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