The largest value of $x$ that satisfies $\sqrt{x+1}=x-2$ can be written as $\dfrac{a+\sqrt{b}}{c}$ where $c$ has no common factor with $a$ and $\sqrt{b}$ and $b$ is not divisible by the square of any integer greater than 1. What is the value of $a+b+c$?

Answer 1

We start by attempting to square both sides of the given equation to get rid of the radical.
$$
(\sqrt{x+1})^2 = (\sqrt{x+1})^2 \implies x+1 = x^2 - 4x + 4
$$
Simplifying, we have
$$
x^2 -5x +3 = 0
$$
By factoring, we find
\begin{align*} (x-3)(x-1)&= 0 \\ x &= 3, 1 \end{align*}
Let's check to find that $x=1$ doesn't work, but $x=\boxed{3}$ does work. Checking this solution in the original equation, we get $\sqrt{3+1}=3-2$ which simplifies to $\sqrt4=1$, so this root works as well.