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anonymous

  • one year ago

Help with algebra?

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  1. anonymous
    • one year ago
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    Rationalize the denominator and simplify. \[\frac{ 3\sqrt{6}+5\sqrt{2} }{ 4\sqrt{6}-3\sqrt{2} }\] I know I need to multiply by the conjugate.\[\frac{ 4\sqrt{6}+3\sqrt{2} }{ 4\sqrt{6}+3\sqrt{2} }\] I tried doing all of the math but I must have made a mistake at some point because the answer I got does not fit.

  2. UsukiDoll
    • one year ago
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    yes we need to multiply by the conjugate. then we need to foil afterwards \[\frac{ 3\sqrt{6}+5\sqrt{2} }{ 4\sqrt{6}-3\sqrt{2} } \cdot \frac{ 4\sqrt{6}+3\sqrt{2} }{ 4\sqrt{6}+3\sqrt{2} }\]

  3. UsukiDoll
    • one year ago
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    hello @MrNood

  4. MrNood
    • one year ago
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    (a+b)(a-b) = a^2-b^2 so that is simpler than using FOIL for the denominator

  5. UsukiDoll
    • one year ago
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    the denominator will always been simpler though. because the middle terms from FOIL are gone leaving F L

  6. anonymous
    • one year ago
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    Ok I know I need to do that and for the numerator I got \[102+18\sqrt{3}+40\sqrt{3}\] and on the denominator I got 78 but that doesn't fit.

  7. UsukiDoll
    • one year ago
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    |dw:1437553510720:dw| try compute that for the numerator.. while I work out the denominator

  8. UsukiDoll
    • one year ago
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    |dw:1437553641257:dw|

  9. UsukiDoll
    • one year ago
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    |dw:1437553716764:dw| well the denominator is correct.

  10. anonymous
    • one year ago
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    Ok well I'm getting 72+\[72+\sqrt{3}+40\sqrt{3}+30\]

  11. anonymous
    • one year ago
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    there should be an 18 before the first square root of 3

  12. UsukiDoll
    • one year ago
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    you can combine \[18 \sqrt{3}+ 40\sqrt{3} \]

  13. UsukiDoll
    • one year ago
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    factor the square root part out you should get \[\sqrt{3}(18+40) \]

  14. UsukiDoll
    • one year ago
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    \[58\sqrt{3} \]

  15. Jhannybean
    • one year ago
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    \[\large 18 \color{red}{\sqrt{3}}+ 40\color{red}{\sqrt{3}}\] The red portion appearing in both terms means that they can be simplified further, remember that for the future :)

  16. anonymous
    • one year ago
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    Oh that actually helps a lot I just didn't know I could add those together. It still doesnt fit though? Should the final answer be\[\frac{ 58\sqrt{3}+102 }{ 78 }\]

  17. UsukiDoll
    • one year ago
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    maybe we can split it into fractions and reduce further... I see all even terms

  18. UsukiDoll
    • one year ago
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    \[\frac{ 58\sqrt{3}+102 }{ 78 } \rightarrow \frac{58\sqrt{3}}{78}+\frac{102}{78}\]

  19. Jhannybean
    • one year ago
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    Remember that the divisor, 78 can always be distributed to everything in the numerator as long as you're not changing what is in the numerator, only simplifying it.

  20. UsukiDoll
    • one year ago
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    \[\frac{29\sqrt{3}}{39} +\frac{17}{13}\]

  21. anonymous
    • one year ago
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    Ohh ok so what I did was divide 58, 102, and 78 by 2 so it does fit.\[\frac{ 51+29\sqrt{3} }{ }\]

  22. anonymous
    • one year ago
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    With 39 on the bottom

  23. anonymous
    • one year ago
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    I think that is the answer so thank you

  24. UsukiDoll
    • one year ago
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    \[\frac{ 51+29\sqrt{3} }{ 39 }\] that's right. :)

  25. UsukiDoll
    • one year ago
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    I just went too far in my reduction.. but by multiplying 3 on both sides of the other fraction it comes up to that

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