# Algebraically why must a single square root be done on all terms rather than individually?

Let's assume we know that $$x+9=10$$. I understand this is illegal: $$\sqrt[]{x} + \sqrt[]{9} = \sqrt[]{10}.$$ And this is correct: $$\sqrt[]{x + 9} = \sqrt[]{10}.$$

Is there an intuitive way to understand why this must be the case?

• I assume you're asking "when I take the square root of both sides of an equation, why can't I distribute it across any additions?" Here's a good reason why not: $\sqrt{9}+\sqrt{16}\neq \sqrt{9+16}$, as you can check. Jul 27, 2023 at 18:13
• Technically, a function which satisfies $f(x+y)=f(x)+f(y)$ is called an additive function. There really aren't many of these. It is also true that for $f(x)=x^2$ that $x^2+9^2\neq 10^2$ and $f(x)=x+1$ then $(x+1)+(9+1)\neq 10+1.$ If you apply the square to $10$ then since $10=x+9,$ you are apply the square root to $x+9.$] Jul 27, 2023 at 18:19
• On the real numbers, an additive function is essentially only the functions $f(x)=ax,$ where $a$ is a real number. (There are lots of other additive functions, but they are very strange bizarro-world function which we can assert exist, but we can't write them down, and they are discontinuous everywhere and non-integrable even using more advanced integrals.) Jul 27, 2023 at 18:25
• In my opinion, we want this to be true when we're learning math for the same reason that a child will write the letter "S" backward. Overgeneralization. The child thinks an "S" is anything with two curves linked together. After learning about the distributive property we think everything has to distribute. Jul 28, 2023 at 5:15
• Perhaps it's better for building intuition to, say, consider multiplication and ask why can you distribute it over addition. You start from the idea that if you do the same thing to both sides of an equation, the equality stays true; say $x + 9 = 10$ can be transformed into $3(x + 9) = 3 \cdot 10$ or $\sqrt{x + 9} = \sqrt{10}$. But then multiplication has a special property that means $3(x + 9) = 3x + 3 \cdot 9$; so, perhaps the question to ask is - what's so special about multiplication? Jul 28, 2023 at 13:08

Note that $$\sqrt{x} + \sqrt{y} = \sqrt{x+y} \iff (\sqrt{x} + \sqrt{y})^2 = (\sqrt{x+y})^2 \iff (\sqrt{x} + \sqrt{y})^2 = x + y$$.

In order to intuitively understand it, let's substitute $$a = \sqrt{x}, b = \sqrt{y}$$

So your question is kind of equivalent to $$(a+b)^2 = a^2 + b^2$$ ?

Let's try to understand this with the area of a square:

Clearly there is more area in $$(a+b)^2$$ than in $$a^2 + b^2$$.

Specifically the difference is $$2ab$$, i.e. $$2\sqrt{x}\sqrt{y}$$, a difference which is significant in most cases.

• Or to directly note that $(\sqrt{x}+\sqrt{y})^2=x+y+2\sqrt{x}\sqrt{y}$. Jul 28, 2023 at 20:57

An equation is a scale, the kind with two pans, that balances when what's in the one pan weighs the same as what's in the other pan. You can do anything you want to do to one pan, and the scale will still balance, so long as you do exactly the same thing to the other pan. So, you can do anything you want to one side of an equation, and it will still be a true equation, so long as you do exactly the same thing to the other side. In particular, you can take the square root of what's on one side of the equation, but only if you take the square root of what's on the other side. And if what's on one side is $$x+9$$, then taking the square root of what's on that side means doing $$\sqrt{x+9}$$.

