Q. 95.0( 6 Votes )

Find the values of p for which the quadratic equation (2p + 1)x2 – (7p + 2)x + (7p – 3) = 0 has real and equal roots.

Answer :

Given equation is (2p + 1)x2 – (7p + 2)x + (7p – 3) = 0

Comparing with standard quadratic equation ax2 + bx + c = 0


a = (2p + 1) b = – (7p + 2) c = (7p – 3)


Given that the roots of equation are real and equal


Thus D = 0


Discriminant D = b2 – 4ac = 0


[ – (7p + 2)]2 – 4.(2p + 1).(7p – 3) = 0 using (a + b)2 = a2 + 2ab + b2


(49p2 + 28p + 4) – 4(14p2 + p – 3) = 0


49p2 + 28p + 4 – 56p2 – 4p + 12 = 0


– 7p2 + 24p + 16 = 0


7p2 – 24p – 16 = 0


7p2 – 28p + 4p – 16 = 0


7p(p – 4) + 4(p – 4) = 0


(7p + 4)(p – 4) = 0


(7p + 4) = 0 or (p – 4) = 0



The values of p are for which roots of the quadratic equation are real and equal.


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