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05-11-2006, 03:16 AM
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!

royhaas
05-11-2006, 07:50 AM
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.

soroban
05-11-2006, 08:36 AM
Hello, 4 little piggies mom!

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


34)\;\;\sqrt{3}\cdot\left(5\sqrt{3}\,+\,\sqrt{3}\r ight)
Add: \,\sqrt{3}\cdot(6\sqrt{3})

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



44)\;\;\frac{6}{10\,+\,\sqrt{2}}
Rationalize: \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}



\L\frac{14}{60\,-\,\sqrt{578}}
That radical looks suspiciously reducible . . .

Sure enough: \,\sqrt{578}\:=\:\sqrt{289\cdot2}\:=\:\sqrt{289}\c dot\sqrt{2}\:=\:17\sqrt{2}

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

Rationalize: \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}

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