Question

The roots of each of the following quadratic equations are real and equal, find k.
(1) 3y² + ky +12 = 0
(2) kx (x – 2) + 6 = 0

Answer 1

(1) 3y² + ky +12 = 0
The roots of the given quadratic equation are real and equal. So, the discriminant will be 0.
$$\begin{gathered} b^2-4ac=0 \\ \Rightarrow k^2-4\times 3\times 12=0 \\ \Rightarrow k^2-144=0 \\ \Rightarrow k^2=144 \\ \Rightarrow k=\pm 12 \end{gathered}$$

(2) kx (x – 2) + 6 = 0
$\Rightarrow k{x}^2-2kx+6=0$
The roots of the given quadratic equation are real and equal. So, the discriminant will be 0.
$$\begin{gathered} b^2-4ac=0 \\ \Rightarrow {\left(-2k\right)}^2-4\times k\times 6=0 \\ \Rightarrow 4k^2-24k=0 \\ \Rightarrow 4k\left(k-6\right)=0 \\ \Rightarrow 4k=0\mathrm{or}k-6=0 \\ \Rightarrow k=0\mathrm{or}k=6 \end{gathered}$$
But k cannot be equal to 0 since then there will not be any quadratic equation.
So, k = 6