Question

The equation$x^2+2(m-1)x+m+5=0$ has an equal and real root. Find the value of $m.$

Answer 1

$x^2+2(m-1)x+m+5=0$ compare the given equation with $a{x}^2+bx+c=0$

$a=1,b=2(m-1),\&c=m+5$

For positive roots,$b^2-4ac=0$

$$\begin{gathered} \Rightarrow 4{(m-1)}^2–4(m+5)=0 \\ \Rightarrow 4{(m-1)}^2=4(m+5) \\ \Rightarrow m^2-2m+1=m+5 \\ \Rightarrow m^2-3m-4=0 \\ \Rightarrow (m-4)(m+1)=0 \\ \Rightarrow m=4andm=-1. \end{gathered}$$

Hence, the required value of $m$ is $4,\&-1.$