Question

Find the value of k for each quadratic equation has equal root (k+4)x²+(k+1)X+1=0

Answer 1

Answer:

k = 5,-3

Step-by-step explanation:

(k+4)x²+(k+1)x+1 = 0

It has equal roots.

Therefore, D = 0

D = b²-4ac

= (k+1)²-4×(k+4)×1 = 0

k²+1+2k-4k-16 = 0

k²-2k-15 = 0

k²-5k+3k-15 = 0

k(k-5)+3(k-5) = 0

(k+3)(k-5) = 0

Therefore, k = 5,-3

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