Question

Find the value of k for which the quadratic equation (k + 4)x2 + (k+1)x + 1 = 0 has two real equal roots​

Answer 1

(k+4)x2 +(k+1)x+1=0 D=b2 -4ac =(k+1)2-4(k+4)(1) =k2 +2k+1-4k-16 =k2 -2k-15 For equal roots, D=0 D=0 K2 -2k-15=0 k2 -5k+3k-15=0 k(k-5)+3(k-5)=0 (k+3)(k-5)=0 k+3=0 OR k-5=0 k = -3, k = 5

Answer 2

Answer:

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