anonymous
  • anonymous
Simplify the radical (squ96) I got this 2*48 3*32 4*24 6*16 8*12 12*8 16*6 @jdoe0001
Mathematics
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SOLVED
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chestercat
  • chestercat
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jim_thompson5910
  • jim_thompson5910
hint: \[\large \sqrt{96} = \sqrt{16*6}\]
anonymous
  • anonymous
@jim_thompson5910 u lost me but if @jdoe0001 can walk me though it
jim_thompson5910
  • jim_thompson5910
you factor 96 into two numbers where one is a perfect square

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jim_thompson5910
  • jim_thompson5910
then you use the rule \[\large \sqrt{x*y} = \sqrt{x}*\sqrt{y}\] to simplify
jdoe0001
  • jdoe0001
|dw:1378759684695:dw| thus \(\bf 96 = 2\times 2\times 2\times 2\times 2\times 3 \implies 2^2\times 2^2 \times 3 \implies (2\times 2)^2 \times 3 \implies 4^2\times 3\\ \quad \\ \sqrt{96} \implies \sqrt{4^2\times 3} \)
anonymous
  • anonymous
why didn't you get the last 2?
jdoe0001
  • jdoe0001
ohh yea.... shoot. missed that
jdoe0001
  • jdoe0001
\(\bf 96 = 2\times 2\times 2\times 2\times 2\times 3 \implies 2^2\times 2^2 \times 2 \times 3 \implies (2\times 2)^2 \times 2\times 3 \\\implies 4^2\times 2 \times 3\\ \quad \\ \sqrt{96} \implies \sqrt{4^2\times 2\times 3}\)
anonymous
  • anonymous
okay I understand that but why did you choose 2?
jdoe0001
  • jdoe0001
well, sure... it can be rewritten also using 2, like \(\bf 96 = 2\times 2\times 2\times 2\times 2\times 3 \implies 2^2\times 2^2 \times 2 \times 3 \\ \quad \\ \sqrt{96} \implies \sqrt{2^2\times 2^2 \times 2 \times 3 }\)
anonymous
  • anonymous
okay but dont there have to a # on the outside of the squ
jdoe0001
  • jdoe0001
yes, the numbers you take out, are the ones whose exponent match the radical squared root is "root 2", so any values with an exponent of "2" come out so what would that give you?
anonymous
  • anonymous
4squ6
jdoe0001
  • jdoe0001
yeap

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