In the following equation:

$$\sqrt{2x + 1} + 1 = x$$

You are supposed to isolate the radical:

$$\sqrt{2x + 1} = x - 1$$

And then proceed by squaring both sides.

If you start by solving the equation this way, you will eventually complete the square and get an answer of: $$4$$

However, why must the radical be isolated before squaring both sides?

Why can't you do, for example...

$$(\sqrt{2x + 1} + 1)^2 = x^2$$

I know this would lead you down the wrong path, but I don't know why. It doesn't make sense to me because I can (once I isolate the radical) square both sides when one side $$x-1$$

involves addition/subtraction. Is there some special property of radicals that makes them have to be completely alone before they can be squared?

Thank you.

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    $\begingroup$ There's nothing mathematically incorrect with $$(\sqrt{2x+1} +1)^2 = x^2$$ The reason you isolate the radical first is because then squaring both sides eliminates the radical. $\endgroup$ Commented Aug 31, 2016 at 22:56
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    $\begingroup$ Try it and see. If you don't isolate the radical and you square correctly you will have an expression that still has a radical in it. $\endgroup$ Commented Aug 31, 2016 at 22:57
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    $\begingroup$ As John Coleman said, you will get a radical again, and still square it, yet another radical and so on... $\endgroup$
    – user345851
    Commented Aug 31, 2016 at 22:58
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    $\begingroup$ When I first saw the title, I thought this was talking about radical people on two sides of a war... $\endgroup$
    – user541686
    Commented Sep 1, 2016 at 5:32
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    $\begingroup$ @Mehrdad And indeed the classic "square both sides" military strategy doesn't work unless all radicals have been isolated! ....wait. $\endgroup$ Commented Sep 2, 2016 at 17:15

9 Answers 9


You can of course write $$ (\sqrt{2x+1}+1)^2=x^2, $$ but when you multiply out you get $$ 2x+2\sqrt{2x+1}+2=x^2, $$ and there is still a radical in your new equation. The point in isolating the radical is that after that, as you square the equation, you get rid of it completely.


Well, it doesn't lead to the wrong path: if you square the equation right away you get that


And because $\sqrt{2x+1}=x-1$, you get the equation


This is $x^2-4x=0$ which has solutions $x=0,4$, but the solution $x=0$ doesn't do it because $\sqrt{1}$ is taken to be $1$ (and not $-1$).

(Sure, I'm replacing the square root by clearing $x-1$: everything you do to solve the equation will end up being equivalent. Point is that it is not wrong, just a bit more farfetched. It is interesting to note that mindlessly squaring forces one to note "clearing the square root" is necessary, and one can then realize one could have started doing this in the first place.)

  • $\begingroup$ +1 wow, fantastic answer to a question that I almost ended up ignoring. $\endgroup$
    – user541686
    Commented Sep 2, 2016 at 8:34

You can do what ever you darned well want. But you have to do it correctly. You can do this:

$\sqrt{2x + 1} + 1 = x$

$(\sqrt{2x + 1} + 1)^2 = x^2$

$(2x + 1) + 2\sqrt{2x + 1} + 1 = x^2$

but now what? ... it's true but you've just made things more difficult.

What you can NOT do under any circumstances is this:

$(\sqrt{2x + 1} + 1)^2 = x^2$

$(2x + 1) + 1 = x^2$

The point is if you want to solve it, then you want to get rid of the radical and you can't do that is you square a sum with other terms.

$(a + \sqrt{b})^2 = a^2 + 2a\sqrt{b} + b$

So that doesn't do anything to get rid of it. But it's not wrong. It's just... not what you want.


" Is there some special property of radicals that makes them have to be completely alone before they can be squared?"

Not really, everything has to be completely alone before you square it if you don't want the square to involve other ... things in it.

Actually, your question is a bit like asking "Why must we isolate before we divide:"

"$3x + 2 = 11$"

"Why do we isolate the $3x$

"$3x = 11 -2$

"Why don't we just divide first:

""$(3x +2)/3 = 11/3$

The answer is we can do what we darned well like:

$x + \frac 23 = \frac {11}3$

$x = \frac {11}3 - \frac 23$.

Nothing wrong with that... but nothing right with it either.

We isolate terms, for whatever operation, for the purpose of isolating them so we can work directly with them.

That's all.

  • $\begingroup$ Once you've got $(2x+1) + 2\sqrt{2x+1} + 1 = x^2$, you can put $x-1$ in place of $\sqrt{2x+1}$. You get a quadratic equation, and some extraneous roots. So it's not impossible to do it that way; it's just more complicated. $\qquad$ $\endgroup$ Commented Aug 31, 2016 at 23:10
  • $\begingroup$ @MichaelHardy You didn't see my post, did you? =) $\endgroup$
    – Pedro
    Commented Aug 31, 2016 at 23:13
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    $\begingroup$ I never said it was impossible. Just that it gets you where it takes you which is probably not where you want to go. And you can't really replace $x-1$ with $\sqrt{2x+1}$ when you never isolated the radical first. $\endgroup$
    – fleablood
    Commented Aug 31, 2016 at 23:14

There is (often) another way to solve these equations that isn't really easier but at least I think it is interesting.

