Question

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

• A. $k=4$
• B. $k<0$
• C. $k=4$ or $k<0$
• D. $k=0$ or 4

Answer 1

Answer: (D)

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