Question

√27x^3y^4
OVER
√3xy^2

Answer 1

Like so?\[\frac{\sqrt{27}x^3y^4}{\sqrt{3}xy^2}\]

Answer 2

@jabberwock yessirr!

Answer 3

Okidoki. So let's separate this a little bit to make what we're going to do easier to see:
\[\frac{\sqrt{27}}{\sqrt{3}}\frac{x^3}{x}\frac{y^4}{y^2}\]
If we only have multiplication or division of square roots, we can combine the square roots like this:
\[\sqrt{\frac{27}{3}}\frac{x^3}{x}\frac{y^4}{y^2}\]

Answer 4

The numerical part simplifies to the square root of 9, which is 3
\[3\frac{x^3}{x}\frac{y^4}{y^2}\]Good so far?

Answer 5

\[\frac{a^n}{a^m}=a^{n-m}\]

Answer 6

yesss soooo sorry @jabberwock continue!

Answer 7

Alright. Now consider just the part with the x's:
\[\frac{x^3}{x}=\frac{x*x*x}{x}=x^2\]

Answer 8

why did that just turn into x squared?

Answer 9

Ah, one of the x's up top cancels with one of the x's on the bottom

Answer 10

So you just have x*x after that

Answer 11

ohhhh that makes sense! haha okay what next?

Answer 12

Same reasoning with the y's
\[\frac{y^4}{y^2}=\frac{y*y*y*y}{y*y}=y^2\]

Answer 13

We are cancelling two y's up top with two y's on the bottom

Answer 14

okayy cool. so we have y squared and x squared.. so the final would be 3x^2y^2???

Answer 15

So, quick note:
\[x=x^1\]
Look at the x's again:
\[\frac{x^3}{x^1}=x^{3-1}=x^2\]Look at the y's:
\[\frac{y^4}{y^2}=y^{4-2}=y^2\]

Answer 16

wow okay, just simple subtraction!

Answer 17

So as a shortcut, you can just subtract when you have exponents with the same base on the top and the bottom. For example:
\[\frac{x^{100}}{x^{75}}=x^{25}\]
And yes, that final answer you gave looks right :)

Answer 18

Exactly!

Answer 19

Sorry, looks like I'm lagging

Answer 20

haha its fine! thanks soo much! i have more tho!

Answer 21

last one!

Answer 22

No worries. K, shoot.

Answer 23

\[3\sqrt{24}+\sqrt{54}-\sqrt{6}\]

Answer 24

Okidoki. This is slightly more complicated. Let's just deal with the first term.
\[3\sqrt{24}\]Let's look for a PERFECT SQUARE that divides into 24

Answer 25

Ideas?

Answer 26

ohhhh! √6 x √4

Answer 27

Exactly! Know what to do with it?

Answer 28

well do i simplify the √6 even more? and the other would be 2..

Answer 29

2 isn't a perfect square, so we can't use that.

Answer 30

We're only going to pull out perfect squares. If we could have pulled out a 4 or a 9, we would have done that, too.

Answer 31

\[3\sqrt{6}(\sqrt{4})=3\sqrt{6}(2)=6\sqrt{6}\]

Answer 32

Okay with that?

Answer 33

i dont understand that..

Answer 34

ohhhh nevermind i get it

Answer 35

Cool cool. Want to try the next term?

Answer 36

uhmm could it be √6+√9?

Answer 37

so it'd be √6+3?

Answer 38

\[\sqrt{6}\sqrt{9}=3\sqrt{6}\]Multiplication instead of addition.

Answer 39

alright so is that as far as we can simplify?

Answer 40

We now have
\[6\sqrt{6}+3\sqrt{6}-\sqrt{6}\]
Think of the square root of 6 like a variable. You can combine them kind of like like-terms.

Answer 41

Let me know if you need help with that part.

Answer 42

so we divide both sides by 3?

Answer 43

There's only one side, because this isn't an equation (there's no equal sign)

Answer 44

How would you simplify
6x+3x-x?

Answer 45

8x.

Answer 46

so it would be 8√6!

Answer 47

Yep. So simplify
\[6\sqrt6+3\sqrt{6}-\sqrt{6}\]

Answer 48

Exactly :)

Answer 49

so what would it be when we put it all together?

Answer 50

That's it. 8root6

Answer 51

okay thanks so much!

Answer 52

No problem!