Why do absolute values appear in the simplification of square roots?

[Rably]

Homework Statement

Simplify.

a) [tex]\sqrt{x^6}[/tex]
b) [tex]8 \sqrt{x^7y^{10}} - 10 \sqrt{x^7y^{10}}[/tex]

For b, it's y^10. I can't make it look right for some reason.
Mod note: Fixed the exponent.

Homework Equations

The Attempt at a Solution

I can simplify all of them but I don't know when or where I need to put in absolute value symbols to the solution. I know the solution because my book shows me, but I don't understand why the absolute values are where they are.

For instance, the solution to a is [tex]|x^3|[/tex] but I can only get to [tex]x^3[/tex] without becoming confused.
For b, i can get to [tex]-2x^3y^5 \sqrt{x}[/tex] but the solution is [tex]-2x^3|y^5| \sqrt{x}[/tex]
So how do you know when an absolute value is required?

 

[SammyS]

Homework Statement

Simplify.

a) [tex]\sqrt{x^6}[/tex]
b) [tex]8 \sqrt{x^7y^{10}} - 10 \sqrt{x^7y^{10}}[/tex]
For b, it's y^10. I can't make it look right for some reason.

Use y^{10}

Homework Equations

The Attempt at a Solution

I can simplify all of them but I don't know when or where I need to put in absolute value symbols to the solution. I know the solution because my book shows me, but I don't understand why the absolute values are where they are.

For instance, the solution to a is [tex]|x^3|[/tex] but I can only get to [tex]x^3[/tex] without becoming confused.
For b, i can get to [tex]-2x^3y^5 \sqrt{x}[/tex] but the solution is [tex]-2x^3|y^5| \sqrt{x}[/tex]
So how do you know when an absolute value is required?

[itex]\displaystyle \sqrt{u^2}=|u|[/itex]

Also, remember that if n is a positive integer, then [itex]u^{2n}\ge0\,,[/itex] so there is no need to use absolute value.

 

[Rably]

So in my scenario, x has to be positive because it's still underneath the square root so I don't need to put an absolute value around x^3, whereas y can be either positive or negative because it's not still beneath the square root so an absolute value is required?

 

[Mark44]

So in my scenario, x has to be positive because it's still underneath the square root

No. x can be any real number.

Like SammyS said,
$$ \sqrt{x^2} = |x|$$

so
$$ \sqrt{x^6} = \sqrt{(x^3)^2} = |x^3|$$

This is also the same as |x|³.

so I don't need to put an absolute value around x^3, whereas y can be either positive or negative because it's not still beneath the square root so an absolute value is required?

 

[SammyS]

So in my scenario, x has to be positive because it's still underneath the square root so I don't need to put an absolute value around x^3, whereas y can be either positive or negative because it's not still beneath the square root so an absolute value is required?

Which scenario ?

If x is to an even power, that result is non-negative. So that can be under a radical -- actually one signifying a square root -- no matter what value x has.

 

[Rably]

Which scenario ?

If x is to an even power, that result is non-negative. So that can be under a radical -- actually one signifying a square root -- no matter what value x has.

I was referring to question b. Do you mean if x has an even power in the initial equation or in the solution?

 

[Rably]

No. x can be any real number.

Like SammyS said,
$$ \sqrt{x^2} = |x|$$

so
$$ \sqrt{x^6} = \sqrt{(x^3)^2} = |x^3|$$

This is also the same as |x|³.

Oh wow, I think it makes sense now. Seeing the step between x^6 and |x^3| was really helpful. I wasn't writing that step down, I was simply skipping to the final step. Thanks a bunch.