# If the sum and product of the roots of the equation kx2 + 6x + 4k = 0 are real, then k =

Question:

If the sum and product of the roots of the equation $k x^{2}+6 x+4 k=0$ are real, then $k=$

(a) $-\frac{3}{2}$

(b) $\frac{3}{2}$

(c) $\frac{2}{3}$

(d) $-\frac{2}{3}$

Solution:

The given quadric equation is $k x^{2}+6 x+4 k=0$, and roots are equal

Then find the value of $c$

Let $\alpha$ and $\beta$ be two roots of given equation

And, $a=k, b=6$ and,$c=4 k$

Then, as we know that sum of the roots

$\alpha+\beta=\frac{-b}{a}$

$\alpha+\beta=\frac{-6}{k}$

And the product of the roots

$\alpha \cdot \beta=\frac{c}{a}$

$\alpha \beta=\frac{4 k}{k}$

$=4$

According to question, sum of the roots = product of the roots

$\frac{-6}{k}=4$

$4 k=-6$

$k=\frac{-6}{4}$

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

Therefore, the value of $c=\frac{-3}{2}$

Thus, the correct answer is $(a)$