User Name Remember Me? Password

 Algebra Pre-Algebra and Basic Algebra Math Forum

 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: 879 Thanks: 60 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: 879 Thanks: 60 Math Focus: सामान्य गणित $\displaystyle \sqrt{3}\times \sqrt{3}=3$ $\displaystyle \because \sqrt{3} = 3^{\frac{1}{2}}$ $\displaystyle \therefore \sqrt{3}\times \sqrt{3} = 3^{\frac{1}{2}}\times 3^{\frac{1}{2}} = 3^{\frac{1}{2}+\frac{1}{2}}=3^{1}=3$ Tags question, surd Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post mathsheadache Algebra 1 October 30th, 2014 05:22 PM Monokuro Trigonometry 2 May 6th, 2014 07:14 AM mathematics101010 Algebra 6 August 4th, 2013 04:00 AM prescott2006 Calculus 0 March 9th, 2008 11:00 AM mathematics101010 Elementary Math 4 December 31st, 1969 04:00 PM

 Contact - Home - Forums - Cryptocurrency Forum - Top      