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 June 22nd, 2014, 01:24 AM #1 Member   Joined: Jun 2014 From: Math Forum Posts: 67 Thanks: 4 rational  June 22nd, 2014, 05:53 AM #2 Global Moderator   Joined: Dec 2006 Posts: 21,036 Thanks: 2274 $\displaystyle \sqrt[\large3]{45+29\sqrt2} + \sqrt[\large3]{45-29\sqrt2} = (3 + \sqrt2) + (3 - \sqrt2) = 6$ June 22nd, 2014, 06:05 AM #3 Senior Member   Joined: Feb 2010 Posts: 714 Thanks: 151 Let $\displaystyle A= \sqrt{x} + \sqrt{y}$ where $\displaystyle x=45+29\sqrt{2}$ and $\displaystyle y=45-29\sqrt{2}$. Note that $\displaystyle x+y=90$, $\displaystyle xy=2025-841 \cdot 2=343=7^3$ Then $\displaystyle A^3 = (\sqrt{x} + \sqrt{y})^3$ $\displaystyle A^3 = x + 3\sqrt{x^2y}+3\sqrt{xy^2}+y$ $\displaystyle A^3 = (x+y) + 3\sqrt{xy}(\sqrt{x} + \sqrt{y})$ $\displaystyle A^3 = 90 + 21A$ $\displaystyle A^3 - 21A - 90=0$ This equation has two complex roots and one real real. Using the rational roots theorem we find that the only real root is 6. Thus $\displaystyle \sqrt{45+29\sqrt{2}} + \sqrt{45-29\sqrt{2}}=6$ and is therefore a rational number. Tags rational 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 LucasPR Abstract Algebra 13 May 1st, 2013 11:32 PM drewm Algebra 4 July 2nd, 2011 04:02 PM stuart clark Algebra 1 May 19th, 2011 12:14 AM lepeisi Abstract Algebra 3 March 30th, 2011 01:00 PM advancedfunctions Calculus 2 April 6th, 2010 12:41 PM

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