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Question
Find the value(s) of p for which the quadratic equation (2p + 1)x2 – (7p + 2)x + (7p – 3) = 0 has equal roots. Also find these roots.
Solution
The quadratic equation given is (2p + 1)x2 – (7p + 2)x + (7p – 3) = 0
Comparing with ax² + bx + c = 0, we have
a = 2p + 1, b = -(7p + 2), c = (7p - 3)
D = b2 - 4ac
⇒ 0 = [-(7p + 2)]2 -4(2p + 1)(7p - 3)
0 = 49p2 + 4 + 28p - 4(14p2 - 6p + 7p - 3)
0 = 49p2 + 4 + 28p - 56p2 - 4p + 12
0 = -7p2 + 24p + 16
0 = -7p2 + 28p - 4p + 16
0 = -7p(p - 4) -4(p - 4)
0 = (-7p - 4)(p - 4)
⇒ -7p - 4 = 0 or p - 4 = 0
Hence, the value of p = `(-4)/(7)` or p = 4.
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