solve

Asked by palaksophia | 12th May, 2009, 11:34: AM

Expert Answer:

(4-k)x2 + (2k+4)x + (8k+1)=0

a = (4-k) b= 2k+4     c = 8k+1

For the equation to be a perfect square

b2 = 4ac   (2k+4)2  = 4(4-k)(8k+1)

k2+4+4k = 31k-8k2+4

9k2  =  27 k

k = 0 ; 3

 

Answered by  | 15th May, 2009, 11:04: AM

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