anonymous
  • anonymous
Simplify 5√3•4√3=
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
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DanJS
  • DanJS
string of multiplication \[5 * \sqrt{3} * 4 * \sqrt{3}\]
anonymous
  • anonymous
And
DanJS
  • DanJS
remember a square root is the same as raised to 1/2 power root 3 times root 3 will cancel the square root, \[\large \sqrt{x}*\sqrt{x} = [\sqrt{x}]^2 = x^{(1/2)*2} = x\]

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DanJS
  • DanJS
so you just get 5 * 4 * 3
anonymous
  • anonymous
OH so that's all
anonymous
  • anonymous
How about this one √8•6√5
anonymous
  • anonymous
DanJs can you please help me with the rest
DanJS
  • DanJS
yeah, you can combine those roots since they are multiplied 6*root(8*5) 6* root(40) simplify the root 40
anonymous
  • anonymous
How do I simplify root 40
DanJS
  • DanJS
6* root(4 * 10) = 6* root(4) * root(10) = 6 * 2 * root 10 12*root(10)
DanJS
  • DanJS
break it up into any squares it may contain , here 4 * 10 = 2^2 * 10,
DanJS
  • DanJS
you can pull that 2 out front root 2^2 = 2
DanJS
  • DanJS
break the numbers into their prime factors, and take out any perfect squares you can, if any
anonymous
  • anonymous
Thank you I also have two more I really don't understand anything about roots
anonymous
  • anonymous
If u can help with these
1 Attachment
retirEEd
  • retirEEd
To make the first equation simpler to understand, let ignore the root by letting x=root(5) What would the answer be for 3x+ 11x-x ?
anonymous
  • anonymous
14x-x
retirEEd
  • retirEEd
yes but what is 14x - 1x ?
anonymous
  • anonymous
13x
retirEEd
  • retirEEd
correct now recall x = root(5) so substitute x back into your equation to get the final answer
anonymous
  • anonymous
How do I do that
retirEEd
  • retirEEd
where you see an x, replace it with a root(5)
anonymous
  • anonymous
So is my answer 13√5
retirEEd
  • retirEEd
a simpler way to do the first equation is to factor out the root(5) such that 3root(5) + 11root(5) - root(5) = (3 +11 -1)root(5) = 13root(5)
retirEEd
  • retirEEd
yes
anonymous
  • anonymous
Ok thank you so much what about the second equation
retirEEd
  • retirEEd
simplify the first two terms and way DanJS showed you and sum the terms. Look for a common root for all three terms.
anonymous
  • anonymous
I don't get it
anonymous
  • anonymous
Root 3
retirEEd
  • retirEEd
break the numbers into their prime factors, and take out any perfect squares you can, if any per DanJS
anonymous
  • anonymous
I don't know how to break the numbers into their prime factors
retirEEd
  • retirEEd
do you know what prime factors are?
anonymous
  • anonymous
Yeah I do but what about the 3
retirEEd
  • retirEEd
3 is a prime number
anonymous
  • anonymous
So what's next
retirEEd
  • retirEEd
In number theory, the prime factors of a positive integer are the prime numbers that divide that integer exactly.
retirEEd
  • retirEEd
What are the prime factors for 12?
anonymous
  • anonymous
2 6 2 3
retirEEd
  • retirEEd
Your close what are the prime factors of 6 ?
anonymous
  • anonymous
2 3
retirEEd
  • retirEEd
Yes, so the prime factor of 12 is 2, 2 and 3 such that if I take 2 x 2 x 3 I get 12. This is the same as 2^2 x 3 = 12 break the numbers into their prime factors, and take out any perfect squares you can, if any per DanJS The perfect square is the 2^2 or 4
anonymous
  • anonymous
So what's my final answer
retirEEd
  • retirEEd
so to simplify root(12) = root(4) times root(3) the square root of 4 is 2 so root(12) is 2root(3) Now you have to simplify root(27) by the same method then add all 3 terms with the common root
anonymous
  • anonymous
Can you demonstrate it please
retirEEd
  • retirEEd
figure out the prime factors in 27 find a perfect square and add the terms
anonymous
  • anonymous
27 9 3 3 3
DanJS
  • DanJS
yes, when you are adding together terms containing roots, the stuff under the root has to be the same you cant add root(2) to root(5) for example,
DanJS
  • DanJS
so really you may not be able to simplify at all maybe, follow what you did with the factoring first
DanJS
  • DanJS
\[\sqrt{12}=\sqrt{3*4}=\sqrt{3*2*2} = \sqrt{3*2^2} = \sqrt{3}*\sqrt{2^2} = 2*\sqrt{3}\]
DanJS
  • DanJS
27 = 9 * 3 = 3 * 3 * 3 = 3^2 * 3 root(27) = root(3^2)*root(3) = 3*root 3
DanJS
  • DanJS
all are multiples of root 3 now, you can total them up
anonymous
  • anonymous
So what's my answer going to be
DanJS
  • DanJS
\[2\sqrt{3}-3\sqrt{3}+\sqrt{3}\] 2 -3 + 1 or 0*root(3) = 0
DanJS
  • DanJS
simplifies to zero
anonymous
  • anonymous
Yeah so what do I write
anonymous
  • anonymous
Just zero

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