My Math Forum Equation With An Irrational Number

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 November 30th, 2017, 03:16 PM #1 Senior Member   Joined: Oct 2013 From: New York, USA Posts: 619 Thanks: 83 Equation With An Irrational Number If 2x + x*(square root of 2) = 6, what is x?
 November 30th, 2017, 03:21 PM #2 Senior Member   Joined: Feb 2016 From: Australia Posts: 1,722 Thanks: 600 Math Focus: Yet to find out. $x = \dfrac{6}{(2 + \sqrt{2})}$?
 November 30th, 2017, 03:35 PM #3 Math Team   Joined: Jan 2015 From: Alabama Posts: 3,261 Thanks: 894 In a little more detail: $\displaystyle 2x+ x\sqrt{2}= (2+ \sqrt{2})x= 6$ Divide both sides by $\displaystyle 2+ \sqrt{2}$ If you don't like that $\displaystyle \sqrt{2}$ in the denominator, rationalize the denominator by multiplying by $\displaystyle \frac{2- \sqrt{2}}{2- \sqrt{2}}$: $\displaystyle \frac{6}{2+ \sqrt{2}}\frac{2- \sqrt{2}}{2- \sqrt{2}}= \frac{6(2- \sqrt{2})}{(2+ \sqrt{2})(2- \sqrt{2}}= \frac{12- 6\sqrt{2}}{4- 2}= 6- 3\sqrt{2}$ Thanks from EvanJ
November 30th, 2017, 03:55 PM   #4
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Quote:
 Originally Posted by EvanJ If 2x + x*(square root of 2) = 6, what is x?
2x = 6 - sqrt(2)x
6 - 2x = sqrt(2)x
4x^2 - 24x + 36 = 2x^2
x^2 - 12x + 18 = 0

x = 6 +- 3sqrt(2)

November 30th, 2017, 04:47 PM   #5
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Quote:
 Originally Posted by Denis 2x = 6 - sqrt(2)x 6 - 2x = sqrt(2)x 4x^2 - 24x + 36 = 2x^2 x^2 - 12x + 18 = 0 x = 6 +- 3sqrt(2)
Squaring introduces an extraneous root.

 November 30th, 2017, 04:54 PM #6 Math Team   Joined: Jan 2015 From: Alabama Posts: 3,261 Thanks: 894 I first misread the problem thinking that the "x" was inside the square root in which case squaring would be the way to go. But it is not, this is just a simple linear equation for x.

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