# Square root problem

1. Sep 7, 2009

### temaire

1. The problem statement, all variables and given/known data

Can someone explain to me how $$|-4|= \sqrt{(-4)^{2}}$$

I'm wondering why you can't cancel out the square root sign and the square above the -4, to leave you with -4.

3. The attempt at a solution

I know this has something to do with the absolute value of -4, being 4, but I just don't completely understand why you can't cancel out the square root sign and the square.

2. Sep 8, 2009

### jhooper3581

|-4| ≠ √(-4²) (because |-4| = 4 and √(-4²) = √(-1(4²)) = √(-1(16)) = √(-16) = 4i, where i = √(-1) is the imaginary number). Therefore, trying to explain how |-4| = √(-4²) is meaningless.

3. Sep 8, 2009

### kbaumen

It's not meaningless. In R, a square root cannot be a negative number, hence the absolute value sign.

4. Sep 8, 2009

### njama

Because |-4|=|$\pm$4|=4. Why it is like that?

Because √(-4)2=√16=4 but also √42=√16=4.

You need to consider both positive and negative value for 4.

Last edited: Sep 8, 2009
5. Sep 8, 2009

### jhooper3581

kbaumen, I suspect that the original poster had edited his/her post. Before he/she edited his/her post, it was √(-4²), and go to this http://hotmath.com/images/gt/lessons/genericalg1/parabola_width.gif" [Broken], and it's obvious that y > 0 for any x values that is x ≠ 0), and |-4| = 4.

Last edited by a moderator: May 4, 2017
6. Sep 8, 2009

### VietDao29

By writing $$\sqrt{x} = r$$, it means that, r is the principal square root of x.

The principal square root of a non-negative number x is the number that satisfies the 2 following properties:
• Firstly, it's a non-negative number, i.e r >= 0. (1)
• And secondly, its square is x (i.e, it's one of the 2 square roots of x). (2)

For any positive real x (i.e x > 0), there are 2 distinct square roots of x:
• The principal one: $\sqrt{x} > 0$
• And the other one: $-\sqrt{x} < 0$

For x = 0, the 2 square roots becomes 1, which is: $$\sqrt{0} = 0$$

For x < 0, there's no real square root of it. There exists complex ones, but just don't worry about it for now. :)

$$\sqrt{(-4) ^ 2} \neq -4$$ because -4 is negative, and hence cannot be the principal square root of (-4)2.

-------------------

Using the 2 properties of the principle square root above, namely (1), and (2), we derive the following formula:
$$\sqrt{\alpha ^ 2} = |\alpha|$$.

Which can be easily proven. You can give it a try, if you want. :)

7. Sep 8, 2009

### kbaumen

In that case I agree with you, but when I replied it was already $$\sqrt{(-4)^2}$$. Perhaps I should've read your post more carefully.

Last edited by a moderator: May 4, 2017
8. Sep 8, 2009

### Staff: Mentor

|-4| $\neq \pm 4$! The absolute value of a number has a single value that is always nonnegative!
This doesn't make any sense. For real numbers, the square root function's domain is nonnegative real numbers. The number 4 does not have a negative value.

9. Sep 8, 2009

### njama

Sorry about it I meant |-4|=|$\pm 4$|=4. I will fix it immediately.

10. Sep 9, 2009

### Elucidus

Comment:

$$\sqrt{(-4)^2} \neq \sqrt{-1(4^2)}$$

The left side equals 4 while the right equals 4i (not 4). $(-4)^2 = (-1 \cdot 4)^2 = (-1)^2(4)^2 = (1)(16) = 16.$

In general

$$\sqrt{x^2} = \left| x \right|$$ as has been mentioned.

--Elucidus

Last edited: Sep 9, 2009
11. Sep 9, 2009

### Staff: Mentor

In addition to Elucidus's comment, there is a factoring property for the square root (and other roots) that is often misused.

If a and b are nonnegative, then $\sqrt{ab}~=~\sqrt{a}\sqrt{b}$.

12. Sep 9, 2009

### kbaumen

Notice that he wrote

$$\sqrt{(-4^2)}$$

$$\sqrt{(-4)^2}$$.

Thou shalt not misquote.

13. Sep 9, 2009

### Elucidus

Indeed, I mistook that quote. Apologies. The OP stated $$\sqrt{(-4)^2}$$ and I mistakingly thought jhooper3581 was citing that expression. He did not and I missed that.

--Elucidus

14. Sep 12, 2009

### kbaumen

So did I at first.

15. Sep 12, 2009

### temaire

So would it be correct to say that although there is a positive and a negative square root to a number, we only consider the positive one since the square root is a function and functions can only have one y value for every x value?

16. Sep 12, 2009

### njama

No that is simply not correct. Consider y = x2. In fact, this is your problem.

Here is a illustration. I hope that after all those posts you will understand what these people are saying to you.

Last edited by a moderator: May 4, 2017
17. Sep 12, 2009

### temaire

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Actually I believe I am correct, well at least to some degree. I think that you just misunderstood me. I said that since the square root is a function, you can only have one y value for each x value, thus we only consider the principal square root. What are you trying to show with your illustration?

Last edited by a moderator: May 4, 2017
18. Sep 12, 2009

### njama

Ok, if your function is $y=\sqrt{x}$, then there is unique value for y, by choosing different $x \in [0, +\infty]$. And do you know what makes the function with that property?

The function is neither odd nor even.

19. Sep 12, 2009

### VietDao29

Well, what do you mean? Which function are you referring to? Of course the square root function is neither odd, nor even, because its domain is R+, which is clearly not symmetric to 0. But I'm wondering if it does have anything to do here?!?!

As far as my knowledge goes, this is correct. :)

20. Sep 12, 2009

### temaire

Thanks VeitDao, you've been a great help.