Find that value of k in the following quadratic equation whose roots are real and equal:

(k + 4) x² + (k + 1) x + 1 = 0

When we compare the above quadratic equation with the generalized one we get,

ax² + bx + c = 0

a = k + 4

b = k + 1

c = 1

Since the quadratic equations have real and equal roots,

b² – 4ac = 0 for real and equal roots

⟹ (k + 1) ² – (4 × (k + 4) × 1) = 0

⟹ k² + 2k + 1 – 4k - 16 = 0

⟹ k² - 2k – 15 = 0

⟹ k² - 2k – 15 = 0

On factorizing the above equation,

Sum = -2

Product = -15

Therefore the two numbers satisfying the above conditions are 3 and -5.

k² – 5k + 3k – 15 = 0

k(k – 5) + 3(k – 5) = 0

(k + 3) (k – 5) = 0

Solving first part,

k + 3 = 0

k = -3

Solving second part,

k – 5 = 0

k = 5