Question

If equation $x^2-(2+m)x+1(m^2-4m+4)=0$ has coincident roots, then :

• A. $m=0$
• B. $m=6$
• C. $m=2$
• D. $m=\frac{2}{3}$

Answer 1

Answer: (B), (D)

$m=6$
$m=\frac{2}{3}$

Given: $x^2-(2+m)x+1(m^2-4m+4)=0$ has coincident roots

Therefore $b^2-4ac=0$

here $a=1,b=-(2+m),c=(m^2-4m+4)$

$\Rightarrow [-(2+m){]}^2-4\times 1\times (m^2-4m+4)=0$

$\Rightarrow 4+m^2+4m-4m^2-16m-16=0$

$\Rightarrow -3m^2+20m-12=0$

$\Rightarrow 3m^2-20m+12=0$

$\Rightarrow 3m^2-2m-18m+12=0$

$\Rightarrow m(3m-2)-6(3m-2)=0$

$\Rightarrow (m-6)(3m-2)=0$

$\therefore m=\frac{2}{3},6$