Find the value of p for which the quadratic equation = 0 has equal roots. Also, find these roots.

Given quadratic equation, (2p + 1) x² - (7p + 2) x + (7p – 3) = 0

Comparing the given equation with ax² + bx + c = 0, we get

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

We know that a quadratic equation ax² + bx + c = 0 has equal roots (i.e., coincident roots), if b² – 4ac = 0.

∴ (-(7p + 2))² – 4(2p + 1) (7p - 3) = 0

⇒ 49p² + 28p + 4 – 56p² + 24p – 28p + 12 = 0

⇒ -7p² + 24p + 16 = 0

⇒ 7p² - 24p - 16 = 0

By factorization method,

⇒ 7p² – 28p + 4p – 16 = 0

⇒ 7p (p – 4) + 4(p – 4) = 0

⇒ (p – 4) (7p + 4) = 0

⇒ (p - 4) = 0 (or) (7p + 4) = 0

⇒ p = 4 (or) p = -4/7

For equal roots, roots =

⇒ Roots =

Case 1: When p = 4

⇒ Roots = (7(4) + 2)/2(2(4) + 1)

= 30/18

= 5/3, 5/3

Case 2: When p = -4/7

⇒ Roots = (7(-4/7) + 2)/2(2(-4/7) + 1)

= -2/(-2/7)

= 7, 7

The value of p is 4 or -4/7. The roots are 5/3, 5/3 or 7, 7.