Question 1

Find the zeroes of the following quadratic polynomials and verify the relationship between the zeroes 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) $3 x^{2}-x-4$

Accepted Answer

(i) $\quad x^{2}-2 x-8=(x-4)(x+2)$

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

Therefore, the zeroes of $x^{2}-2 x-8$ are 4 and -2 .
Sum of zeroes $=4-2=2=\frac{-(-2)}{1}=\frac{-( ext { Coefficient of } x)}{ ext { Coefficient of } x^{2}}$
Product of zeroes $=4 imes(-2)=-8=\frac{(-8)}{1}=\frac{ ext { Constant term }}{ ext { Coefficient of } x^{2}}$
(ii) $\quad 4 s^{2}-4 s+1=(2 s-1)^{2}$

The value of $4 s^{2}-4 s+1$ is zero when $2 s-1=0$, i.e., $s=\frac{1}{2}$ Therefore, the zeroes of $4 s^{2}-4 s+1$ are $1 / 2$ and $1 / 2$.

Sum of zeroes $=\frac{1}{2}+\frac{1}{2}=1=\frac{-(-4)}{4}=\frac{-( ext { Coefficient of } s)}{\left( ext { Coefficient of } s^{2} ight)}$
Product of zeroes $=\frac{1}{2} imes \frac{1}{2}=\frac{1}{4}=\frac{ ext { Constant term }}{ ext { Coefficient of } s^{2}}$
(iii) $6 x^{2}-3-7 x=6 x^{2}-7 x-3=(3 x+1)(2 x-3)$

The value of $6 x^{2}-3-7 x$ is zero when $3 x+1=0$ or $2 x-3=0$, i.e.,
$x=\frac{-1}{3} \quad$ or $\quad x=\frac{3}{2}$

Therefore, the zeroes of $6 x^{2}-3-7 x$ are $\frac{-1}{3}$ and $\frac{3}{2}$.
Sum of zeroes $=\frac{-1}{3}+\frac{3}{2}=\frac{7}{6}=\frac{-(-7)}{6}=\frac{-( ext { Coefficient of } x)}{ ext { Coefficient of } x^{2}}$
Product of zeroes $=\frac{-1}{3} imes \frac{3}{2}=\frac{-1}{2}=\frac{-3}{6}=\frac{ ext { Constant term }}{ ext { Coefficient of } x^{2}}$
(iv) $4 u^{2}+8 u=4 u^{2}+8 u+0$
$=4 u(u+2)$

The value of $4 u^{2}+8 u$ is zero when $4 u=0$ or $u+2=0$, i.e., $u=0$ or $u=-2$

Therefore, the zeroes of $4 u^{2}+8 u$ are 0 and -2 .

Sum of zeroes $=0+(-2)=-2=\frac{-(8)}{4}=\frac{-( ext { Coefficient of } u)}{ ext { Coefficient of } u^{2}}$
Product of zeroes $=0 imes(-2)=0=\frac{0}{4}=\frac{ ext { Constant term }}{ ext { Coefficient of } u^{2}}$
(v) $t^{2}-15$
$$
\begin{aligned}
& =t^{2}-0 t-15 \
& =(t-\sqrt{15})(t+\sqrt{15})
\end{aligned}
$$

The value of $t^{2}-15$ is zero when $t-\sqrt{15}=0$ or $t+\sqrt{15}=0$, i.e., when $t=\sqrt{15}$ or $t=-\sqrt{15}$

Therefore, the zeroes of $t^{2}-15$ are $\sqrt{15}$ and $-\sqrt{15}$.

Sum of zeroes $=\sqrt{15}+(-\sqrt{15})=0=\frac{-0}{1}=\frac{-( ext { Coefficient of } t)}{\left( ext { Coefficient of } t^{2} ight)}$
Product of zeroes $=(\sqrt{15})(-\sqrt{15})=-15=\frac{-15}{1}=\frac{ ext { Constant term }}{ ext { Coefficient of } x^{2}}$
(vi) $3 x^{2}-x-4$
$$
=(3 x-4)(x+1)
$$

The value of $3 x^{2}-x-4$ is zero when $3 x-4=0$ or $x+1=0$, i.e., when $x=\frac{4}{3}$ or $x=-1$

