# What is the square root of x^2?

1. Oct 30, 2012

### tahayassen

It can't be x, because you get a positive number when x is negative.

2. Oct 30, 2012

### arildno

It is |x|

3. Oct 30, 2012

### tahayassen

What if your have (x^2)^0.5? Doesn't that equal x?

4. Oct 30, 2012

### micromass

No. That is only x if x is positive.

5. Oct 30, 2012

### Staff: Mentor

$$\sqrt{x^2} = |x|$$

6. Oct 31, 2012

### haruspex

The square root function (√) is defined (usually) to return the principal square root, i.e. the non-negative one. But the "square root" of x2 has two possible values: ±√(x2) = ±|x| = ±x.
As for (x2)0.5, I would say that does not define a function in the strict sense, so it returns a ± result. Prepared to be shouted down on that one, though.

7. Oct 31, 2012

### micromass

I guess it depends on the definition of "exponentiation". In complex analysis, the operation $x^{1/2}$ is indeed multivalued. So in that context, you are right.
When working with real number, I do think that the definition of $x^{1/2}$ is the principal square root.
But again, these are just definitions so it's not very interesting.

8. Oct 31, 2012

### Staff: Mentor

This is not true. $\sqrt{x^2} = |x|$.

Can you cite a single resource that makes the claim that, for example √4 = -2?

Although we say that every positive number has two square roots, a positive one and a negative, when we refer to "the square root" of something, we're talking about the principal (or positive) square root.

9. Nov 1, 2012

### haruspex

That's what I said. The distinction I'm drawing is between taking a square root in a calculation, in which you have to allow for both signs, and the definition of the square root function (as indicated by √), which must be single valued by definition.

10. Nov 1, 2012

### Vadar2012

In high school, I was taught √(x^2) = ±x. It was a long time ago, but will never forget it. Always lost marks for writing anything else.

11. Nov 1, 2012

### pwsnafu

Ouch that sucks. Vibes.
Your teacher should have known better: if it was equal to ±x, the quadratic formula wouldn't need it.

12. Nov 1, 2012

### arildno

High schools all over the world are, for some reason, terrified of the absolute value sign, and thus teaches wrongly.
I do not know why they are so scared...

13. Nov 1, 2012

### Staff: Mentor

It seems to me that we are saying different things. What I said (quoted above) is that √(x2) has a single value, which depends on whether x is positive or negative.

What you seem to be saying is that √(x2) has two values, ±x. What you said is quoted below.

An example might help clear this up.

Solve for x: x2 = 4

Taking the square root of both sides, we get
√(x2) = √4
|x| = 2
If x < 0, then x = -2
If x > 0, then x = 2

A fine point here is that |x| ≠ ±x. If that were the case, then the graph of y = |x| would not represent a function, since each nonzero x value would have two y values. The resulting graph would be the combined graphs of y = x and y = -x.

Instead, the graph of y = |x| has a V shape, and is exactly the same as the graph of y = √(x).

14. Nov 1, 2012

### tahayassen

Why did you only take the positive square-root of both sides? Shouldn't you take both the positive and negative square-roots of both sides since both are valid solutions?

15. Nov 1, 2012

### arildno

No, why?

Taking the NEGATIVE square root operation is another, EQUALLY valid operation, but with absolutely no new information gained.

Your answer would then be:
-|x|=-2

16. Nov 1, 2012

### tahayassen

No what?

Edit: -√4 * -√4 = 4

17. Nov 1, 2012

### tahayassen

Wow. Everything I was taught in high school is wrong!

18. Nov 1, 2012

### arildno

That is correct.

19. Nov 1, 2012

### tahayassen

$${ ({ { x }^{ 2 } }) }^{ \frac { 1 }{ 2 } }=|x|\\ \\ \sqrt { x } =\quad |x|\\ -\sqrt { x } =\quad -|x|$$

How would I get -|x| from using the exponent method?

edit: Never mind. I'm an idiot.

20. Nov 1, 2012

### Staff: Mentor

You can't go from the first equation to the second. The second line should be
$$\sqrt { x^2 } = |x|$$

and similarly for the third line.

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