<s>[INST] Solving $\sqrt{x+5} = x - 1$ I'm currently learning about radicals and simplifying them, and I came across this problem on the internet and tried to solve it:
$$\sqrt{x+5} = x - 1$$
So I used this logic:
$$
\begin{align}
\sqrt{x+5} &= x - 1 \\
x + 5 &= (x-1)^2 \\
x + 5 &= (x-1)(x-1) \\
x + 5 &= x^2 - 2x + 1 \\
0 &= x^2 - 3x - 4 \\
0 &= (x-4)(x+1) \\
\end{align}
$$
Therefore, $x = -1$ and $x = 4$ satisfy the equation $0 = (x-4)(x+1)$.
But then I tried to plug them in the original problem $\sqrt{x+5}=x-1$:
$$
\begin{align}
\sqrt{4 + 5} &= 4 - 1 \\
\sqrt{9} &= 3 \\
3 &= 3
\end{align}
$$
So using 4 works as expected, but when using $-1$:
$$
\begin{align}
\sqrt{-1 + 5} &= -1 - 1 \\
\sqrt{4} &= -2 \\
2 &\ne -2
\end{align}
$$
At what stage am I going wrong?
And according the WolframAlpha, the solution is $x = 4$.
 [/INST] *

*Function on the left hand side is strictly increasing (i.e always increasing).

*Function on the right hand side is strictly increasing (i.e always increasing) too.

*We draw a graph to find out where too look for the roots. You can use this or any other tool to draw two graphs for left and right hand side functions. We'll see that there is one interception at $x=4$ and there will be no other interceptions because functions striclty increase.
</s>