Question

find the quadratic polynomial the sum and product of two zeros are0- 4 upon3 respectively and hence find the zero​

Answer 1

Answer:

The required polynomial is $\frac{1}{3}{(3x^2-4)$

The zeros are $\frac{2}{\sqrt3}\:$and $\:\frac{-2}{\sqrt3}$

Step-by-step explanation:

Let the zeros be $\alpha\:and\:\beta$

Given:

$\alpha+\beta=0$

$\implies\:\alpha=-\beta$.........(1)

$\alpha\beta=\frac{-4}{3}$

$(-\beta)\beta=\frac{-4}{3}$ (using (1))

${-\beta}^2=\frac{-4}{3}$

${\beta}^2=\frac{4}{3}$

${\beta}=$±$\frac{2}{\sqrt3}$

Then

$\alpha=\frac{2}{\sqrt3}$

$\beta=\frac{-2}{\sqrt3}$

The required qiadratic polynomial is

$x^2-(\alpha+\beta)x+\alpha\beta$

$x^2-(0)x+(\frac{-4}{3})$

$x^2-\frac{4}{3}$

$\frac{1}{3}{(3x^2-4)$