Question

The equations $k{x}^2+x+k=0$ 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

Answer: (C)

Let $\alpha$ be the common root
$\Rightarrow k{\alpha }^2+\alpha +k=0$
$k{\alpha }^2+k\alpha +1=0$
On solving, we get,
$\frac{{\alpha }^2}{1-k^2}=\frac{\alpha }{k^2-k}=\frac{1}{k^2-k}$
$\frac{{\alpha }^2}{\alpha }=\frac{1-k^2}{k^2-k}$ and $\frac{\alpha }{1}=\frac{k^2-k}{k^2-k}$
$\Rightarrow \alpha =\frac{1-k^2}{k^2-k}=1\Rightarrow k^2-k=1-k^2$
$\Rightarrow 2k^2-k-1=0\Rightarrow k=-\frac{1}{2},1$
For $k=1,$ equations are identical, thus not possible
Hence, $k=-\frac{1}{2}$