The equation x^2+ kx + 8 = k has no real solutions for x. Show that k satisfies k^2 + 4k < 32.
If a quadractic equation ax^2 + bx + c = 0 has no real solutions, this means that the discriminant is less than 0, i.e. b^2-4ac<0. Let's put our equation in this form: x^2 + kx + 8 = k rearranges to x^2 + kx + (8-k) = 0. So in this case a=1, b=k, c = 8-k. Therefore no real solutions means that k^2 - 41(8-k) < 0 (substituting our values of a, b and c) and then k^2 - 32 + 4k < 0 (multiplying out) and then k^2 + 4k < 32. (rearranging) And we're done!
TD
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