Question

If the sum of the squares of zeroes of the polynomial $5x^2+3x+k=0$ is $\frac{-11}{25}$, find the value of k.

• A. 3
• B. 4
• C. 2
• D. -2

Answer 1

Answer: (C)

2

The sum of zeroes of the polynomial $a{x}^2+bx+c=0$ is $\frac{-b}{a}$.

Here the given equation is $5x^2+3x+k=0$.

Therefore, the sum of the zeroes is $\frac{-3}{5}$.

Let the zeroes be $\alpha$ and $\beta$.

$\Rightarrow$ $\alpha +\beta =\frac{-3}{5}$ and $\alpha \beta =\frac{k}{5}$.

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

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

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

$\Rightarrow$ $k=2$

Therefore,Option C is correct.