Question

Simplify: √48x^9y^2
Answer: 4x^4y√3x

Not sure how to get from point A to B, though.

Answer 1

the main idea of simplifying a square root is if you have "pair" inside the square root
you can take out the pair, and put *one* of the pair outside the square root

Example:
\[ \sqrt{2 \cdot 2} = 2 \]
if there is nothing left inside the square root, we can ignore it.

if there is still something inside the square root we have to keep it. Example
\[ \sqrt{2 \cdot 2 \cdot 3} = 2\sqrt{3} \]

Answer 2

for letters, like x and y, you look for pairs.

remember that x^9 is short hand for x*x*x*x*x*x*x*x*x (x times itself 9 times)
hopefully you see there are lots of pairs you can take out.
there are 4 pairs: (x*x) * (x*x) * (x*x) * (x*x) * x with one x left over.
you pull out the 4 pairs, and replace them with *one* x from each pair:
x*x*x*x * sqrt(x)
of course people write this as
\[ x^4 \sqrt{x} \]

Answer 3

for numbers, break the number into prime factors
48= 2*24= 2*2*12 = 2*2*2*6 = 2*2*2*2*3
now do the same thing as for letters. pull out the pairs.

Answer 4

Ok, so I sort of get that--separately. I now understand why x^9 became x^4 and why y^2 became y. I'm still confused about everything else. The x that stays inside is the one left over, right? And I'm assuming that the 3 left inside is from when we found the prime factors of 48 (2*2*2*2*4), because like you said, the non-pairs remain? But what about the 4? How do you get 4 from 2*2*2*2? Like this?: (2*2)(2*2) and then lump the 2's in pairs together and multiply them so that it'd be 2*2=4?

Answer 5

48 is (2*2*2*2*3) (not (2*2*2*2*4)

Answer 6

That was a typo =p

Answer 7

let's look at the 48
\[ \sqrt{48} = \sqrt{( 2 \cdot 2) \cdot ( 2 \cdot 2) \cdot 3} \]
you pull out the pairs, but keep only one 2 from each pair:
\[ \sqrt{( 2 \cdot 2) \cdot ( 2 \cdot 2) \cdot 3} \\ 2 \cdot \cancel{2} \cdot 2 \cdot \cancel {2} \sqrt{3}\]

Answer 8

that gives you
\[ 2 \cdot 2 \cdot \sqrt{3} = 4 \sqrt{3} \]

Answer 9

Ooh, I see! Wow. Okay. Thanks a bunch for taking the time to explain this. I was just not getting it at all. I'm going to need a lot more practice, but I'm pretty sure I got the basis of solving this down. Thanks a lot!

Answer 10

Oh, wait, I have a question. Suppose they'd give an exponent such as, oh, x^50 or something (is that even possible?). How would I solve that without writing down 50 x's or something?

Answer 11

if you have 50 x's how many pairs do you have ?

Answer 12

Ah, okay, I see what you mean. I'd try to use common sense, then. =p Haha. Thanks, again!

Answer 13

if you had
\[ \sqrt{x^{51} }\]
you know you want an even number of x's (so you can divide by 2), so write it as
\[ \sqrt{x^{50} \cdot x} \]
and "pull out" the x^50 (x times itself 50 times) and replace it with x times itself 25 times (tossing 1/2 the x's)
\[ x^{25}\sqrt{x} \]
and leaving 1 x left inside the square root.

Answer 14

if that is unclear, play with smaller numbers and do it the hard way, and think about it.

Answer 15

Seriously, thanks a lot for helping me out!