Answer :

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

ax^{2} + bx + c = 0

a = k + 4

b = k + 1

c = 1

Since the quadratic equations have real and equal roots,

b^{2} – 4ac = 0 for real and equal roots

⟹ (k + 1) ^{2} – (4 × (k + 4) × 1) = 0

⟹ k^{2} + 2k + 1 – 4k - 16 = 0

⟹ k^{2} - 2k – 15 = 0

⟹ k^{2} - 2k – 15 = 0

On factorizing the above equation,

Sum = -2

Product = -15

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

k^{2} – 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

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