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Question -

For what value of k, (4 – k) x² + (2k + 4) x + (8k + 1) = 0, is a perfect square.



Answer -

(4 – k) x² + (2k + 4) x + (8k + 1) = 0
Here, a = 4 – k, b = 2k + 4, c = 8k + 1

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