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Question
Solve the following quadratic equation.
(2x + 3)2 = 25
Solution 1
(2x + 3)2 = 25
⇒ 4x2 + 12x + 9 = 25
⇒ 4x2 + 12x + 9 − 25 = 0
⇒ 4x2 + 12x − 16 = 0
Dividing whole equation by 4,
⇒ x2 + 3x − 4 = 0
⇒ x2 + 4x − x − 4 = 0
⇒ x(x + 4) − 1(x + 4) = 0
⇒ (x + 4)(x − 1) = 0
⇒ x = 1, −4.
Solution 2
(2x + 3)2 = 25
∴ (2x + 3)2 – 25 = 0
∴ (2x + 3)2 – (5)2 = 0
∴ (2x + 3 – 5)(2x + 3 + 5) = 0 ...[∵ a2 – b = (a – b)(a + b)]
∴ (2x – 2)(2x + 8) = 0
By using the property, if the product of two numbers is zero, then at least one of them is zero, we get,
∴ 2x – 2 = 0 or 2x + 8 = 0
∴ 2x = 2 or 2x = –8
∴ x = `2/2` or x = `(–8)/2`
∴ x = 1 or x = –4
∴ The roots of the given quadratic equation are 1 and –4.
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