# How to square both the sides of an equation?

Question: $x^2 \sqrt{(x + 3)} = (x + 3)^{3/2}$

My solution: $x^4 (x + 3) = (x + 3)^3$

$=> (x + 3)^2 = x^4$

$=> (x + 3) = x^2$

$=> x^2 -x - 3 = 0$

$=> x = (1 \pm \sqrt{1 + 12})/2$

I understand that you can't really square on both the sides like I did in the first step, however, if this is not the way to do it, then how can you really solve an equation like this one (in which there's a square root on the LHS) without substitution?

• You can square both sides if they are both non-negative. – Arnaud D. Apr 20 '16 at 15:54
• @ArnaudD.: You can always square both sides. – TonyK Apr 20 '16 at 16:12
• @ArnaudD.: No, but you get a valid equation. $a = b \implies a^2=b^2$. – TonyK Apr 20 '16 at 16:20
• Going from the second to third line of your solution you take the square root of both sides. This is not allowed, as it may eliminate solutions. In this case it does not because you already have $x+3 \ge 0$ from the problem statement. If you started with $(x+3)^2=x^4$ it could be that $x+3 = -x^2$ It turns out that only has complex solutions, but you need to make an argument the square root is allowed. – Ross Millikan Apr 20 '16 at 22:00
• @user331377: that is correct because a number has two square roots. Consider $a^2=b^2$. This is true when $a=b$, but also when $a=-b$. If you take the square root you lose one solution. – Ross Millikan Apr 21 '16 at 3:45

Why can't you square both sides like you did? You absolutely can. If you do this you just need to make sure you didn't introduce any extraneous solutions. You can make sure of this by checking each solution you get. Admittedly that may be a little difficult with answers like $x = (1 \pm \sqrt{13})/2$, but it is what it is.

What I'm not sure of is how your $x^4$ became an $x^2$ when you divided both sides by $x+3$. Actually now that I look closer that $x^2$ just looks like a typo. Anyway.. I would generally not divide like that because then you lose solutions. In this case you lost the solution $x = -3$. It's better to factor rather than divide.

Recall that $y^{3/2} = y\sqrt{y}$, so in particular we have $(x+3)^{3/2} = (x+3)\sqrt{x+3}$. If you want to do this without squaring both sides, I'd proceed like this:

\begin{align} x^2\sqrt{x+3} &= (x+3)^{3/2}\\[0.3cm] x^2\sqrt{x+3} - (x+3)^{3/2} &= 0\\[0.3cm] x^2\sqrt{x+3} - (x+3)\sqrt{x+3} &= 0\\[0.3cm] \sqrt{x+3}\left(x^2 - (x+3)\right) &= 0\\[0.3cm] \sqrt{x+3}(x^2-x-3) &= 0 \end{align} So either $\sqrt{x+3} = 0$ or $x^2 - x - 3 = 0$. The first gives $x = -3$ and the second gives $x = (1 \pm \sqrt{13})/2$.

• It was a typo, my mistake! – MathEnthusiast Apr 20 '16 at 16:03
• Is there a rule of thumb for when we can divide both sides like that and when we can not? Also, squaring is perfectly fine as long as we check at the end, right? And if that seems like a lot, we can always factor, correct? – MathEnthusiast Apr 20 '16 at 16:05
• No worries, I figured it was a typo when I saw it was correctly $x^2$ in the next line. – tilper Apr 20 '16 at 16:05
• Additionally, we'll never lose a solution if we take the exponent of both the sides to the power of an odd number, correct? – MathEnthusiast Apr 20 '16 at 16:08
• Honestly I would say never divide like that. Yes, squaring is fine as long as you check your answer at the end (which I recommend always doing anyway) and as long as it makes sense to square (sometimes it just creates more work and isn't needed). I don't know that we can always factor but it's definitely something to keep in mind. From an advanced math point of view we can technically always factor, but this looks like it's for a precalc or college algebra class, in which case we would want to view things like $x^2 - x - 3$ as something we can't factor. – tilper Apr 20 '16 at 16:09

We have $$x^2\sqrt{x+3} = \sqrt{x+3}\cdot \vert x+3 \vert \implies \sqrt{x+3}\left(x^2-\vert x+3 \vert\right) = 0$$ In the previous statement, we made use of the fact that $$(x+3)^{3/2} = \sqrt{x+3}\cdot \vert x+3 \vert$$ Hence, we have either

• $\sqrt{x+3} = 0 \implies x = -3$
• $x^2-x-3 = 0$ and $x+3 > 0$. This implies $x=\dfrac{1\pm\sqrt{13}}2$
• $x^2+x+3 = 0$ and $x+3<0$. This gives us no solution.
• Is there absolutely no way of solving questions like this by squaring both the sides (it just seems way way easier)? Or, what about a question like this one: slader.com/textbook/…. How would we solve this? – MathEnthusiast Apr 20 '16 at 15:59
• He just squared both sides which seems wrong. – MathEnthusiast Apr 20 '16 at 15:59
• @Leg: if $x+3$ is negative, then $\sqrt{x+3}$ is undefined. So it makes no sense to say $(x+3)^{3/2} = \sqrt{x+3}\cdot|x+3|$. You are just muddying the waters for the OP. – TonyK Apr 20 '16 at 16:23
• @TonyK if it is negative than it is defined. The solution would be a complex number. – The Great Duck Jun 14 '16 at 19:56

You can square it like that, and the equality will still hold - remember these expressions are equal, so squaring them mean they are still equal. This can, however, produce spurious solutions - if you do this you should check that the values you get do indeed solve the given equation.

Note however, that $\sqrt{x+3} = (x+3)^{1/2}$, and have another look at the equation. Don't forget that if you divide by anything, you have to make sure it isn't $0$...

edit: the other comments do more involving the different cases arising from different values of $x$ and are quite clear

• I get (1 \pm \sqrt{13})/2 which is not the correct answer: slader.com/textbook/… – MathEnthusiast Apr 20 '16 at 16:01
• All right, thanks for the correction! – MathEnthusiast Apr 20 '16 at 16:06
• I deleted my comment because what I said was incorrect. However, the solutions in the link are still wrong. The error is in the very end with the back-substitution to $x$. The second-to-last line is incorrect. – tilper Apr 20 '16 at 16:12

You have to specify the equation domain of validity: $D=[-3,+\infty)$.

On this domain, \begin{align*}x^2 \sqrt{(x + 3)} = (x + 3)^{3/2}&\iff x^4(x + 3)= (x + 3)^3\iff(x+3)\Bigl[x^4-(x+3)^2\Bigr]=0 \\ &\iff(x+3)(x^2-x-3)(x^2+x+3)=0 \\&\iff(x+3)(x^2-x-3)=0 \qquad\text{on }\,D \end{align*} Now the second equation: $\;p(x)=x^2-x-3=0$ has two roots in $D$: $\dfrac{1\pm\sqrt{13}}2$, both of which are greater than $-3$.

Thus, this equation has exactly three roots.

• Why the down vote? – Bernard Apr 21 '16 at 11:46
• You've solved for the roots of $p(x)$ incorrectly. – MathematicsStudent1122 Jun 14 '16 at 19:01
• Silly me! Why didn't I check my computations? Thanks for pointing it. – Bernard Jun 14 '16 at 19:39

You can square both sides of the equation the way you did. But there is a problem in the third line of your working.

You had $x^{4}(x+3)=(x+3)^{3}$ and then divided both sides by $(x+3)$ to get $x^{4}=(x+3)^{2}$.

This is a problem because you are losing solutions to the overall equation.

It is better to use factorisation to solve the equation for $x$.