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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.
Solution
x2 + (p – 3)x + p = 0
Comparing with ax2 + bx + c = 0, we have
a = 1, b = (p – 3), c = p
Since, the roots are real and equal, D = 0
`⇒ b^2 - 4ac = 0`
`=> (p - 3)^2 - 4(1)(p) = 0`
`=> p^2 + 9 - 6p - 4p = 0`
`=> p^2 - 10p + 9 = 0`
`=> (p - 1)(p - 9) = 0`
`=> p = 1 or p = 9`
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