Question

The value of $k$ so that that the equations $x^2+kx+(k+2)=0$ and $x^2+(1-k)x+3-k=0$ have exactly one common root is

• A. $\frac{1}{2}$
• B. $\frac{-1}{2}$
• C. 1
• D. None of these

Answer 1

Answer: (D)

$x^2+kx+(k+2)=0$ ....(1)
$x^2+(1-k)x+3-k=0$ ...(2)
$(1)-(2)$
$\Rightarrow (x+1)(2k-1)=0$
$\Rightarrow (x+1)(2k-1)=0$
Hence, $x=-1$ is a common root
Now putting $x=-1$ in any equation, we get no value of $k$.
If $k=\frac{1}{2}$, then both the equations are identical.
Hence, both roots common.
Hence, no value of $k$ is possible. $\Rightarrow (D)]$