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)

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