For what value of k (4-k)x^2+(2k+4)x+(8k+1)=0 is a perfect square?
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For a perfect square, put D = b^2 – 4ac = 0
[2k + 4]^2 – 4 [4 – k] [8k + 1] = 0
4k^2 + 16 + 16k – 4 [32k + 4 – 8k^2 – k] = 0
4k^2 + 16 + 16k – 128k – 16 + 32k^2 + 4k = 0
36k^2 – 108 k = 0
k^2 – 3k = 0
k(k – 3) = 0
k = 0 or k = 3