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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