Question

Find zeros of quadratic polynomial whose sum and product respectively of the zeroes are $\frac{-8}{3},\frac{4}{3}$ .

• A. $-2,\frac{1}{3}$
• B. $-2,\frac{-2}{3}$
• C. $-3,\frac{1}{3}$
• D. $-4,\frac{-2}{3}$

Answer 1

Answer: (B)

$-2,\frac{-2}{3}$

Given,

Sum of the zeros (S) $=-\frac{8}{3}$

Product of the zeros (P) $=\frac{4}{3}$

The, the required polynomial is given,

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

$\Rightarrow p\left(x\right)=k(x^2+\frac{8}{3}x+\frac{4}{3})$

$\Rightarrow p\left(x\right)=\frac{k}{3}(3x^2+8x+4)$

where $k$ is a non zero real number

Zeros of the polynomial are

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

$\Rightarrow \frac{k}{3}(3x^2+8x+4)=0$

$\Rightarrow 3x^2+8x+4=0$

$\Rightarrow 3x^2+6x+2x+4=0$

$\Rightarrow 3x(x+2)+2(x+2)=0$

$\Rightarrow (x+2)(3x+2)=0$

$\Rightarrow x=-2$ or $x=\frac{-2}{3}$

Thus, the zeros are $-2$ and $\frac{-2}{3}$