Therefore, the zeroes of $3 x^{2}-x-4$ are $4 / 3$ and -1 .
Sum of zeroes $=\frac{4}{3}+(-1)=\frac{1}{3}=\frac{-(-1)}{3}=\frac{-( ext { Coefficient of } x)}{ ext { Coefficient of } x^{2}}$
Product of zeroes $=\frac{4}{3}(-1)=\frac{-4}{3}=\frac{ ext { Constant term }}{ ext { Coefficient of } x^{2}}$

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) $\quad \sqrt{2}, \frac{1}{3}$
(iii) $0, \sqrt{5}$
(iv) 1,1
(v) $-\frac{1}{4}, \frac{1}{4}$
(vi) 4,1

Answer 2:
(i) $\frac{1}{4},-1$

Let the polynomial be $\mathrm{ax}^{2}+\mathrm{bx}+\mathrm{c}$ and its zeroes be $\alpha$ and $\beta$.
$$
\begin{aligned}
& \alpha+\beta=\frac{1}{4}=\frac{-b}{a} \
& \alpha \beta=-1=\frac{-4}{4}=\frac{c}{a} \
& ext { If } a=4, ext { then } b=-1, c=-4
\end{aligned}
$$

Therefore, the quadratic polynomial is $4 x^{2}-x-4$.
(ii) $\quad \sqrt{2}, \frac{1}{3}$

Let the polynomial be $\mathrm{ax}{ }^{2}+\mathrm{bx}+\mathrm{c}$ and its zeroes be $\alpha$ and $\beta$.
$$
\begin{aligned}
& \alpha+\beta=\sqrt{2}=\frac{3 \sqrt{2}}{3}=\frac{-b}{a} \
& \alpha \beta=\frac{1}{3}=\frac{c}{a} \
& ext { If } a=3, ext { then } b=-3 \sqrt{2}, c=1
\end{aligned}
$$

Therefore, the quadratic polynomial is $3 x^{2}-3 \sqrt{2} x+1$.
(iii) $0, \sqrt{5}$

Let the polynomial be $\mathrm{ax}{ }^{2}+\mathrm{bx}+\mathrm{c}$ and its zeroes be $\alpha$ and $\beta$.
$$
\begin{aligned}
& \alpha+\beta=0=\frac{0}{1}=\frac{-b}{a} \
& \alpha imes \beta=\sqrt{5}=\frac{\sqrt{5}}{1}=\frac{c}{a}
\end{aligned}
$$

If $a=1$, then $b=0, c=\sqrt{5}$
Therefore, the quadratic polynomial is $x^{2}+\sqrt{5}$.
(iv) 1,1

Let the polynomial be $\mathrm{ax}{ }^{2}+\mathrm{bx}+\mathrm{c}$ and its zeroes be $\alpha$ and $\beta$.
$$
\begin{aligned}
& \alpha+\beta=1=\frac{1}{1}=\frac{-b}{a} \
& \alpha imes \beta=1=\frac{1}{1}=\frac{c}{a} \
& ext { If } a=1, ext { then } b=-1, c=1
\end{aligned}
$$

Therefore, the quadratic polynomial is $x^{2}-x+1$.
(v) $-\frac{1}{4}, \frac{1}{4}$

Let the polynomial be $\mathrm{ax}{ }^{2}+\mathrm{bx}+\mathrm{c}$ and its zeroes be $\alpha$ and $\beta$.
$$
\begin{aligned}
& \alpha+\beta=\frac{-1}{4}=\frac{-b}{a} \
& \alpha imes \beta=\frac{1}{4}=\frac{c}{a} \
& ext { If } a=4, ext { then } b=1, c=1
\end{aligned}
$$

Therefore, the quadratic polynomial is $4 x^{2}+x+1$.
(vi) 4,1

Let the polynomial be $\mathrm{ax}{ }^{2}+\mathrm{bx}+\mathrm{c}$ and its zeroes be $\alpha$ and $\beta$.
$$
\begin{aligned}
& \alpha+\beta=4=\frac{4}{1}=\frac{-b}{a} \
& \alpha imes \beta=1=\frac{1}{1}=\frac{c}{a} \
& ext { If } a=1, ext { then } b=-4, c=1
\end{aligned}
$$

Therefore, the quadratic polynomial is $x^{2}-4 x+1$.

\section*{Mathematics}
(Chapter - 2) (Polynomials)
(Class - X)

\section*{Exercise 2.3}

Accepted Answer

(No solution provided.)