Question

Without solving the following quadratic equation, find the value of ‘p’ for which the given equations have real and equal roots: x² + (p – 3)x + p = 0.

Answer 1

x² + (p – 3)x + p = 0

Here a = 1, b = (p - 3), c = p

∵ Equation has real and equal roots

∴ b² - 4ac = 0

⇒ (p - 3)² - 4(1)(p) = 0

⇒ (p - 3)² - 4p = 0

⇒ p² + 9 - 6p - 4p = 0

⇒ p² - 10p + 9 = 0

⇒ p² - 9p - p + 9 = 0

⇒ p(p - 9) -1(p - 9) = 0

⇒ (p - 1)(p - 9) = 0

∴ p = 1, 9