• Mar 28th 2010, 05:07 PM
knowledgekick
Done
• Mar 28th 2010, 05:54 PM
Prove It
Quote:

Originally Posted by knowledgekick
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.

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

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

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

$= -3\sqrt{6}$.

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.