# For each of the following, find a quadratic

Question:

For each of the following, find a quadratic polynomial whose sum and product respectively of the zeroes are as given. Also, find the zeroes of these

polynomials by factorisation.

(i) $\frac{-8}{3}, \frac{4}{3}$

(ii) $\frac{21}{8}, \frac{5}{16}$

(iii) $-2 \sqrt{3},-9$

(iv) $\frac{-3}{2 \sqrt{5}},-\frac{1}{2}$

Solution:

(i) Given that, sum of zeroes $(S)=-\frac{8}{3}$

and product of zeroes $(P)=\frac{4}{3}$

$\therefore$ Required quadratic expression, $f(x)=x^{2}-S x+P$

$=x^{2}+\frac{8}{3} x+\frac{4}{3}=3 x^{2}+8 x+4$

Using factorisation method, $=3 x^{2}+6 x+2 x+4$

$=3 x(x+2)+2(x+2)=(x+2)(3 x+2)$

Hence, the zeroes of $f(x)$ are $-2$ and $-\frac{2}{3}$.

(ii) Given that, $S=\frac{21}{8}$ and $P=\frac{5}{16}$

$\therefore$ Required quadratic expression, $f(x)=x^{2}-S x+P$

$=x^{2}-\frac{21}{8} x+\frac{5}{16}=16 x^{2}-42 x+5$

Using factorisation method $=16 x^{2}-40 x-2 x+5$

$=8 x(2 x-5)-1(2 x-5)=(2 x-5)(8 x-1)$

Hence, the zeroes of $f(x)$ are $\frac{5}{2}$ and $\frac{1}{8}$

(iii) Given that, $S=-2 \sqrt{3}$ and $P=-9$

$\therefore$ Required quadratic expression,

$f(x)=x^{2}-S x+P=x^{2}+2 \sqrt{3} x-9$

$=x^{2}+3 \sqrt{3} x-\sqrt{3} x-9$    [using factorisation method]

$=x(x+3 \sqrt{3})-\sqrt{3}(x+3 \sqrt{3})$

$=(x+3 \sqrt{3})(x-\sqrt{3})$

Hence, the zeroes of $f(x)$ are $-3 \sqrt{3}$ and $\sqrt{3}$.

(iv) Given that, $S=-\frac{3}{2 \sqrt{5}}$ and $P=-\frac{1}{2}$

$\therefore$ Required quadratic expression,

$f(x)=x^{2}-S x+P=x^{2}+\frac{3}{2 \sqrt{5}} x-\frac{1}{2}$

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

Using factorisation method, $=2 \sqrt{5} x^{2}+5 x-2 x-\sqrt{5}$

$=\sqrt{5} x(2 x+\sqrt{5})-7(2 x+\sqrt{5})$

$=(2 x+\sqrt{5})(\sqrt{5} x-1)$

Hence,the zeroes of $f(x)$ are $-\frac{\sqrt{5}}{2}$ and $\frac{1}{\sqrt{5}}$