My Math Forum Surds

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 June 26th, 2012, 08:06 AM #1 Senior Member   Joined: Sep 2011 Posts: 395 Thanks: 0 Surds I've shown part of my solution, as it gets pretty convoluted after what I've shown. I can't tell if I'm on the right track with this one. Can anyone help me out? Many thanks. Q. Given that $(a+\sqrt{3})(b-\sqrt{3})=7+3\sqrt{3}$ & that a & b are positive integers, find the value of a & b. Attempt: $(a+\sqrt{3})=\frac{7+3\sqrt{3}}{b-\sqrt{3}}$ => $a+\sqrt{3}=\frac{7+3\sqrt{3}}{b-\sqrt{3}}.\frac{b+\sqrt{3}}{b-\sqrt{3}$ => $a+\sqrt{3}=\frac{7b+7\sqrt{3}+3b\sqrt{3}+3(3)}{b^2-3}$ => $a+\sqrt{3}=\frac{7b+(7+3b)\sqrt{3}+9}{b^2-3}$ => $\frac{7b+7\sqrt{3}+3b\sqrt{3}+3(3)}{b^2-3}(b-\sqrt{3})=7+3\sqrt{3}$ Ans.: (From text book): a = 2, b = 5
 June 26th, 2012, 08:24 AM #2 Member   Joined: Jun 2012 From: UK Posts: 39 Thanks: 0 Re: Surds Usual method would be, starting from your original equation, multiply out the brackets. Then for the two sides to be equal, the coefficients of root(3) on either side must be equal; and the two terms not involving root(3) must also be equal. This will give you two equations, which you can then go on to solve.
 June 26th, 2012, 09:03 AM #3 Global Moderator     Joined: Oct 2008 From: London, Ontario, Canada - The Forest City Posts: 7,807 Thanks: 1045 Math Focus: Elementary mathematics and beyond Re: Surds (a + ?(3))(b - ?(3)) = ab - a?(3) + b?(3) - 3 = 7 + 3?(3). Rational part: ab - 3 = 7 Irrational part: b - a = 3. b = a + 3 a(a + 3) - 3 = 7 a² + 3a - 10 = 0 (a - 2)(a + 5) = 0, a = 2, -5. (a, b) = (2, 5), (-5, -2).
 June 26th, 2012, 10:01 AM #4 Senior Member   Joined: Sep 2011 Posts: 395 Thanks: 0 Re: Surds Ok. Thanks guys.

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