• You could apply this argument to $$2*(x+9)$$ too, but that function does distribute. Jul 28, 2023 at 9:41
• @Brondahl (+1) I think from a pedagogical point of view we should apply this argument to all functions applied to both sides of the equation! Interestingly, when people add or subtract a term to both sides, even students who struggle with algebra usually don't add/subtract it to each term on both sides. The fact that multiplying/dividing an equation can be done term-by-term seems to confuse people that other operations must be done the same way. When introducing the idea, it's arguably better to write $2(LHS\dots) = 2(RHS\dots)$ and then expand, just to explain why we do it term by term Jul 28, 2023 at 11:58
• @Brondahl First you do 2* to both sides of the equation. Then, if you like, you can rewrite 2*(x+9) as 2*x + 2*9. If you want. You don't have to do the second part. Jul 28, 2023 at 20:11
• @user253751 This is true, but often people forget that they are doing 2xLHS = 2xRHS... because they didn't understand the distributive property, or know it is. Jul 28, 2023 at 23:15
• @GerryMyerson Oooohhhh ... you think the confusion/request for justification from OP is "why can't I do 'take sqrt of each thing on either side of the equals separately' as a legitimate operation on the equation?". I see... that makes a great deal more sense of the question in the first place! Jul 29, 2023 at 9:43

A false equation like $$\sqrt{9}+\sqrt{16}=\sqrt{9+16}$$ is, of course, false, and generally speaking we do not need to go on with elaborate explanations for why evidently false statements are false.

And yet there could still be an interesting intuitive reason for this falsity; here's one.

We know by the Pythagorean Theorem that if $$a,b>0$$ are the lengths of the two legs of a right triangle and $$c>0$$ is the length of the hypotenuse then $$a^2+b^2=c^2$$ If it were true that $$\sqrt{x+y}=\sqrt{x}+\sqrt{y}$$ for all $$x,y \ge 0$$, then it would follow that $$a + b = c$$ Therefore, the length of the hypotenuse is the sum of the lengths of the two legs. But that's a clear violation of our intuition that the unique shortest path between two points is a straight line.

• But maybe this proof demands knowledge of a more advanced proof, i.e Pythagoras' Theorem ? Jul 28, 2023 at 9:50
• Certainly. It is way too elaborate to balance the evident falsity of the equation $\sqrt{9}+\sqrt{16}=\sqrt{9+16}$. But this was not intended at all as a proof: the OP asked for some intuition. I edited my answer to make that clearer. Jul 28, 2023 at 13:33
• Intuition to find quickly two numbers which when squared and added will total to a number with an integer square root ? Not for most people, I feel - although I accept some people can think in that way. Jul 28, 2023 at 13:55
• It's certainly a good thing that there is a broad distribution of mathematical intuition :-) Jul 28, 2023 at 14:00
• :-) :-) :-) :-) :-) :-) :-) Jul 28, 2023 at 14:10

Rather than thinking about why the square root doesn't distribute over the addition you might be better off thinking about how rare that kind of distributing is.

Suppose $$f(x+y) = f(x) + f(y)$$ for some unknown operation $$f$$.

Then setting $$x = y = 0$$ tells you $$f(0)$$ must be $$0$$. Then $$f(2x) = f(x) + f(x) = 2f(x)$$ and so on. With a little more work and assuming the operation $$f$$ is smooth enough, you can conclude that $$f(ax) = af(x)$$ for some fixed constant $$a$$, which turns out to be $$f(1)$$.

That is just the old fashioned distributive law for arithmetic.

You can see the same thing geometrically by looking at the graph of $$f$$. It must be a straight line through the origin.

The graphs of $$\sqrt{x}$$, $$x^2$$, $$a^x$$, $$\log x$$ and $$\sin x$$ don't look at all like that, which should warn you away from the freshman's dream.

Maybe there is a geometric way to approach this in some very illustrative way, but the algebraic reason is something I can explain.

First, think about what a square root actually is. $$\sqrt{x}$$ is the number for which $$\left(\sqrt{x}\right)^2 = x$$.

If we take your example with $$x$$ and $$9$$, we can then ask ourselves whether the nice looking equation $$\left(\sqrt{x}+\sqrt{9}\right)^2 = \left(\sqrt{x+9}\right)^2 = x+9 = \left(\sqrt{x}\right)^2 + \left(\sqrt{9}\right)^2$$ holds or not.

More generally, does $$(a+b)^2 = a^2 + b^2$$ hold ?

