Q) 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

Ans: 

Step 1: First, let’s compare the given equation with standard quadratic equation ax² + bx + c = 0, we get:
a = 1
b = 2 (m – 1)
c = (m + 5)

Step 2: We know that a standard quadratic equation to have real and equal roots, its discriminant should be zero i.e. D = 0

∴ b² – 4ac = 0

By substituting values of a, b & c, we get:

(2 (m – 1))² – 4 x 1 x (m + 5) = 0

∴ 4 (m – 1)² – 4 (m + 5) = 0

∴ (m – 1)² – (m + 5) = 0

∴ (m² – 2 m + 1) – m – 5 = 0

∴ m² – 2 m + 1 – m – 5 = 0

∴ m² – 3 m – 4 = 0

∴ m² – 4 m + m – 4 = 0

∴ m (m – 4) + 1 (m – 4) = 0

∴ (m – 4) (m + 1) = 0

∴ m = 4, m = – 1

Therefore, the value of m is 4 and – 1.

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