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

Question

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?

Answer 1

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.

Answer 2

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 $\sqrt{x+9}$.

Answer 3

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.

Answer 4

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.

For more, read about the Cauchy's functional equation.

Answer 5

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 $x\neq0$.

Answer 6

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 ($10$): $$\sqrt{x+9}=\sqrt{10}$$

Answer 7

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 $\sqrt{9} + \sqrt{16} = 3 + 4 = 7$.

So we found a contradiction.

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

Answer 8

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.

Answer 9

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

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

Answer 10

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.

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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.