The answer is "Yes, but only if $$a=0$$ and/or $$b=0$$". So, in your example, the expression holds when $$x = 0$$ (since 9 can never be 0).

Otherwise $$(a+b)^2 \neq a^2 + b^2$$, which would be necessary for $$\sqrt{x}+\sqrt{9} = \sqrt{x+9}$$ to hold for all $$x\neq0$$.

When you do something to both sides of the equation, you have to do it to the entire sides.

Perhaps it's better to think of this in terms of the replacement rule "if two expressions are equal, you can replace one by the other."

Start with: $$\sqrt{x+9}=\sqrt{x+9}$$ Replace one $$x+9$$ with something we know it's equal to ($$10$$): $$\sqrt{x+9}=\sqrt{10}$$

• You could apply this argument to $$2*(x+9)$$ too, but that function does distribute. Jul 28, 2023 at 9:39
• But this talk of distribution of one operation over another seems non-intuitive and much more advanced than the matter at hand of $\sqrt{a+b} = \sqrt{a} + \sqrt{b}$ or not, where $a, b \in R$ . . . Jul 28, 2023 at 10:14
• But it is disallowed to jump directly from $x+9=10$ to $2*x+2*9 = 2*10$. One must go from $x+9=10$ to $2*(x+9) = 2*10$ to $2*x+2*9=2*10$. Jul 28, 2023 at 12:01
• @Trunk The matter of "does $$\sqrt{x+9}=\sqrt{x}+\sqrt{9}$$?" is literally the statement of "distribution of one operation over another". These are exactly the same thing. Jul 28, 2023 at 12:08
• @Brondahl Yes, and in 16/64, you can cancel the sixes and get the correct result 1/4. Saying that there are cases where distributing an operation yields the correct result does nothing to dispute the fact that distributing operations is in general not valid, and if you want to do it in a specific case, then you need a specific justification for that case. Jul 29, 2023 at 20:51

Building on the scale argument:

You have the equation

$$x + 9 = 10 \tag{1}$$

Now let's introduce a new variable, or a shortcut notation:

$$a = x+9 \tag{2}$$

so combining $$(1)$$ and $$(2)$$ we obtain

$$a = 10$$

Now surely you'll agree that if I take the square root on both sides it is:

$$\sqrt{a} = \sqrt{10}$$

Use equation $$(2)$$ again to plug in the definition for $$a$$

$$\sqrt{x+9} = \sqrt{10}$$

Now, to see that $$\sqrt{a+b} \neq \sqrt{a} + \sqrt{b}$$, let's assume it was true for a moment.

We can choose $$a=9$$ and $$b=16$$.

Then $$\sqrt{9+16}=\sqrt{25}=5$$ but $$\sqrt{9} + \sqrt{16} = 3 + 4 = 7$$.

Therefore we can not distribute the squareroot (or any root) across a sum.

• This is all mathematically rigorous. But to most of us it would not be intuitive. And to all 12-13 year olds even less so. Jul 28, 2023 at 13:33
• This is the way I taught it to many struggling students and what I found to be most effective in making them understand why it doesn't work that way. Especially the "plug in some numbers and see" allows them to see for themselves if their assumption is correct. Jul 28, 2023 at 14:22
• I love how this explanation explicitly teaches the skill of chunking parts of expressions. Jul 28, 2023 at 22:00

If we have $${x+9}={10},$$that means the left and right side of the equation represent the same number (in this case, 10). So if we have some number, which is obviously equal to itself, it follows intuitively that the square root of that number would equal the square root of that same number. This should clarify why $${x+9}={10}$$ allows us to infer that $$\sqrt{x+9}=\sqrt{10}$$

As for why we can't take the square roots of all terms in an equation, it's simply not how the square root works. It's like asking why can't we break a number into two arbitrary summands, and then expect the sum of the square roots of those two to be equal to the square root of the original number.

We have a special name for functions that work like this:

$$f(a) + f(b) = f(a+b)$$ we call them "linear".

The square root isn't linear. Some operations that are linear include "3*". So you can do

$$3*a + 3*b = 3*(a+b)$$ but

$$\sqrt[]{a} + \sqrt[]{b} = \sqrt[]{a+b}$$ isn't one of them. This isn't because $$\sqrt[]{x}$$ is anything special -- most functions aren't linear!

