Find the values of k for which the quadratic equation (k+4) x² + (k+1) x + 1 = 0 has equal roots. Also, find these roots.

Given quadratic equation, (k + 4) x² + (k + 1) x + 1 = 0

Comparing the given equation with ax² + bx + c = 0, we get

a = (k + 4), b = (k + 1), c = 1

We know that a quadratic equation ax² + bx + c = 0 has equal roots (i.e., coincident roots), if b² – 4ac = 0.

∴ (k + 1)² – 4(k + 4) (1) = 0

⇒ k² + 2k + 1 – 4k – 16 = 0

⇒ k² – 2k – 15 = 0

By factorization method,

⇒ k² – 5k + 3k – 15 = 0

⇒ k(k – 5) + 3(k – 5) = 0

⇒ (k – 5) (k + 3) = 0

⇒ (k - 5) = 0 (or) (k + 3) = 0

⇒ k = 5 (or) k = -3

For equal roots, roots =

⇒ Roots =

Case 1: When k = 5

⇒ Roots = - (5 + 1)/2(5 + 4)

= -6/18

= -1/3, -1/3

Case 2: When k = -3

⇒ Roots = -(-3 + 1)/2(-3 + 4)

= 2/2

= 1, 1

Ans. The value of k is 5 or -3.The roots are -1/3, -1/3 or 1, 1.