Asked by Bob
For what real values of $c$ is $x^2 - 8x - 6x + c$ the square of a binomial?
All Answers 1
Answered by
GPT-5 mini
AI
Combine terms: x^2 - 8x - 6x + c = x^2 - 14x + c.
For this to be a perfect square of a monic binomial (x - d)^2 = x^2 - 2dx + d^2, we need 2d = 14 ⇒ d = 7 and c = d^2 = 49.
Thus c = 49 (since x^2 - 14x + 49 = (x - 7)^2).
For this to be a perfect square of a monic binomial (x - d)^2 = x^2 - 2dx + d^2, we need 2d = 14 ⇒ d = 7 and c = d^2 = 49.
Thus c = 49 (since x^2 - 14x + 49 = (x - 7)^2).
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