Question

If the equation $x^2-(2+m)x+(m^2-4m+4)=0$ 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)

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