anonymous
  • anonymous
Simplify: √48x^9y^2 Answer: 4x^4y√3x Not sure how to get from point A to B, though.
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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phi
  • phi
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 \[ \sqrt{2 \cdot 2 \cdot 3} = 2\sqrt{3} \]
phi
  • phi
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 \[ x^4 \sqrt{x} \]
phi
  • phi
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.

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anonymous
  • anonymous
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?
phi
  • phi
48 is (2*2*2*2*3) (not (2*2*2*2*4)
anonymous
  • anonymous
That was a typo =p
phi
  • phi
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: \[ \sqrt{( 2 \cdot 2) \cdot ( 2 \cdot 2) \cdot 3} \\ 2 \cdot \cancel{2} \cdot 2 \cdot \cancel {2} \sqrt{3}\]
phi
  • phi
that gives you \[ 2 \cdot 2 \cdot \sqrt{3} = 4 \sqrt{3} \]
anonymous
  • anonymous
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!
anonymous
  • anonymous
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?
phi
  • phi
if you have 50 x's how many pairs do you have ?
anonymous
  • anonymous
Ah, okay, I see what you mean. I'd try to use common sense, then. =p Haha. Thanks, again!
phi
  • phi
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.
phi
  • phi
if that is unclear, play with smaller numbers and do it the hard way, and think about it.
anonymous
  • anonymous
Seriously, thanks a lot for helping me out!

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