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Question
Without solving the following quadratic equation, find the value of ‘p’ for which the given equation has 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
For real and equal roots
D = b2 - 4ac = 0
(p - 3)2 - 4 x 1 x p = 0
p2 - 6p + 9 - 4p = 0
p2 - 10p + 9 = 0
⇒ p2 - p - 9 (p - 1) = 0
p(p - 1) - 9 (p - 1) = 0
⇒ (p - 1) (p - 9) = 0
⇒ p = 1 or p = + 9
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