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January 12th, 2015, 08:05 AM   #1
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Can my surd question be right?

Question is
$\displaystyle \frac{-6}{4-2\sqrt{3}}$

So, $\displaystyle \frac{-6}{4-2\sqrt{3}}$ x $\displaystyle \frac{4+2\sqrt{3}}{4+2\sqrt{3}}$

= $\displaystyle \frac{-24-12\sqrt{3}}{16+8\sqrt{3}{-8\sqrt{3}-4}}$

=$\displaystyle \frac{-24-12\sqrt{3}}{12}$

$\displaystyle =\frac{-24}{12}$-$\displaystyle \frac{12\sqrt{3}}{12}$

Im confused because $\displaystyle -2\sqrt{3}$x$\displaystyle +2\sqrt{3}$=$\displaystyle -4$?

Because the square roots would cancel out leaving just $\displaystyle -4$

..or would it be -4x3?
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January 12th, 2015, 08:30 AM   #2
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$\displaystyle (4-2\sqrt{3})\times (4+2\sqrt{3}) = 4\times 4+4\times 2\sqrt{3}-2\sqrt{3}\times 4 -2\sqrt{3}\times 2\sqrt{3}= 16 +8\sqrt{3}-8\sqrt{3}-12=4$
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January 12th, 2015, 08:36 AM   #3
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$\displaystyle \sqrt{3}\times \sqrt{3}=3$
$\displaystyle \because \sqrt{3} = 3^{\frac{1}{2}}$
$\displaystyle \therefore \sqrt{3}\times \sqrt{3} = 3^{\frac{1}{2}}\times 3^{\frac{1}{2}} = 3^{\frac{1}{2}+\frac{1}{2}}=3^{1}=3$
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