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 January 12th, 2015, 08:05 AM #1 Newbie   Joined: Oct 2014 From: London Posts: 27 Thanks: 0 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?
 January 12th, 2015, 08:30 AM #2 Math Team     Joined: Jul 2013 From: काठमाडौं, नेपाल Posts: 901 Thanks: 61 Math Focus: सामान्य गणित denominator $\displaystyle (4-2\sqrt{3})\times (4+2\sqrt{3}) = 4\times 4+4\times 2\sqrt{3}-2\sqrt{3}\times 4 -2\sqrt{3}\times 2\sqrt{3}= 16 +8\sqrt{3}-8\sqrt{3}-12=4$
 January 12th, 2015, 08:36 AM #3 Math Team     Joined: Jul 2013 From: काठमाडौं, नेपाल Posts: 901 Thanks: 61 Math Focus: सामान्य गणित $\displaystyle \sqrt{3}\times \sqrt{3}=3$ $\displaystyle \because \sqrt{3} = 3^{\frac{1}{2}}$ $\displaystyle \therefore \sqrt{3}\times \sqrt{3} = 3^{\frac{1}{2}}\times 3^{\frac{1}{2}} = 3^{\frac{1}{2}+\frac{1}{2}}=3^{1}=3$

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