For much of your mathematics education, you have played with linear functions. This is for a good reason -- math is really good at solving linear problems! So you are given functions and operations that behave nicely.

Functions like $$x^2$$ and $$\sqrt[]{x}$$ are not as nice as $$*2$$ is.

Note that a lot of work is done to make functions and operations that don't look linear into linear functions, so you can solve them using linear techniques.

In the future, you are going to learn ways to make functions that don't look linear into linear functions.

Logarithms is an example of this. If you have

$$a^x b^y = c^z$$ and you take the logarithm of both sides:

$$\ln{a^x b^y} = \ln(c^z)$$ you get

$$x \ln{a} + y \ln{b} = z \ln{c}$$

If I’m understanding you right, you’re familiar and comfortable with distributivity—the idea that, say, $$2\left(x+9\right)=2\left(x\right)+2\left(9\right)$$. This is your “normal”. And you want to know how you can be comfortable with square roots being “special” and not working this way.

My recommendation is to flip this on its head: Build the intuition that distributivity is special.

Look at “$$x+9=10$$” (or any equation). Mentally replace it with two boxes; not just icons as shown below, but actual, physical containers.

$${\color{Green}█}={\color{Brown}█}$$

The equation tells us that the contents of these boxes are the same. They may or may not be identical, but they’re the same in some sense that matters to us right now: same monetary value, same weight, same amount of pain if dropped on your foot…

Now, you modify one of the boxes in some way that alters its contents. Maybe you put something else into it, or you turn it on its side, or you drive a car over it, or you set it on fire. If you do exactly the same thing to the other box, you expect to have the same effect on its contents, right? So the equation still holds true:

$$🔥{\color{Green}█}=🔥{\color{Brown}█}$$

But now let’s say you know that the thing in the boxes is made up of smaller bits. You can disassemble the pieces, and alter each one on its own… but will the result be the same as if you’d altered the whole lot?

For some alterations, the answer is yes. Grinding up into a fine powder, for instance.

For others—in fact probably for most things—the answer is no. Light a few thousand matches one at a time, and you will just get a lot of fire and burnt matches. Setting thousands of matches on fire all at once will give you a nice little explosion (Hyneman, J. & Savage, A. (2009), episode 117).

$$🔥{\color{Green}▏}+🔥{\color{Green}▏}+🔥{\color{Green}▏}+🔥{\color{Green}▏}+🔥{\color{Green}▏}+🔥{\color{Green}▏}+🔥{\color{Green}▏}+🔥{\color{Green}▏}\ne🔥{\color{Brown}█}$$

In maths as in life, most things have a different effect on one big thing than they do on its individual components. Square roots are completely ordinary in this way. Multiplication-over-addition is unusual for not giving a different result when done separately to each term.

• Great new look at this question. Jul 29, 2023 at 16:59
• I wonder if there's any sort of physical transformation that can't be abstracted into multiplication and yet still distributes. Grinding into a powder, for instance, is a division into many small parts, and division is inverse multiplication Jul 30, 2023 at 19:00
• @No Name So you are saying that $A/(B+C) \neq A/B + A/C$ can be undone by regarding it as a multiplication of reciprocals, i.e. $1/(1/A) * 1/(B+C) = 1/(B/A + C/A)$ ? I wonder if that is a fair interpretation of the distributivity expression though. Jul 31, 2023 at 8:45
• @Trunk Honestly, I hadn't thought of that. And pulverization is more like $(B+C)/A = B/A + C/A$, which is true. Nice catch, though Jul 31, 2023 at 16:50
• I have been thinking of social politics world situations where couples/groups/classes do a lot better than the sum of what would occur to the individuals involved. Unionization is not done for fashion. A group of hyenas can drive a singular lion away from a kill. But one or two hyenas can't and two lions can hold off a good group of hyenas. The insurance for a couple will always be less than the total payable by each person as an individual. Jul 31, 2023 at 17:13