Question

Find the values of $k$ for the following quadratic equation, so that it has two equal roots.

$2x^2+kx+3=0$

Answer 1

Compute the required value:

On comparing the given equation with $a{x}^2+bx+c=0$, we get,

$a=2,b=k$ and $c=3$

As we know, Discriminant : $D=b^2-4ac$

$$\begin{gathered} =k^2-4\left(2\right)\left(3\right) \\ =k^2-24 \end{gathered}$$

For equal roots,

Discriminant $D=0$

$$\begin{gathered} \Rightarrow k^2-24=0 \\ \Rightarrow k^2=24 \\ \Rightarrow k=\pm \sqrt{24} \\ =\pm 2\sqrt{6} \end{gathered}$$

Hence, the value of $k$ is $\pm 2\sqrt{6}$.