Question

For what value of k,the quadratic equation (4-k)x^2+(2k+4)x+8k+1=0 is a perfect square.

Answer 1

Answer:

i dont now

Step-by-step explanation:

Answer 2

comparing the equation with ax^2 + bx + c = 0

therefore we get a = (4-k)  ,  b= (2k + 4 )  ,  c = 8k+ 1

therefore delta = 0

0 = b^2 - 4ac

0= (2k + 4 ) ^2 - 4 (4 - k) (8k+1)

0 = 4k^2 + 16k + 16 - 128k + 32k^2 - 16k = 4k

0= 12k^2 - 36k

12k^2 = 36k

k^2 = 3k

k = root of 3k

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