Question

If the equation $x^{2}-\left(2+m\right)x+\left(m^{2}-4m+4\right)=0$ in $x$ has equal roots, then the values of $m$ are

• A. $\frac{2}{3}, 1$
• B. $\frac{2}{3}, 6$
• C. $0,1$
• D. $0, 2$
• E. $\frac{2}{3}, 0$

Answer 1

Answer: (B)

$x^{2}-(2+m) x+\left(m^{2}-4 m+4\right)=0$
Since, $x$ has equal roots, then
$B^{2}-4 AC=0$
$(m+2)^{2}-4\left(m^{2}-4 m+4\right)=0$
$\Rightarrow \,\left(m^{2}+4+4 m\right)-\left(4 m^{2}-16 m+16\right)=0$
$\Rightarrow \,-3 m^{2}+20 m-12=0$
$\Rightarrow \, 3 m^{2}-20 m+12=0$
$\Rightarrow \, 3 m^{2}-18 m-2 m+12=0$
$\Rightarrow \, 3 m(m-6)-2(m-6)=0$
$\Rightarrow \,(m-6)(3 m-2)=0$
$\Rightarrow \, m=\frac{2}{3}, 6$