Find the largest value of $c$ such that $\frac{c^2 + 6c -27}{c-3} +2c = 8$.

Answer 1

First, we notice that $\frac{c^2 + 6c - 27}{c - 3}$ can be rewritten as $c+\frac{18}{c-3}$. Thus, our equation becomes $c+\frac{18}{c-3}+2c=8$. We simplify and combine like terms to find that $c+\frac{18}{c-3}+2c=3c+\frac{18}{c-3}=8$. Hence $3c^2-9c+18=8c-24$, which rearranges to $3c^2-17c+18=0$. By the quadratic formula, we find that $c=\frac{17 \pm \sqrt{17^2 - 4(3)(18)}}{6} = \frac{17 \pm \sqrt{13}}{6}$. Since $c$ must be positive, the maximum value is $\frac{17 + \sqrt{13}}{6} = \boxed{\frac{17 + \sqrt{13}}{6}}$.