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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.
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Solution
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
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