\begin{align} \sqrt{2x + 1} + 1 &= x \\ (\sqrt{2x + 1} - 1)(\sqrt{2x + 1} + 1) &= (\sqrt{2x + 1} - 1)x \\ 2x &= (\sqrt{2x + 1} - 1)x \\ \sqrt{2x + 1} - 1 &= 2 &\text{(Need to check $x=0$ is not a solution.)}\\ \sqrt{2x + 1} &= 3 \\ 2x + 1 &= 9 \\ x &= 4 \end{align}

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    $\begingroup$ The problem with this one is that $\sqrt{2x + 1}$ - 1 can be zero so -- as always -- you need to check your solution is indeed correct. In very broad terms, once a radical is involved you need to check back. $\endgroup$
    – chx
    Commented Sep 1, 2016 at 10:02
  • $\begingroup$ @chx 8 - Yes, you do need to check it. $\endgroup$ Commented Sep 1, 2016 at 21:56
  • $\begingroup$ @chx $\sqrt{2x+1}-1=0$ in fact corresponds to $x=0$. Of course, squaring both sides also adds extraneous roots. $\endgroup$ Commented Sep 3, 2016 at 21:26
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    $\begingroup$ you are still isolating square root beetwen lines 4 and 5 $\endgroup$
    – RiaD
    Commented Dec 1, 2017 at 17:23

The main problem with your approach is that squaring will still leave a radical, whereas isolating the radical won't. So it's mostly about simplicity, rather than doing some compulsory transformation.

However, squaring is not sufficient. You can only do $$ (\sqrt{2x+1}+1)^2=x^2 $$ under the assumption that $x\ge0$, or you may add some spurious solutions.

There is a different approach, though. You can set $t=\sqrt{2x+1}$, with $t\ge0$, and so $2x+1=t^2$ and $$ x=\frac{t^2-1}{2} $$ so your equation becomes $$ t+1=\frac{t^2-1}{2} $$ that simplifies to $$ t^2-2t-3=0 $$ The roots are $t=-1$ and $t=3$, but the negative root must be discarded. Therefore we get $\sqrt{2x+1}=3$, that easily gives $x=4$.

The alternative method of isolating the radical is, however, simpler.

We get $\sqrt{2x+1}=x-1$, which, under the condition $x-1\ge0$, can be squared: $$ 2x+1=x^2-2x+1 $$ and so $x^2-4x=0$. The roots are $0$ and $4$, but only the latter satisfies $x\ge1$.


The reason that the radical must stand alone is that the square root and the square are inverse operations of one another. So in order for the radical to disappear (i.e. for the square root operation and the squaring operation to cancel one another), the squaring must be applied to the pure radical, not the radical plus some other stuff.

  • $\begingroup$ It must not stand all alone, it may be a factor of a product, but not a sum's summand. $\endgroup$ Commented Aug 31, 2016 at 23:01
  • $\begingroup$ @MichaelHoppe True. But squaring a product amounts to squaring each factor, and thus the the radical itself is squared in that case. No such luck for sums. $\endgroup$
    – Arthur
    Commented Aug 31, 2016 at 23:02
  • $\begingroup$ I added a comment to this effect before noticing your answer. This point is crucial. $\endgroup$ Commented Sep 2, 2016 at 17:23

There is no need to isolate the radical. However, you usually want to isolate the radical in order to simplify computations. From $$\sqrt{2x+1}+1=x=\frac{\left(\sqrt{2x+1}\right)^2-1}{2}\,,$$ we have $$\left(\sqrt{2x+1}\right)^2-2\,\sqrt{2x+1}-3=0\,.$$ Thus, $$\left(\sqrt{2x+1}-3\right)\,\left(\sqrt{2x+1}+1\right)=0\,.$$ Since $\sqrt{2x+1}+1>0$, we get $$\sqrt{2x+1}-3=0\,.$$ Thus, $x=4$. (Well, the last part still requires isolation of the radical: $\sqrt{2x+1}=3$.)


Why can't you do, for example... $$(\sqrt{2x + 1} + 1)^2 = x^2$$

Just in case you're overlooking this: It's critically important to realize that exponents do not distribute over additions. That is, if you think that the left-hand side above simplifies to $(2x+1) + 1$, that's false.

As others have pointed out, when you properly expand a binomial square in this way as per $(a+b)^2 = a^2 + 2ab + b^2$, you'll still have a radical in the expression. Only by isolating everything under the radical do the square-power and square-root elegantly cancel out.

Understanding binomial squares is one of the most important basic facts at this level of algebra/precalculus.


$$(\sqrt{2x + 1} + 1)^2 = x^2$$ $$\implies2x+2\sqrt{2x+1}+2=x^2$$ $$\implies2\sqrt{2x+1}=x^2 -2x -2$$ $$\implies4(2x+1)=(x^2 -2x -2)^2$$ $$\implies4(2x+1)=x^4 +4x^2 + 4 - 4x^3 + 8x -4x^2$$ $$\implies0=x^4 +4x^2 + 4 - 4x^3 + 8x -4x^2 -8x -4 $$ $$\implies 0=x^4 - 4x^3$$ $$\therefore x= 4$$

Do anything as long as LHS = RHS

  • $\begingroup$ You are giving also a wrong solution: if $x=0$, the left-hand side is $2$, and the right-hand side is $0$. $\endgroup$
    – egreg
    Commented Sep 1, 2016 at 12:56

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