Question

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

Answer 1

The given equation is (3k + 1)x² + 2(k + 1)x + 1 = 0

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

∴ D = b² – 4ac 

= [2(k + 1)]² – 4 × (3k + 1) × 1 

= 4(k² + 2k + 1) – 4(3k + 1)

= 4k² + 8k + 4 – 12k – 4  

= 4k² – 4k 

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

∴ 4k² – 4k = 0 

⇒ 4k(k – 1) = 0

⇒ k = 0 or k – 1 = 0 

⇒ k = 0 or k = 1 

Hence, 0 and 1 are the required values of k.