B Square root of y^6

1. Aug 29, 2016

cmajor47

I'm trying to decide if simplifying sqrt(y^6) requires use of the absolute value bars. For example, the rule "nth root(u^n) = abs(u) when n is even" can be used to simplify sqrt(y^6) as sqrt[(y^3)^2]=abs(y^3). However, the rules of rational exponents can also be used to simplify sqrt(y^6) as (y^6)^1/2=y^(6/2)=y^3. So are the absolute value bars necessary when simplifying sqrt(y^6), or not?
I know that the domain of sqrt(y^6) is [0,inf), so does this allow for the absolute value bars to be unnecessary?

2. Aug 29, 2016

Staff: Mentor

The absolute values are necessary, for the same reason that $\sqrt{x^2} = |x|$.
For your problem, if you wrote $\sqrt{y^6} = y^3$, the left side is always nonnegative, for any real y, but the right side can be negative when y is negative.

3. Aug 29, 2016

Ssnow

If you specify that $y\geq 0$, then $\sqrt{y^6}=y^3$ is correct ...

4. Aug 29, 2016

micromass

Staff Emeritus
Here you are using the property $(a^b)^c = a^{bc}$ which is true only for positive $a$.

5. Oct 12, 2016

Deepak suwalka

The absolute value bars really become necessary when we have variables under the √ sign.
$\sqrt{a^2}\;\;=$|a|
Similarly, $\sqrt{y^6}$= |$y^3$|
But we want to make sure that the value under root is positive,
if y=2 then
$\sqrt{y^6}= +y^3 \;\; or\;\;\sqrt{2^6}=+8$
But when we have equation such as
$x^2=23+2$
$x^2=25$
$x=\pm\sqrt{25}$
Then we get,
$x=\pm 5$

I hope it' ll help.

Last edited: Oct 12, 2016
6. Oct 12, 2016

pwsnafu

This is false. $x^2 = 25 \implies x=\pm\sqrt{25} \implies x = \pm 5$.
$\sqrt{25} = 5$, never -5.

7. Oct 12, 2016

Deepak suwalka

yes, i know. I have misstekenly typed, btw thanks

8. Oct 12, 2016

Staff: Mentor

The quote from @pwsnafu threw me off for a bit. At first I thought that "This is false" referred to the implication that immediately followed what he wrote. To be clear, "This is false" refers to these two lines that Deepak wrote: