​​​​​​(4-k)x^2 + (2k+4)x + (8k+1) = 0

Asked by saitipparthi79.10spicertl | 14 May, 2020, 09:20: AM
Question: For what value of k, ​​​​​​(4-k)x2 + (2k+4)x + (8k+1) = 0 is a perfect square?
Solution: (4-k)x2 + (2k+4)x + (8k+1) = 0 is a perfect square, which means the roots of the quadratic equation are equal.
Therefore, its discriminant is zero.
Therefore, b2-4ac=0
(2k+4)2 - 4(4-k)(8k+1) = 0
4k2 + 16 + 16k - 4(31k + 4 - 8k2) = 0
4k2 + 16 + 16k - 124k - 16 + 32k2 = 0
-36k2 - 108k = 0
36k(k + 3) = 0
k=0 or k=-3
Answered by Renu Varma | 14 May, 2020, 11:38: AM

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