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 $(a)$