+1439 Let f(a,b) = 2a - 3b^2 + 7 - 4ab + 8a^2. If $f(k,-3) = -10$, then what is $k$?
+1632 Substitution: f(k,-3) = 2k - 3(-3)^2 + 7 - 4(k)(-3) + 8(k)^2 = = 2k - 3*9 + 7 + 12k + 8k^2 = 8k^2 + 14k - 20 = -10
8k^2 + 14k - 10 = 0; divide by 2: 4k^2 + 7k - 5 = 0
Use quadratic formula: \(k = {-b \pm \sqrt{b^2-4ac} \over 2a}\); substituting, you have \(k = {-7\pm\sqrt{49+80}\over8}={-7\pm\sqrt{129}\over8}\), two possible values.
