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Math Help - Intermediate Algebra Help Please

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    Intermediate Algebra Help Please

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    Last edited by knowledgekick; April 2nd 2010 at 10:20 AM.
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  2. #2
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    Quote Originally Posted by knowledgekick View Post
    Can you guys help me out with these questions? I'll list the ones I'm having problems with...

    Sorry about the formatting. I tried typing it as best I could. I did / for fractions and typed out any symbol I couldn't paste.

    1) √54 - √216=

    2) √200x to the 7th power - √468x - √128x to the 7th power + √1,573x=

    3) 6√3(√5+2√3)=

    4) (√3+10)(√3+10)=

    5) 4th route √17/3=

    6) cubed route √2/121k squared=

    7) 16/8-√128=

    8) 13/√7-√13=
    1. You can't add or subtract square roots unless they have the same base, just like in algebra you can't add or subtract anything unless you have like terms. So here you have to simplify the surds and hope you get the same base.

    \sqrt{54} - \sqrt{216} = \sqrt{9\cdot 6} - \sqrt{36\cdot 6}

     = \sqrt{9}\cdot\sqrt{6} - \sqrt{36}\cdot\sqrt{6}

     = 3\sqrt{6} - 6\sqrt{6}

     = -3\sqrt{6}.


    2. Your question is unreadable.

    Is it \sqrt{200x^7} - \sqrt{468x} - \sqrt{128x^7}+\sqrt{1573x}?

    Or is it \sqrt{200}x^7 - \sqrt{468}x - \sqrt{128}x^7+\sqrt{1573}x?


    3. Just expand the brackets...

    6\sqrt{3}(\sqrt{5} + 2\sqrt{3}) = 6\sqrt{3}\sqrt{5} +6\sqrt{3}\cdot 2\sqrt{3}

     = 6\sqrt{15} + 12\sqrt{9}

     = 6\sqrt{15} + 12\cdot 3

     = 6\sqrt{15} + 36

     = 6(\sqrt{15} + 6).


    4. FOIL it out

    (\sqrt{3} + 10)(\sqrt{3} + 10) = \sqrt{3}\sqrt{3} + 10\sqrt{3} + 10\sqrt{3} + 100[/tex]

     = 9 + 20\sqrt{3} + 100

     = 109 + 20\sqrt{3}.



    The rest of the questions are unreadable.
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