Question

For what values of k, the roots of the quadratic equation (k+4)x2+(k+1)x+1 = 0 are equal?
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Answer 1

Answer-

$\huge\red{Step\:by\:step\: explanation}$

a = k+4 , b = k+1 and, c = 1

∴ D = 0

⇒ (k+1)2-4(k+4) = 0

⇒ k2+2k+1-4k-16 = 0

⇒ k2-2k-15 = 0

⇒ k2-5k+3k-15 = 0

⇒ k(k-5)+3(k-5) = 0

⇒ (k+3)(k-5) = 0

∴ k=-3 or k=5

Hope it helps

Answer 2

Let,a = k+4 , b = k+1 and, c = 1

∴ D = 0

→ (k+1)2-4(k+4) = 0

→ k2+2k+1-4k-16 = 0

→k2-2k-15 = 0

→k2-5k+3k-15 = 0

→ k(k-5)+3(k-5) = 0

→(k+3)(k-5) = 0

∴ k=-3 or k=5

$<marquee>Hope it helps!!$