Question

The equation $k{x}^2+x+k=0$ and $k{x}^2+kx+1=0$ have exactly one root in common for

• A. $k=-\frac{1}{2},1$
• B. `$k=1$
• C. $k=-\frac{1}{2}$
• D. $k=\frac{1}{2}$

Answer 1

The correct option is C $k=-\frac{1}{2}$

Let $\alpha$ be the common root.

Then $k{\alpha }^2+\alpha +k=0$ and $k{\alpha }^2+k\alpha +1=0$

Solving, we get $\frac{{\alpha }^2}{1-k^2}=\frac{\alpha }{k^2-k}=\frac{1}{k^2-k}\Rightarrow \frac{1-k^2}{k^2-k}=\frac{k^2-k}{k^2-k}=1$

$\Rightarrow 2k^2-k-1=0\Rightarrow k=-\frac{1}{2},1$

For $k=1$, equations become identical, thus not possible.

Hence, $k=-\frac{1}{2}.$