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- Mar 28th 2010, 05:07 PMknowledgekickIntermediate Algebra Help Please
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- Mar 28th 2010, 05:54 PMProve It
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.

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

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

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

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

2. Your question is unreadable.

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

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

3. Just expand the brackets...

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

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

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

$\displaystyle = 6\sqrt{15} + 36$

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

4. FOIL it out

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

$\displaystyle = 9 + 20\sqrt{3} + 100$

$\displaystyle = 109 + 20\sqrt{3}$.

The rest of the questions are unreadable.