# 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 $x = 4$.

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Hah, just realized I could have inputted the LaTeX directly in the post instead of fiddling around with the CodeCogs Equation Editor. –  Andreas Grech May 24 '11 at 23:17
I'm impressed by your effort to show your work, and with such care! Yes, being able to use LaTeX on this site is a really nice feature. –  amWhy May 24 '11 at 23:41
Also: just to chime in with user9716, there was/is nothing wrong with the procedure you used for solving the given equation. The thing to remember is that when you then solve for a variable, you arrive at potential solutions: if there is/are a solution, it will be among the potential solutions. So it was smart for you to have checked both values; that's the only way to confirm or discard potential solutions. –  amWhy May 24 '11 at 23:53
This is a classic example of how we lie to students when we first teach them algebra. When you perform a sequence of algebraic manipulations on an equation to solve for $x$, you are really showing an implication: $\sqrt{x+5}=x-1$ implies that $x=-1$ or $x=4$. However we haven't shown the reverse implication: $x=-1$ or $x=4$ implies $\sqrt{x+5}=x-1$, and in fact this reverse implication is false because squaring both sides is not a reversible step. –  Grumpy Parsnip May 25 '11 at 3:50
Solve the equation: $x=1$. Step 1) Square both sides: $x^2 = 1$. Step 2) Roots are $+1$ and $-1$... –  Aryabhata May 25 '11 at 15:29

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 :)

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+1 Thanks, that's what I was confused about before, because I thought they where solutions, not potential solutions. –  Andreas Grech May 28 '11 at 9:05

The problem is: $$x+5 = (x - 1)^2$$ doesn't imply $$\sqrt{x + 5} = x - 1$$ but $$\sqrt{x + 5} = \vert x - 1\vert$$

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But what is the reason it implies the absolute value? –  Andreas Grech May 24 '11 at 23:20
Roots return always positive numbers, so $\sqrt{a^2}$ is $a$ if $a\ge 0$, $-a$ otherwise. –  AlbertH May 24 '11 at 23:31

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

$$A^n-B^n=(A-B)(A^{n-1}+A^{n-2}B+\cdots +AB^{n-2}+B^{n-1}).$$

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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.

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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 $2 \neq -2$

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There is no error. –  Did May 25 '11 at 12:10
I didn't understand your answer. Why do you have 3 lines showing 4 = 4 and two lines showing 0 = 0? –  Andreas Grech May 25 '11 at 12:38
He went line by line in your question under "So I used this logic" and substituted $-1$ for $x$ in each line, then evaluated each side of the equation. So the last $4=4$ comes from $x+5=x^2-2x+1$ –  Ross Millikan May 25 '11 at 13:10

$\sqrt{x+5} = x - 1$ Because, you need to put condition $x\geq 1$ So, If you tried to plug $x = -1 < 1$, which not satified.

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

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

3. 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.

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