Question

If the sum of squares of the zeroes of the polynomials $6x^2+x+k$ is $\frac{25}{36}$ find the value of k?

Answer 1

Let $\alpha$ and $\beta$ be the zeroes of the polynomial

$\Rightarrow {\alpha }^2+{\beta }^2=\frac{25}{36}$(given)

We have ${(\alpha +\beta )}^2={\alpha }^2+{\beta }^2+2\alpha \beta$ .....$\left(1\right)$

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

We have $\alpha =\frac{-1}{6},\alpha \beta =\frac{k}{6}$

Substituting the above in $\left(1\right)$ we get

${\left(\frac{-1}{6}\right)}^2-2\times \frac{k}{6}=\frac{25}{36}$

$\Rightarrow \frac{1}{36}-\frac{2k}{6}=\frac{25}{36}$

$\Rightarrow -\frac{2k}{6}=\frac{25-1}{36}=\frac{24}{36}=\frac{4}{6}$

$\Rightarrow -2k=4$ or $k=-2$