Question

If the sum of the zeros of the quadratic polynomial $k{x}^2+2x+3k$ is equal to the product of its zeros, then $k=$

$\left(a\right)-\frac{1}{3}$
$\left(b\right)\frac{1}{3}$
$\left(c\right)\frac{2}{3}$
$\left(d\right)-\frac{2}{3}$

Answer 1

We know that, in a quadratic polynomial $a{x}^2+bx+c$

Sum of the zeros $=-\frac{b}{a}$

Product of zeros $=\frac{c}{a}$

It is given that, In $k{x}^2+2x+3k$

Sum of the zeros $=$ Product of zeros

$\Rightarrow -\frac{2}{k}=\frac{3k}{k}$

$\Rightarrow -\frac{2}{k}=3$

$\Rightarrow -\frac{2}{3}=k$

$\therefore k=\frac{2}{3}$

Hence, Option $c$ is correct.