Why Can't You Cancel Out the Square Root and Square in a Square Root Problem?

[temaire]

Homework Statement

Can someone explain to me how [tex]|-4|= \sqrt{(-4)^{2}}[/tex]

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

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.

 

[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.

 

[kbaumen]

|-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.

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

 

[njama]

Because |-4|=|[itex]\pm[/itex]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.

 

[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" , and it's obvious that y > 0 for any x values that is x ≠ 0), and |-4| = 4.

 

[VietDao29]

Homework Statement

Can someone explain to me how [tex]|-4|= \sqrt{(-4)^{2}}[/tex]

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

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.

By writing [tex]\sqrt{x} = r[/tex], 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: [itex]\sqrt{x} > 0[/itex]
• And the other one: [itex]-\sqrt{x} < 0[/itex]

For x = 0, the 2 square roots becomes 1, which is: [tex]\sqrt{0} = 0[/tex]

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

[tex]\sqrt{(-4) ^ 2} \neq -4[/tex] 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:
[tex]\sqrt{\alpha ^ 2} = |\alpha|[/tex].

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

 

[kbaumen]

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" , and it's obvious that y > 0 for any x values that is x ≠ 0), and |-4| = 4.

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

 

[Mark44]

Because |-4|=[itex]\pm[/itex]4. Why it is like that?

|-4| [itex]\neq \pm 4[/itex]! The absolute value of a number has a single value that is always nonnegative!

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

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

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.

 

[njama]

|-4| [itex]\neq \pm 4[/itex]! 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.

Sorry about it I meant |-4|=|[itex]\pm 4[/itex]|=4. I will fix it immediately.

 

[Elucidus]

|-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.

Comment:

[tex]\sqrt{(-4)^2} \neq \sqrt{-1(4^2)}[/tex]

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

In general

[tex]\sqrt{x^2} = \left| x \right|[/tex] as has been mentioned.

--Elucidus

 

[Mark44]

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 [itex]\sqrt{ab}~=~\sqrt{a}\sqrt{b}[/itex].

 

[kbaumen]

Comment:

[tex]\sqrt{(-4)^2} \neq \sqrt{-1(4^2)}[/tex]

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

In general

[tex]\sqrt{x^2} = \left| x \right|[/tex] as has been mentioned.

--Elucidus

Notice that he wrote

[tex]\sqrt{(-4^2)}[/tex]

instead of

[tex]\sqrt{(-4)^2}[/tex].

Thou shalt not misquote.

 

[Elucidus]

Notice that he wrote

[tex]\sqrt{(-4^2)}[/tex]

instead of

[tex]\sqrt{(-4)^2}[/tex].

Thou shalt not misquote.

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

--Elucidus

 

[kbaumen]

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

--Elucidus

So did I at first.

 

[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?

 

[njama]

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?

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.

http://img225.imageshack.us/img225/42/graphofquadratic.png

 

[temaire]

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.

http://img225.imageshack.us/img225/42/graphofquadratic.png

[/URL]

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?

 

[njama]

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

Spoiler

The function is neither odd nor even.

 

[VietDao29]

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

Spoiler

The function is neither odd nor even.

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??

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?

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

 

[temaire]

Thanks VeitDao, you've been a great help.

 

[njama]

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??

Here is why:

...you can only have one y value for each x value, thus we only consider the principal square root...