Answer :
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
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