Square root of both sides

Question

If you have the equation: $x^2=2$ You get: $x=\pm \sqrt{2}$

But what do you do actually do? What do you multiply both sides with to get this answer? You take the square root of both sides, but the square root of what? If you understand what i mean?

You could think of it as a length in Pythagoras' theorem and that it is applied practically. — Niklas Rosencrantz, May 22 '16 at 18:04
Taking a square root and solving a quadratic equation are two different things. — Jasper, May 22 '16 at 20:33
It's acceptable because there are two and only two possible numbers where $x^2 = 2$. If $x^2 = 2$ the only possible values $x$ can be are $\sqrt 2$ and $-\sqrt 2$. If $x$ equals anything else, then $x^2 \ne 2$. — fleablood, May 22 '16 at 22:08
@Jasper Are they? Taking a square root is solving an equation of the form $x^2 = a$; solving a quadratic is done by rewriting the equation in the form $(x-a)^2 = b$ and taking the square root of both sides. — sTertooy, May 22 '16 at 22:11

score 13 · Answer 1 · edited May 22 '16 at 21:38

13

As $2=\left(\sqrt 2\right)^2$, we can transform $x^2=2$ into $x^2-\left(\sqrt 2\right)^2=0$, or $\left(x+\sqrt 2\right)\left(x-\sqrt 2\right)=0$. Now we use that a product is zero iff at least one of its factors is zero.

Alternatively, if you want to procede by applying something to both sides of the equation, this something is "taking the square root". This way $x^2=2$ implies that $\sqrt{x^2}=\sqrt 2$. Now you need to know that $\sqrt{x^2}=|x|$ for all real numbers $x$.

edited May 22 '16 at 21:38

Mutantoe

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answered May 22 '16 at 18:08

Hagen von Eitzen

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I prefer the first one. +1 – callculus42 May 22 '16 at 18:13
I prefer the second one. It's something that should be obvious to every university student, and yet most have been taught so poorly that they don't get it. – Martin Argerami May 23 '16 at 05:43
@MartinArgerami: I agree wholeheartedly with your second sentence. But not the first one! =) The reason is that the first method does not require knowledge of the square-root function and is in some sense more fundamental. It depends on the property Hagen mentioned which in turn depends on existence of inverses for nonzero reals. It holds even when "$\sqrt{2}$" is interpreted as purely notation for "some real number $s$ such that $s^2 = 2$". We choose the positive one in the standard convention only because of our historical preference, but it doesn't matter for algebraic factorization. – user21820 May 28 '16 at 06:32
I agree to a certain extent, but in the end it amounts to a trick, so I wouldn't use the word "fundamental". – Martin Argerami May 28 '16 at 15:03

score 10 · Answer 2 · answered May 22 '16 at 18:53

The basic idea of manipulating equations is that 'doing the same thing to both sides' will yield the same output on both sides because you started with the same input on both sides -- that the inputs were the same is precisely what the original equation said. However, this property of each input always giving the same output is not a given, so it's important to make sure that whatever it is you're doing to each side really does have this property. In other words, it needs to be a function.

That's all very well for something like 'add 3', ie $f(x)=x+3$, as this is a function that is defined for all real numbers, but taking the square root requires a bit more care to pin down the exact function we want to apply to both sides. We can take the positive square root, resulting in $|x|=\sqrt 2$, but that doesn't quite give us $x$, as we don't know whether $x$ is positive or negative. We do however know that if $x$ is positive then $x=|x|=\sqrt 2$, whereas if $x$ is negative then $x=-|x|=-\sqrt 2$, which we can combine into a single expression using the $\pm$ sign.

So, in short, in solving $x^2=2$, you're really solving two separate cases and combining them into a single expression.

Computer Scientist here relearning many of my missed math fundamentals -- I never got the ah-ha moment for rules about operations on both sides. Because of this answer it finally clicked! Basically, make it a pure function and test that every property per each input on each side (e.g. positive, negative, 0) returns the same result. — Josh Hibschman, May 14 '21 at 15:02

Martin Argerami · Answer 3 · 2016-05-23T21:42:50.503

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You don't "multiply" by anything. One way to see it is that the equality can be written as $x^2-2=0$, so $$ (x-\sqrt2)(x+\sqrt2)=0. $$ For the product to zero, at least one of the factors has to be zero, so either $x=\sqrt2$ or $x=-\sqrt2$.

