Question

If the sum of the squares of zeros of the polynomial $5x^2+3x+k$ is $-\frac{11}{25}$, find the value of $k$.

Answer 1

Let the zeros of the quadratic equation be $\alpha$ and $\beta$

Sum of the roots , $\alpha +\beta =-\frac{3}{5}$

and product of the zeros , $\alpha \beta =\frac{k}{5}$

Also, given,

${\alpha }^2+{\beta }^2=-\frac{11}{25}$

Now,

$(\alpha +\beta {)}^2={\alpha }^2+{\beta }^2+2\alpha \beta$

$=>(-\frac{3}{5}{)}^2=-\frac{11}{25}+2\times \frac{k}{5}$

$=>\frac{9}{25}=-\frac{11}{25}+\frac{2k}{5}$

$=>\frac{9}{25}+\frac{11}{25}=\frac{2k}{5}$

$=>\frac{20}{25}=\frac{2k}{5}$

$=>k=\frac{20\times 5}{2\times 25}$

$=>k=2$