Question

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

x² + 2(m – 1)x + (m + 5) = 0

Answer 1

x² + 2(m – 1)x + (m + 5) = 0

Equating with ax² + bx + c = 0

a = 1, b = 2(m – 1), c = (m + 5)

Since equation has real and equal roots.

So, D = 0

`=>` b² – 4ac = 0

`=>` [2(m – 1)² – 4 × 1 × (m + 5) = 0

`=>` 4(m – 1)² – 4 (m + 5) = 0

`=>` 4[m² – 2m + 1 – m – 5)] = 0

`=>` m² – 3m – 4 = 0

`=>` (m + 1)(m – 4) = 0

Either m + 1 = 0 or m – 4 = 0

`=>` m = –1 or m = 4

Hence, the values of m are –1, 4.