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anonymous

  • 5 years ago

How would you fine the square root of this number: 3square root of 28?

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  1. anonymous
    • 5 years ago
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    Is this 3*Sqrt(28)?

  2. anonymous
    • 5 years ago
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    I think so I'm doing a worksheet and I'm trying to workout the problem the answer is suppoed to 6 square root of 7

  3. anonymous
    • 5 years ago
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    3sqrt(28)=3sqrt(4*7)=3sqrt((2^2)*7)=3sqrt(2^2)sqrt(7)=3*2*sqrt7=6sqrt7

  4. anonymous
    • 5 years ago
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    How would you do the square root of this problem: -3 square root of 54. The answer is supposed to be -9 square root of 6.

  5. anonymous
    • 5 years ago
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    can you write 54 into product of prime numbers (1,2,3,5,7,11,....) for me?

  6. anonymous
    • 5 years ago
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    2,3,3,3

  7. anonymous
    • 5 years ago
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    my computer went crazy sorry for so many entries

  8. anonymous
    • 5 years ago
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    ok, there you have 54=2x3x3x3. there is a rule like this: sqrt(a^2)=a, a>0 so you have 54=2x3x3^2 sqrt54=sqrt(2x3x3^2)=sqrt(2x3) x sqrt(3^2) you do the rest for me? the thing here is you need to understand how it work, then you can do any problem

  9. anonymous
    • 5 years ago
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    I don't understand what you want me to do?

  10. anonymous
    • 5 years ago
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    -.- \[-3\sqrt{54}=-3\sqrt{2*3*3^{2}}=-3\sqrt{6}\sqrt{3^{2}}=-3*3*\sqrt{6}=-9\sqrt{6}\]

  11. anonymous
    • 5 years ago
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    but you still have to understand the method

  12. anonymous
    • 5 years ago
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    I understand how you factored it down but how do you get the answer of -9 square root of 6?

  13. anonymous
    • 5 years ago
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    √(3^2)=3, you understand this right? so√(54)=√(2*3*3^3)=√(6*3^3)=(√6)*(√3^3)=(√6)*3 then -3*√(54)=-3*(√6)*3=-3*3*(√6)=-9√6 clear?

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