Question

For the equation $x^2+k^2=(2k+2)x,k\in R$, roots are complex, then

• A. $k=\frac{-1}{2}$
• B. $k>\frac{-1}{2}$
• C. $k<\frac{-1}{2}$
• D. $\frac{-1}{2}<k<0$

Answer 1

The correct option is C

$k<\frac{-1}{2}$

Explanation for correct option:

Step1. Finding roots of the equation.

$x^2+k^2=(2k+2)x,k\in R$

Root is given by,

$\mathit{x}\mathbf{}\mathbf{=}\mathbf{}\frac{\mathbf{-}\mathbf{b}\mathbf{\pm }\sqrt{\mathbf{D}}}{\mathbf{2}\mathbf{a}}\mathbf{,}\mathbf{}\mathit{D}\mathbf{}\mathbf{=}\mathbf{}{\mathit{b}}^{\mathbf{2}}\mathbf{-}\mathbf{}\mathbf{4}\mathit{a}\mathit{c}$

Step2. Nature of the root

For complex root,$D<0$

$x^2-(2k+2)x+k^2=0$

Here, $a=1,b=-(2k+2)\hspace{0.17em}\&c=k^2$

Now, $D<0$

$b^2-4ac<0$

Put all the value,

$\begin{array}{rcl}{\{-(2k+2\left)\right\}}^2-4\times 1\times k^2& <& 0\\ {(2k+2)}^2-4k^2& <& 0\\ 4k^2+4+8k-4k^2& <& 0\\ k& <& -\frac{1}{2}\end{array}$

Hence option $\left(C\right)$ is correct answer.