Question

Find $k$ for which the equation $2x^2-2kx+1=0$ has real and equal roots.

Answer 1

Step 1: Determining the discriminant

For the given quadratic equation $2x^2-2kx+1=0$, the values of $a=2,b=-2k$ and $c=1$.

Substituting the values in the discriminant formula $\left(D\right)=b^2-4ac$, we get

$\begin{array}{rcl}& & {D=(-2k)}^2-4\left(2\right)\left(1\right)\\ & & =4k^2-8\end{array}$

Step 2: Determining the values of $k$

Using the concept of nature of roots which says that when the roots of the quadratic equation are real and equal, $D=0$.

$$\begin{gathered} \Rightarrow 4k^2-8=0 \\ \Rightarrow 4k^2=8 \\ \Rightarrow k^2=2 \\ \therefore k=\pm \sqrt{2} \end{gathered}$$

Therefore, the values of $k$ are $-\sqrt{2}$ and $\sqrt{2}$.