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anonymous

  • 4 years ago

How do I simplify 2√6 / 2√6 + 1? The answer is 24 - 2√6 / 23?.. How? D:

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  1. asnaseer
    • 4 years ago
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    use the identity:\[a^2-b^2=(a+b)(a-b)\]to remove the radical expression in the denominator. so, in this case, multiply numerator and denominator by \(2\sqrt{6}-1\)

  2. anonymous
    • 4 years ago
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    that's what I did but should I be multiply 2 x 2 cand √6 x √6?

  3. asnaseer
    • 4 years ago
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    \[\frac{2\sqrt{6}}{2\sqrt{6}+1}=\frac{2\sqrt{6}}{2\sqrt{6}+1}*\frac{2\sqrt{6}-1}{2\sqrt{6}-1}=\frac{2\sqrt{6}(2\sqrt{6}-1)}{(2\sqrt{6}+1)(2\sqrt{6}-1)}\]\[=\frac{(2*\sqrt{6}*2*\sqrt{6}-2\sqrt{6})}{(2\sqrt{6})^2-1^2}=\frac{4*6-2\sqrt{6}}{4*6-1}=\frac{24-2\sqrt{6}}{23}\]

  4. anonymous
    • 4 years ago
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    Although, I don't understand why the exercise is called "simplify". There is nothing simple about the alternate form.

  5. asnaseer
    • 4 years ago
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    having no radicals in the denominator is considered "simpler"

  6. anonymous
    • 4 years ago
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    :)

  7. anonymous
    • 4 years ago
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    why did you do 1^6?

  8. anonymous
    • 4 years ago
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    I meant 1^2

  9. asnaseer
    • 4 years ago
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    \[a(b+c)=ab+ac\]so:\[2\sqrt{6}(2\sqrt{6}-1)=(2\sqrt{6})*(2\sqrt{6})-(2\sqrt{6})*1\]\[=2*2*\sqrt{6}*\sqrt{6}-2\sqrt{6}=4*6-2\sqrt{6}=24-2\sqrt{6}\]

  10. anonymous
    • 4 years ago
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    you multiplied it by 1, why?

  11. asnaseer
    • 4 years ago
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    |dw:1327863505800:dw|

  12. anonymous
    • 4 years ago
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    ohh, so it's not -2√6?

  13. asnaseer
    • 4 years ago
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    it is - look at what I typed up there, you do get \(-2\sqrt{6}\) when you multiply \(2\sqrt{6}\) by -1.

  14. asnaseer
    • 4 years ago
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    I guess I used:\[a(b-c)=ab-ac\]in this case.

  15. anonymous
    • 4 years ago
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    ohh.. sorry :(

  16. asnaseer
    • 4 years ago
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    do you understand the process now?

  17. anonymous
    • 4 years ago
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    yes, I do thank you!

  18. asnaseer
    • 4 years ago
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    yw

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