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Q.
The value of $k$ so that that the equations $x^2+k x+(k+2)=0$ and $x^2+(1-k) x+3-k=0$ have exactly one common root is
Complex Numbers and Quadratic Equations
Solution:
$ x^2+k x+(k+2)=0 $ ....(1)
$x^2+(1-k) x+3-k=0 $ ...(2)
$(1)-(2) $
$\Rightarrow(x+1)(2 k-1)=0$
$\Rightarrow( x +1)(2 k -1)=0$
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. $\Rightarrow(D)]$