Multiplying and dividing Radicals.... Help Pleaseeeeee!

34. (sqrt)3{5(sqrt)s +(sqrt)3} I come up with 8(sqrt)6
Am I on the right track or do I need to buy a whole new train??

44. 6/ {10+(sqrt)2} I get {60 + 6 (sqrt 2)} / 96

Last one 14/{60 -(sqrt)578} I get {840-14(sqrt578)}/{3022}

Any and all help would be great but I'd rather get started in the right directions over just getting the answers...thank again!

Can't tell about the first problem because of "(sqrt)s". The last two have the wrong sign in the numerators in front of the radical term. They can also be simplified a little more.

Hello, 4 little piggies mom!

There's a typo in the first one . . . I'll make a guess.

$\displaystyle 34)\;\;\sqrt{3}\cdot\left(5\sqrt{3}\,+\,\sqrt{3}\right)$

Click to expand...

Add: $\displaystyle \,\sqrt{3}\cdot(6\sqrt{3})$

Multiply: $\displaystyle \,6\cdot\sqrt{3}\cdot\sqrt{3}\:=\:6\cdot3\:=\:18$

$\displaystyle 44)\;\;\frac{6}{10\,+\,\sqrt{2}}$

Click to expand...

Rationalize: \(\displaystyle \L\,\frac{6}{10\,+\,\sqrt{2}}\,\cdot\,\frac{10\,-\,\sqrt{2}}{10\,-\,\sqrt{2}} \;= \;\frac{6(10\,-\,\sqrt{2}}{10^2\,-\,(\sqrt{2})^2}\;=\;\frac{6(10\,-\sqrt{2})}{98} \;= \;\frac{3(10\,-\,\sqrt{2})}{49}\)

\(\displaystyle \L\frac{14}{60\,-\,\sqrt{578}}\)

Click to expand...

That radical looks suspiciously reducible . . .

Sure enough: $\displaystyle \,\sqrt{578}\:=\:\sqrt{289\cdot2}\:=\:\sqrt{289}\cdot\sqrt{2}\:=\:17\sqrt{2}$

So we have: \(\displaystyle \L\,\frac{14}{60\,-\,17\sqrt{2}}\)

Rationalize: \(\displaystyle \L\,\frac{14}{60\,-\,17\sqrt{2}}\,\cdot\,\frac{60\,+\,17\sqrt{2}}{60\,+\,17\sqrt{2}} \;= \;\frac{14(60\,+\,17\sqrt{2})}{60^2\,-\,(17\sqrt{2})^2} \;= \;\frac{14(60\,+\,17\sqrt{2})}{3600\,-\,578}\)

\(\displaystyle \L\;\;= \;\frac{14(60\,+\,17\sqrt{2})}{3022} \;= \;\frac{7(60\,+\,17\sqrt{2})}{1511}\)

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Some advice:

Do not multiply out the numerator . . . leave it in factored form.

Get used to the "rationalizing step": $\displaystyle \,(a\,-\,b)(a\,+\,b)\:=\:a^2\,-\,b^2$
So that: $\displaystyle \,(3\,+\,\sqrt{2})(3\,-\,\sqrt{2})\,$ goes directly to: $\displaystyle \,3^2\,-\,(\sqrt{2})^2\:=\:9\,-\,2\:=\:7$
$\displaystyle \;\;$without going through the FOIL each time.