Question

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

Answer 1

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

Since the given quadratic equation has equal roots, its discriminant should be zero.

∴ D = 0

⇒ 4(k+1)²−4×(3k+1)×1=0

⇒4k²+8k+4−12k−4=0

⇒4k²−4k=0

⇒k(k−1)=0

⇒k=0 or 1

Thus, the values of k are 0 and 1.

For k = 0,

(3k+1)x²+2(k+1)x+1=0

⇒x2+2x+1=0

⇒(x+1)2=0

⇒x=−1, −1

For k = 1,

(3k+1)x²+2(k+1)x+1=0

⇒4x²+4x+1=0

⇒(2x+1)²=0

⇒x=−1/2,−1/2

Thus, the equal roots are − 1 and −1/2.