frac{1}{4}=\frac{-b}{a}$

$
\alpha \times \beta=-1=\frac{-1}{1} \times \frac{4}{4}=\frac{-4}{4}=\frac{c}{a}
$

On comparing, we get
$
\therefore a=4, b=-1, c=-4
$

Putting the values of $\mathrm{a}, \mathrm{b}$ and $\mathrm{c}$ in quadratic polynomial $a x^2+b x+c$, we get
Quadratic polynomial which satisfies above conditions $=4 x^2-x-4$
(ii) $\sqrt{2}, \frac{1}{3}$

Let quadratic polynomial be $a x^2+b x+c$
Let $\alpha$ and $\beta$ be two zeros of above quadratic polynomial.
$
a+\beta=\sqrt{2} \times \frac{3}{3}=\frac{3 \sqrt{2}}{3}=\frac{-b}{a}
$
$\alpha \times \beta=\frac{1}{3}$ which is equal to $\frac{c}{a}$
On comparing, we get
$
\therefore a=3, b=-3 \sqrt{2}, c=1
$

Putting the values of $\mathrm{a}, \mathrm{b}$ and $\mathrm{c}$ in quadratic polynomial $a x^2+b x+c$, we get
Quadratic polynomial which satisfies above conditions $=3 x^2-3 \sqrt{2} x+1$.
(iii) $0, \sqrt{5}$

Let quadratic polynomial be $c x^2+b x+c$

Let $\alpha$ and $\beta$ be two zeros of above quadratic polynomial.
$
\alpha+\beta=0=\frac{0}{1}=\frac{-b}{a}
$

$
\alpha \times \beta=\sqrt{5}=\frac{\sqrt{5}}{1}=\frac{c}{a}
$

On comparing, we get
$
\therefore a=1, b=0, c=\sqrt{5}
$

Putting the values of $\mathrm{a}, \mathrm{b}$ and $\mathrm{c}$ in quadratic polynomial $a x^2+b x+c$, we get Quadratic polynomial which satisfies above conditions $=x^2+\sqrt{5}$
(iv) 1,1

Let quadratic polynomial be $c x^2+b x+c$
Let $\alpha$ and $\beta$ be two zeros of above quadratic polynomial.
$
\begin{aligned}
& \alpha+\beta=1=\frac{-(-1)}{1}=\frac{-b}{a} \\
& \alpha \times \beta=1=\frac{1}{1}=\frac{c}{a}
\end{aligned}
$

On comparing, we get
$
\therefore a=1, b=-1, c=1
$

Putting the values of $\mathrm{a}, \mathrm{b}$ and $\mathrm{c}$ in quadratic polynomial $a x^2+b x+c$, we get
Quadratic polynomial which satisfies above conditions $=x^2-x+1$
(v) $\frac{-1}{4}, \frac{1}{4}$

Let quadratic polynomial be $c x^2+b x+c$

$\text { Let } \alpha \text { and } \beta \text { be two zeros of above quadratic polynomial. }$

$
\begin{aligned}
& \alpha+\beta=\frac{-1}{4}=\frac{-b}{a} \\
& \alpha \times \beta=\frac{1}{4}=\frac{c}{a}
\end{aligned}
$

On comparing, we get
$
\therefore a=4, b=1, c=1
$

Putting the values of $\mathrm{a}, \mathrm{b}$ and $\mathrm{c}$ in quadratic polynomial $a x^2+b x+c$, we get Quadratic polynomial which satisfies above conditions $=4 x^2+x+1$
(vi) 4,1

Let quadratic polynomial be $c x^2+b x+c$
Let $\alpha$ and $\beta$ be two zeros of above quadratic polynomial.
$
\begin{aligned}
& \alpha+\beta=4 \frac{-(-4)}{1}=\frac{-b}{a} \\
& \alpha \times \beta=1=\frac{1}{1}=\frac{c}{a}
\end{aligned}
$

On comparing, we get
$
\therefore a=1, b=-4, c=1
$

Putting the values of $a, b$ and $c$ in quadratic polynomial $a x^2+b x+c$, we get
$
\text { Quadratic polynomial which satisfies above conditions }=x^2-4 x+1
$