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simplify 8√63x^5

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do you have an answer?
no, i'm still very confused on how to even start
\[\LARGE \sqrt{63x^5} \implies \sqrt{9 \times 7 \times x^4 \times x}\] does that help?

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a little, the exponents are the things that trip me up the most though.
\[\LARGE \sqrt{9 \times 7 \times x^4 \times x} \implies \sqrt 9 \times \sqrt{x^4} \times \sqrt 7 \times \sqrt x\]
\[\LARGE \sqrt{x^a} \implies x^{a/2}\]
the one before that was too big for the screen lol but let me see if i can keep working this out :)
okay :D
I got |dw:1343225546659:dw|
  • phi
you can also pull x^4 out of the square root to get \[ 24 x^2 \sqrt{7x} \]
how would it be x^2 outside of the radical and inside it only be x? I need a bit of clarification, I'm not very good at math :/
  • phi
by definition: sqrt(2*2) = sqrt(2)*sqrt(2) = 2 or, in general sqrt(x)*sqrt(x)= x
  • phi
x^5 is short hand for x*x*x*x*x (people don't like typing it out, so they use the short hand) sqrt(x*x*x*x*x) = sqrt(x)*sqrt(x)*sqrt(x)*sqrt(x)*sqrt(x) each pair simplfies to x: x*x*sqrt(x) or, using the short hand x^2 * sqrt(x)
  • phi
Of course, if you do this problem a bunch of times, you start saying things like: if the exponent is even (example: x^4) I can "pull out" x^4 , and divide the power by 2. in other words sqrt(x^4) = x^2
okay! so all we did when we pulled out 4 from x^5 is divide it by two?
  • phi
the exponent 4 is divided by 2 when you pull x^4 out of a square root. if you have x^5 you write it as x^4 times x. then pull out the x^4 and leave the x inside
I got it now! :D thank you!

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