Question

(3k+1)x^2+2(k+1)x+k=0​

Answer 1

Answer:

Given equation is y (3k+1)x² + 2(k+

1)x+ 1 = 0

Also, it is given that the equation has

equal roots.

Then D = 0 ⇒ b² - 4ac = 0

a = 3k+1 b = 2(k+1) c = 1

b² - 4ac = [2(k+1)]² − 4(3k + 1) = 0

4k² + 4 + 8k 12k - 40

⇒ 4k² - 4k = 0

⇒ 4k (k − 1) = 0

⇒k=0 k=1

When k = 0

equation y = x² + 2x + 1 = 0

Roots are x = -1, -1

When k = 1

equation 4x² + 4x + 1 = 0

Roots are x = -1 /2, -1 /2