Question

The equations $kx^{2}+x+k=0$ and $kx^{2}+kx+1=0$ 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, $k=-\frac{1}{2}$