Question

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

Answer 1

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

This is of the form ax² + bx + c = 0, where a = 2p + 1, b = –(7p – 2) and c = 7p – 3 

∴ D = b² – 4ac 

= –[–7p + 2]² – 4 × (2p + 1) × (7p – 3) 

= (49p² + 28p + 4) – 4(14p² + p – 3) 

= 49p² + 28p + 4 – 56p² – 4p + 12 

= –7p² + 24p + 16 

The given equation will have real and equal roots if D = 0. 

∴ –7p² + 24p + 16 = 0  

⇒ 7p² – 24p – 16 = 0 

⇒ 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` 

Hence, 4 and `(-4)/7` are the required values of p.