Question

Find the values of k so that the equation (k+4) x² +(k +1) x +1 =0 has real & equal roots

Answer 1

Answer:

Here is your answer...

Step-by-step explanation:

Discriminant = b^2 - 4ac 

                     = (k+1)^2 - 4(1)(k+4)

                     = k^2 +1+2k -4k - 16 

                     = k^2 - 2k - 15

For equal roots, discriminant = 0

k^2 - 2k - 15 = 0 

k^2 - (5 - 3)k - 15 = 0 

k^2 - 5k + 3k - 15 = 0 

k(k - 5) + 3(k - 5) = 0 

(k - 5)(k + 3) = 0 

k = 5 or k = -3 

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Answer 2

Answer:

Step-by-step explanation:

Since the equation have a real equal root

b²=4ac or b²—4ac=0

(K+1)²=4(k+4)

K²+2k+1=4k+16

K²-2k-15=0

K²—5k+3k-15=0

K(k—5)+3(k-5)=0

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

K=-3, 5