anonymous
  • anonymous
Help with rationalizing denominators equation is 1 over the square root 96 not sure what to do
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
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anonymous
  • anonymous
\[1 FRACTION \sqrt{96}\]
anonymous
  • anonymous
\[1/\sqrt{96} \times \sqrt{96}/\sqrt{96} = \sqrt{96}/9\]
anonymous
  • anonymous
\[1/\sqrt{96}\]\[\sqrt{96}/(\sqrt{96}*\sqrt{96})\]\[\sqrt{96}/96\]

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anonymous
  • anonymous
I understand I have to write the smallest number of a perfect square, but unsure of how to water that down. haha
anonymous
  • anonymous
1 is the numerator. 96 is the denominator
anonymous
  • anonymous
96|2 48|2 24|2 12|2 06|2 03|3 01|- \[\sqrt{2^{5}*3}=4\sqrt{2*3}=4\sqrt{?}\]
anonymous
  • anonymous
2?
anonymous
  • anonymous
\[4\sqrt{6}/96 = \sqrt{6}/\]
anonymous
  • anonymous
\[\sqrt{6}/24 \]
anonymous
  • anonymous
would you mind explaining it?
anonymous
  • anonymous
?
anonymous
  • anonymous
in english haha
anonymous
  • anonymous
to simplify a sqrt, you devide the number by primary numbers: 2,3,5,7,11,13,... until you get 1. The amount of primary is the amount of times you are multiplying it so as i did: 96|2 96/2 = 48 48|2 48/2 = 24 24|2 24/2 = 12 12|2 12/2 = 06 06|2 06/2 = 03 03|3 i can't devide 3 by 2 to get a non-decimal number so, i devide by the next primary which is 3 -> 03/3 = 1 01|- I have five 2's and one 3 therefore, 2^5*3 = 96
anonymous
  • anonymous
\[2^{5}*3^{1} = 2^{2}*2^{2}*2^{1}*3^{1} = 96\]
anonymous
  • anonymous
I somewhat get it a bit better.
anonymous
  • anonymous
Times both the numerator and denomiator by \[\sqrt{96}\] to get \[1*\sqrt{96}/\sqrt{96}*\sqrt{96}\] Then the denominator becomes 96 and the numerator becomes \[\sqrt{96}\] You then can simplify the numerator; \[\sqrt{96} = \sqrt{16*64} =4\sqrt{6}\] As now both the numerator and denominator are divisible by 4 we can divide them both to make \[\sqrt{6}/24\]
anonymous
  • anonymous
what should the answer look like?
anonymous
  • anonymous
ah, I got it. I am just hazy on the whole find 1 thing that you mentioned earlier. how did you determine 16 and 64 for the square roots of 96??
anonymous
  • anonymous
The answer is the smallest surd you can get ; \[\sqrt{6}\] over the smallest integer you can get; 24 which equates to \[\sqrt{6}/24\] ^^^ that's your answer..
anonymous
  • anonymous
got that. so I have 10 over square root of 45 first step, mult 45 by 45
anonymous
  • anonymous
OH DEAR>> I meant 16*6 not *64.. my computer is very slow so i can't see my errors in typing
anonymous
  • anonymous
so is it that 6 is the smallest square root, and 16 is because it goes into 96 6 times--leaving a relation between 6 and 16?
anonymous
  • anonymous
yeah :)
anonymous
  • anonymous
ohhhkay cool. thanks dude!
anonymous
  • anonymous
what if it has a factored numerator?such as \[\sqrt{6}/\sqrt{28}\]

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