Question

Without solving the following quadratic equation, find the value of ′m′ for which the given equation has real and equal roots.
$x^2+2(m-1)x+(m+5)=0$

Answer 1

For real and equal roots of equation

$a{x}^2+bx+c=0,$

Discriminant $\left(D\right)=0$

Where,$D=b^2-4ac$

Since for equation $x^2+2(m-1)x+(m+5)=0$

Discriminant $\left(D\right)=\left[2\right(m-1){]}^2-4\times 1(m+5)=0$

$\Rightarrow 4(m-1{)}^2-4(m+5)=0$

$\Rightarrow 4\left[\right(m-1{)}^2-(m+5)]=0$

$\{\because (a-b{)}^2=a^2+b^2-2ab\}$

$\Rightarrow m^2+1-2m-m-5=0$

$\Rightarrow m^2-3m-4=0$

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

$\Rightarrow m(m-4)+1(m-4)=0$

$\Rightarrow (m-4)(m+1)=0$

$\Rightarrow m-4=0$ or $m+1=0$

$\Rightarrow m=4$ or $m=-1$

$\therefore$ For $m=4$ and $-1$,

given equation will have real and equal roots.