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

Explanation

Given equation is `x^(2)+2(m-1)x+(m+5)=0`
Comparing with `ax^(2)+bx+c=0`, we get a=1, b=2(m-1), c=m+5
The equation will have two real and equal roots if
Discriminant (D)=0
`:.D=b^(2)-4ac=0`
or `[2(m-1)]^(2)-4xx1xx(m+5)=0`
`implies4(m^(2)+1-2m)-4(m+5)=0`
`implies4m^(2)+4-8m-4m-20=0`
`implies4m^(2)-12m-16=0`
`impliesm^(2)-3m-4=0`
`impliesm^(2)-4m+m-4=0`
`impliesm(m-4)+1(m-4)=0`
`implies(m+1)(m-4)=0`
`impliesm=-1orm=4` Hence, value (s) of m are -1 and 4.