Find the values of k for each of the following quadratic equations, so that they have two equal roots.
Find the values of $$k$$ for each of the following quadratic equations, so that they have two equal roots. [ JKBOSE – 2020, 19, 16 Most Imp. ]
(i) $$2 x^2+k x+3=0$$
(ii) $$k x(x-2)+6=0$$
Sol. (i) Consider the equation $$2 x^2+k x+3=0$$
Here, $$\quad a=2, b=k, c=3$$
$$
\quad \mathrm{D}=b^2-4 a c=k^2-4 \times 2 \times 3=k^2-24
$$
For real and equal roots, $$\mathrm{D}=0 \Rightarrow k^2-24=0$$
$$
\begin{aligned}
& \Rightarrow \quad k^2=24 \quad \Rightarrow \quad k= \pm \sqrt{24} \\
& \Rightarrow \quad k= \pm 2 \sqrt{6} .
\end{aligned}
$$
Sol.(ii) Consider the equation $$k x(x-2)+6=0$$
$$
\Rightarrow k x^2-2 k x+6=0
$$
Here, $$\quad a=k, b=-2 k$$ and $$c=6$$
$$
\quad \mathrm{D}=b^2-4 a c=(-2 k)^2-4 \times k \times 6=4 k^2-24 k
$$
For real and equal roots, $$\mathrm{D}=0$$
$$
\begin{array}{lccc}
\Rightarrow & 4 k^2-24 k=0 & \Rightarrow 4 k(k-6)=0 \\
\Rightarrow & k=0 & \text { or } & k-6=0 \\
\Rightarrow & k=0 & \text { or } & k=6 .
\end{array}
$$
But $$k=0$$ is not possible
[From given equation]
$$
\quad k=6 .
$$