edited May 23 '16 at 21:42

answered May 22 '16 at 18:07

Martin Argerami

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Done :) $\ \ \ $ – Martin Argerami May 23 '16 at 21:43

score 4 · Answer 4 · answered May 22 '16 at 18:22

4

$$x^2=2 \Rightarrow \begin{cases}y=x^2\\y=2 \end{cases}$$

answered May 22 '16 at 18:22

Roman83

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This doesn't seem to address the original question. – Mark S. May 22 '16 at 19:36
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@MarkS.: Yes, but this shows that the two roots – Roman83 May 22 '16 at 19:44

score 3 · Answer 5 · answered May 22 '16 at 21:34

The relevant theorem is

If $x$ denotes a real number and $y$ denotes a nonnegative real number, then the following are equivalent:

$x^2 = y$

$x = \sqrt{y}$ or $x = -\sqrt{y}$

Usually, people intuit this theorem through experience and/or mimicry, and are implicitly invoking it whenever they carry out an algebraic step like that.

It's not terribly hard to prove given enough other facts, as the other answers show.

Noe Blassel · Answer 6 · 2016-05-22T18:11:11.823

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Note $\pm\sqrt{2}$ is not a number, so there is no real number $a$ such that $a\cdot2=\pm \sqrt{2}$. Rather writing $x=\pm\sqrt{2}$ is shorthand for $x\in\{-\sqrt{2},+\sqrt{2}\}$, indicating that the set of real numbers $x$ satisfying $x^2=2$ is $\{-\sqrt{2},+\sqrt{2}\}$, as can be seen by $x^2=2 \iff x^2-2=0 \iff (x-\sqrt{2})(x+\sqrt{2})=0 \iff x=\sqrt{2}$ or $x=-\sqrt{2}.$

edited May 22 '16 at 18:11

answered May 22 '16 at 18:07

Noe Blassel

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I see what you say, but it's not an answer to the question though. – Mathematician 42 May 22 '16 at 18:10
well it means that the step from $x^2=2$ to $x=\pm \sqrt{2}$ is not an algebraic step, that is, we are not multiplying by something on both sides. Instead, we are implicitely determining the real solutions (e.g. by factorizing), and writing them in shorthand. – Noe Blassel May 22 '16 at 18:16

score 1 · Answer 7 · answered May 22 '16 at 22:01

At some time we learned:

For every positive real number $b$ then for every natural $n$ there is one unique positive real number $c$ so that $c^n = b$. We call that number $c=\sqrt[n]{b}$. (See footnote[#]).

And because for any real number, $x$ then $(-x)^2 = x^2 \ge 0$ (with equality holding if and only if $x = 0$) we can conclude:

For every positive real number $b$ and every even natural numbers $n$ there are two unique real numbers $c_1$ and $c_2$ such that $c_1^n = c_2^n = b$ and that $c_2 = - c_1$. We call the positive one $\sqrt[n]{b}$ and we say that negative one is the negative n-th root.

So....

Suppose we discover that $x^2 = 2$. In other words $x^2$ and 2 are the same thing. We know that there are two unique numbers, $c = \sqrt{2} > 0$ and $-c= -\sqrt{2} < 0$. Such that $c^2 = \sqrt{2}^2 = 2 = x^2$. Well, as there are precisely two such numbers (which are polar opposites of each other) and we know that $x^2 = 2$ so $x$ must be one of them! So we know that $|x| = \sqrt{2}$ and $x = \pm \sqrt{2}$.

We shorten this reasoning into a single line of "taking the square root of both sides"

$x^2 = 2 \implies x = \pm \sqrt{2}$.

We are allowed to do that because we know that $\pm \sqrt{2}$ are the only possible values for $x$ where $x^2 = 2$.

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[#] Actually, we probably never did actually learn that. We were probably just assumed to take it for granted.

None the less, it is a fact and it is provable (although I won't do it here... the proof is more or less claiming that as $x >0$ gets bigger $x^n$ increases so the correlation between $x> 0 \leftarrow \rightarrow x^n$ is one-to-one. And $f(x) = x^n$ is continuous and doesn't jump about then... as there are $x_i < x_j$ such that $x_i^n < b < x_j^n$ and $x^n$ doesn't jump about, there must be some point $x$ where $x^n$ exactly equals $b$ and $x$ is the only positive number where that is true.... take my word for it.)

Square root of both sides

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