Question

For what value of k are the roots of the quadratic equation $kx\left(x-2\sqrt{5}\right)+10=0$ real and equal? [CBSE 2013]

Answer 1

The given equation is

$$\begin{gathered} kx\left(x-2\sqrt{5}\right)+10=0 \\ \Rightarrow k{x}^2-2\sqrt{5}kx+10=0 \end{gathered}$$

This is of the form $a{x}^2+bx+c=0$, where a = k, b = $-2\sqrt{5}k$ and c = 10.

$\therefore D=b^2-4ac={\left(-2\sqrt{5}k\right)}^2-4\times k\times 10=20k^2-40k$

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

$$\begin{gathered} \therefore 20k^2-40k=0 \\ \Rightarrow 20k\left(k-2\right)=0 \\ \Rightarrow k=0\mathrm{or}k-2=0 \\ \Rightarrow k=0\mathrm{or}k=2 \end{gathered}$$

But, for k = 0, we get 10 = 0, which is not true.

Hence, 2 is the required value of k.