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Q. The value of $k$ for which the equation $kx =( x +1)^2$ has exactly one real solution, is

Complex Numbers and Quadratic Equations

Solution:

$k x=x^2+2 x+1$
$\Rightarrow x^2+(2-k) x+1=0$
$\text { Put } D=0 $
$\Rightarrow(k-2)^2=4 \Rightarrow k-2= \pm 4 \Rightarrow k=0 \text { or } 4$