Question

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

Answer 1

Given Equation is (2p + 1)x^2 - (7p + 2)x + (7p - 3) = 0.

Here, a = (2p + 1), b = -(7p + 2), c = (7p - 3).

Given that the equation has real and equal roots.

⇒ D = b^2 - 4ac = 0

       = [-(7p + 2)]^2 - 4(2p + 1)(7p - 3) = 0

       = 49p^2 + 28p + 4 - 4(14p^2 - 6p + 7p - 3) = 0

       = 49p^2 + 28p + 4 - 56p^2 + 24p - 28p + 12 = 0

       = -7p^2 + 24p + 16 = 0

       = 7p^2 - 24p - 16 = 0

       = 7p^2 - 28p + 4p - 16 = 0

       = 7p(p - 4) + 4(p - 4) = 0

       = (7p + 4)(p - 4) = 0

        p = 4, p = -4/7.

Therefore, the value of p = 4, -4/7.

Hope this helps!