For what values of m, is mx^{2}+mx+8x+9 a perfect square?

Explanation

Rewrite the expression as mx^2 + (m+8)x + 9. For it to be a perfect square, its discriminant must be zero: D = b^2 - 4ac = (m+8)^2 - 4(m)(9) = 0. This gives m^2 + 16m + 64 - 36m = 0 \implies m^2 - 20m + 64 = 0, which factors to (m-4)(m-16) = 0. So, m = 4, 16.

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