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anonymous

  • one year ago

Algebra Help?

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  1. anonymous
    • one year ago
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    \[\frac{ \sqrt{a}+2\sqrt{y} }{ \sqrt{a}-2\sqrt{y} }\]

  2. anonymous
    • one year ago
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    Rationalize the denominator and simplify.

  3. anonymous
    • one year ago
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    Why would it be the conjugate of the numerator and not the denominator? Shouldn't it be multiplied by \[\frac{ \sqrt{a}+2\sqrt{y} }{ \sqrt{a}+2\sqrt{y} }\]

  4. rishavraj
    • one year ago
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    u r correct

  5. rishavraj
    • one year ago
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    it should be \[\frac{ \sqrt{a} + 2\sqrt{y} }{ \sqrt{a} - 2\sqrt{y}} \times \frac{ \sqrt{a} + 2\sqrt{y} }{ \sqrt{a} + 2\sqrt{y}}\]

  6. anonymous
    • one year ago
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    Ok so that's the part I always have trouble with. I'm not the best at multiplying radicals.

  7. rishavraj
    • one year ago
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    \[(a - b) \times (a + b) = a^2 - b^2 \]

  8. rishavraj
    • one year ago
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    so it would be \[\frac{ (\sqrt{a} + 2\sqrt{y})^2 }{ (\sqrt{a})^2 - (2\sqrt{y})^2 }\]

  9. mathstudent55
    • one year ago
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    |dw:1437103101604:dw|

  10. anonymous
    • one year ago
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    Ok so @rishavraj then its simplified to get \[\frac{ a+2y }{ a-2y }\] or would those 2s be a 4s?

  11. mathstudent55
    • one year ago
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    |dw:1437103159002:dw|

  12. rishavraj
    • one year ago
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    \[(2\sqrt{y})^2 = 4y ~~~~ and ~~~(a + b)^2 = a^2 + b^2 + 2ab\]

  13. rishavraj
    • one year ago
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    and \[(\sqrt{a} + 2\sqrt{y})^2 = (\sqrt{a})^2 + (2\sqrt{y})^2 + 2 \times \sqrt{a} \times 2\sqrt{y}\]

  14. anonymous
    • one year ago
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    Why the \[+2\times \sqrt{a}\times \sqrt{y}2\]

  15. anonymous
    • one year ago
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    I mean \[2\sqrt{y}\]

  16. rishavraj
    • one year ago
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    (a + b)^2 = a^2 + b^2 + 2ab

  17. anonymous
    • one year ago
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    OK thanks I got it

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