anonymous
  • anonymous
Help with algebra?
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
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anonymous
  • anonymous
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.
UsukiDoll
  • UsukiDoll
yes we need to multiply by the conjugate. then we need to foil afterwards \[\frac{ 3\sqrt{6}+5\sqrt{2} }{ 4\sqrt{6}-3\sqrt{2} } \cdot \frac{ 4\sqrt{6}+3\sqrt{2} }{ 4\sqrt{6}+3\sqrt{2} }\]
UsukiDoll
  • UsukiDoll
hello @MrNood

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MrNood
  • MrNood
(a+b)(a-b) = a^2-b^2 so that is simpler than using FOIL for the denominator
UsukiDoll
  • UsukiDoll
the denominator will always been simpler though. because the middle terms from FOIL are gone leaving F L
anonymous
  • anonymous
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.
UsukiDoll
  • UsukiDoll
|dw:1437553510720:dw| try compute that for the numerator.. while I work out the denominator
UsukiDoll
  • UsukiDoll
|dw:1437553641257:dw|
UsukiDoll
  • UsukiDoll
|dw:1437553716764:dw| well the denominator is correct.
anonymous
  • anonymous
Ok well I'm getting 72+\[72+\sqrt{3}+40\sqrt{3}+30\]
anonymous
  • anonymous
there should be an 18 before the first square root of 3
UsukiDoll
  • UsukiDoll
you can combine \[18 \sqrt{3}+ 40\sqrt{3} \]
UsukiDoll
  • UsukiDoll
factor the square root part out you should get \[\sqrt{3}(18+40) \]
UsukiDoll
  • UsukiDoll
\[58\sqrt{3} \]
Jhannybean
  • Jhannybean
\[\large 18 \color{red}{\sqrt{3}}+ 40\color{red}{\sqrt{3}}\] The red portion appearing in both terms means that they can be simplified further, remember that for the future :)
anonymous
  • anonymous
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\[\frac{ 58\sqrt{3}+102 }{ 78 }\]
UsukiDoll
  • UsukiDoll
maybe we can split it into fractions and reduce further... I see all even terms
UsukiDoll
  • UsukiDoll
\[\frac{ 58\sqrt{3}+102 }{ 78 } \rightarrow \frac{58\sqrt{3}}{78}+\frac{102}{78}\]
Jhannybean
  • Jhannybean
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.
UsukiDoll
  • UsukiDoll
\[\frac{29\sqrt{3}}{39} +\frac{17}{13}\]
anonymous
  • anonymous
Ohh ok so what I did was divide 58, 102, and 78 by 2 so it does fit.\[\frac{ 51+29\sqrt{3} }{ }\]
anonymous
  • anonymous
With 39 on the bottom
anonymous
  • anonymous
I think that is the answer so thank you
UsukiDoll
  • UsukiDoll
\[\frac{ 51+29\sqrt{3} }{ 39 }\] that's right. :)
UsukiDoll
  • UsukiDoll
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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