Question

Find the value 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

Step-by-step explanation:

here, k=0 is possible for it forms a quadratic equtaion i.e. value of a in a quadratic equation is not zero

Answer 2

Answer:

Step-by-step explanation:

Solution :-

Since, he given equation has equal roots,

D = b² - 4ac = 0

Here, a = (3k + 1), b = 2(k + 1) and c = 1

[2(k + 1)]² - 4(3k + 1)(1) = 0

⇒ 4(k² + 2k + 1) - (12k + 4) = 0

⇒ 4k² + 8k + 4 - 12k - 4 = 0

⇒ 4k² - 4k = 0

⇒ k = 0, 1.

Put k = 0, in the given equation,

⇒ x² + 2x + 1 = 0

⇒ (x + 1)² = 0

⇒ x = - 1.

Again put k = 1, in the given equation,

⇒ 4x² + 4x + 1 = 0

⇒ (2x + 1)² = 0

⇒ x = - 1/2

Hence, the roots are - 1, -1, - 1/2 and - 1/2.