Question

Find the values of $k$, for which the quadratic equations has real and equal roots, in the following equation: $x^2+4kx+(k^2-k+2)=0$

Answer 1

Evaluate the discriminant $\mathit{D}$ for the given equation ${\mathbf{x}}^{\mathbf{2}}\mathbf{+}\mathbf{4}\mathbf{k}\mathbf{x}\mathbf{+}\mathbf{(}{\mathbf{k}}^{\mathbf{2}}\mathbf{-}\mathbf{k}\mathbf{+}\mathbf{2}\mathbf{)}\mathbf{=}\mathbf{0}$

Here, $a=1,b=4k,c=k^2-k+2$

If the roots of the equation are equal and real, the discriminant $D$ is equal to zero

$$\begin{gathered} \Rightarrow b^2-4ac=D \\ \Rightarrow {\left(4k\right)}^2-4\left(1\right)(k^2-k+2)=0 \\ \Rightarrow 16k^2-4k^2+4k-8=0 \\ \Rightarrow 12k^2+4k-8=0 \\ \Rightarrow 3k^2+k-2=0 \\ \Rightarrow 3k^2+3k-2k-2=0 \\ \Rightarrow 3k(k+1)-2(k+1)=0 \\ \Rightarrow (3k-2)(k+1)=0 \\ \Rightarrow 3k-2=0andk+1=0 \\ \Rightarrow k=\frac{2}{3}andk=-1 \end{gathered}$$

Thus, the value of $\mathit{k}$ are $\frac{\mathbf{2}}{\mathbf{3}}$ or $\mathbf{-}\mathbf{1}$.