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 $x = 4$.
 A: The problem is that when you square both sides of an equation you may introduce new roots. The new equation has all the solutions of the original one, but may have some that are not.
From $A^{2}-B^{2}=\left( A-B\right) (A+B)$ you can say that $A-B=0\Rightarrow A^{2}-B^{2}=0$ and $A+B=0\Rightarrow A^{2}-B^{2}=0$. But while $A=B$ is a solution of $A-B=0$, $A=-B$ is not, unless $A=B=0$. In the question $A=\sqrt{x+5}$, $B=x-1$.
As I wrote in this answer, in general the equation $A^n=B^n$ has all the solutions of the equation $A=B$ but may have additional ones. This is a consequence of the algebraic identity 
$$A^n-B^n=(A-B)(A^{n-1}+A^{n-2}B+\cdots +AB^{n-2}+B^{n-1}).$$ 
A: Work backwards. Starting with $x=-1$ and $x=4$, when you work backwards through your steps and rewrite everything to get to the very beginning, you end by taking a square-root. When you take that square-root, you have two possible solutions: $+\sqrt{x+5}$ and $-\sqrt{x+5}\quad$. $x=-1$ solves one, $x=4$ solves the other.
A: Plugging $x= -1$ into each step of your work, you get :
$$\begin{align}2 = -2 \\ 4 = 4 \\ 4 = 4 \\ 4 = 4 \\ 0 = 0 \\ 0 = 0 \end{align}$$
Thus without any thinking, you know that the error is in the first step, because $2^2 = 4 = 4 = (-2)^2$, but $2 \neq -2$
A: At no stage...
Here is what happens, you squared the equation 
$$\sqrt{x+5} =x-1 \,.$$
But then, if the two sides of the equation have the same absolute value, but opposite signs they are not equal in this equation but they become equal at the next step.
This means that the equation $x+5=(x-1)^2$ possibly has more solutions. So the answers you got at the end are no necessarily the solution, they are just the possible solutions.
You have to check which one works.
+1 for showing your work :)
A: The problem is:
$$x+5 = (x - 1)^2$$
doesn't imply
$$\sqrt{x + 5} = x - 1$$
but
$$\sqrt{x + 5} = \vert x - 1\vert$$
A: $\sqrt{x+5} = x - 1$
Because, you need to put condition $x\geq 1$
So, If you tried to plug $x = -1 < 1$, which not satified.
A: *

*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.
