Question

Find the non-zero value of k for which the roots of the quadratic equation $9x^2-3kx+k=0$ are real and equal. [CBSE 2014]

Answer 1

The given equation is $9x^2-3kx+k=0$.

This is of the form $a{x}^2+bx+c=0$, where a = 9, b = −3k and c = k.

$\therefore D=b^2-4ac={\left(-3k\right)}^2-4\times 9\times k=9k^2-36k$

The given equation will have real and equal roots if D = 0.

$$\begin{gathered} \therefore 9k^2-36k=0 \\ \Rightarrow 9k\left(k-4\right)=0 \\ \Rightarrow k=0\mathrm{or}k-4=0 \\ \Rightarrow k=0\mathrm{or}k=4 \end{gathered}$$

But, k ≠ 0 (Given)

Hence, the required value of k is 4.