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 = ?

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

Answer 1

Let $f\left(x\right)=k{x}^2+2x+3k$

Here $a=k,b=2andc=3k$

Sum of zeroes, $\alpha +\beta =-b/a=-\left(2\right)/k=-2/k$

Product of zeroes, $\alpha \beta =c/a=3k/k=3$

Given,

Sum of zeroes = Product of zeroes

$-2/k=3$

$k=-2/3$

(d) $-2/3$ is correct option

So, value of $k=-2/3$