Question

Without solving the following quadratic equation, find the value of m for which the given equation has equation has real and equal roots.

`x^2 + 2(m - 1)x + (m + 5) = 0`

Answer 1

Given Quadratic Equation is

`x^2 + 2(m - 1)x + (m + 5) = 0`

Here a = 1,b = 2(m - 1) and c = (m + 5)

Discriminant isgiven by `D^2 = b^2 - 4ac`

For Real and equal roots, D = 0

`=> b^2 - 4ac = 0`

`=> [2(m - 1)]^2 - 4(m + 5) = 0`

`=> m^2 + 1 - 2m - m - 5 = 0`

`=>  m^2 - 3m - 4 = 0`

Factorising we get (m + 1)(m - 4) = 0

=> m = -1 or m = 4