Question

Sum and product of the zeroes a quadratic polynomial are given. Find the zeroes of the polynomial by factorisation.

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

• A. $\frac{\sqrt{5}}{3},\frac{-\sqrt{5}}{2}$
• B. $\frac{\sqrt{5}}{5},\frac{-\sqrt{5}}{2}$
• C. $\frac{\sqrt{5}}{7},\frac{-\sqrt{5}}{2}$
• D. $\frac{\sqrt{5}}{6},\frac{-\sqrt{5}}{2}$

Answer 1

The correct option is B $\frac{\sqrt{5}}{5},\frac{-\sqrt{5}}{2}$

Given,

Sum of the zeros (S) $=\frac{-3}{2\sqrt{5}}$

Product of the zeros (P) $=\frac{-1}{2}$

The, the required polynomial is given,

$p\left(x\right)=k(x^2-Sx+P)$

$\Rightarrow p\left(x\right)=k(x^2+\frac{3}{2\sqrt{5}}x-\frac{1}{2})$

$\Rightarrow p\left(x\right)=\frac{k}{2\sqrt{5}}(2\sqrt{5}{x}^2+3x-\sqrt{5})$

where $k$ is a non zero real number

Zeros of the polynomial are

$p\left(x\right)=0$

$\Rightarrow \frac{k}{2\sqrt{5}}(2\sqrt{5}{x}^2+3x-\sqrt{5})=0$

$\Rightarrow 2\sqrt{5}{x}^2+3x-\sqrt{5}=0$

$\Rightarrow 2\sqrt{5}{x}^2+5x-2x-\sqrt{5}=0$

$\Rightarrow \sqrt{5}x(2x+\sqrt{5})-(2x+\sqrt{5})=0$

$\Rightarrow (\sqrt{5}x-1)(2x+\sqrt{5})=0$

$\Rightarrow x=\frac{1}{\sqrt{5}}=\frac{\sqrt{5}}{5}$ or $x=\frac{-\sqrt{5}}{2}$

Thus, the zeros are $\frac{\sqrt{5}}{5}$ and $\frac{-\sqrt{5}}{2}$