## anonymous one year ago Help with algebra?

1. 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.

2. UsukiDoll

yes we need to multiply by the conjugate. then we need to foil afterwards $\frac{ 3\sqrt{6}+5\sqrt{2} }{ 4\sqrt{6}-3\sqrt{2} } \cdot \frac{ 4\sqrt{6}+3\sqrt{2} }{ 4\sqrt{6}+3\sqrt{2} }$

3. UsukiDoll

hello @MrNood

4. MrNood

(a+b)(a-b) = a^2-b^2 so that is simpler than using FOIL for the denominator

5. UsukiDoll

the denominator will always been simpler though. because the middle terms from FOIL are gone leaving F L

6. 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.

7. UsukiDoll

|dw:1437553510720:dw| try compute that for the numerator.. while I work out the denominator

8. UsukiDoll

|dw:1437553641257:dw|

9. UsukiDoll

|dw:1437553716764:dw| well the denominator is correct.

10. anonymous

Ok well I'm getting 72+$72+\sqrt{3}+40\sqrt{3}+30$

11. anonymous

there should be an 18 before the first square root of 3

12. UsukiDoll

you can combine $18 \sqrt{3}+ 40\sqrt{3}$

13. UsukiDoll

factor the square root part out you should get $\sqrt{3}(18+40)$

14. UsukiDoll

$58\sqrt{3}$

15. anonymous

$\large 18 \color{red}{\sqrt{3}}+ 40\color{red}{\sqrt{3}}$ The red portion appearing in both terms means that they can be simplified further, remember that for the future :)

16. anonymous

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

maybe we can split it into fractions and reduce further... I see all even terms

18. UsukiDoll

$\frac{ 58\sqrt{3}+102 }{ 78 } \rightarrow \frac{58\sqrt{3}}{78}+\frac{102}{78}$

19. anonymous

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

$\frac{29\sqrt{3}}{39} +\frac{17}{13}$

21. anonymous

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

With 39 on the bottom

23. anonymous

I think that is the answer so thank you

24. UsukiDoll

$\frac{ 51+29\sqrt{3} }{ 39 }$ that's right. :)

25. 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