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November 30th, 2017, 04:16 PM   #1
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Equation With An Irrational Number

If 2x + x*(square root of 2) = 6, what is x?
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November 30th, 2017, 04:21 PM   #2
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$x = \dfrac{6}{(2 + \sqrt{2})}$?
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November 30th, 2017, 04:35 PM   #3
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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 $\displaystyle \frac{2- \sqrt{2}}{2- \sqrt{2}}$:

$\displaystyle \frac{6}{2+ \sqrt{2}}\frac{2- \sqrt{2}}{2- \sqrt{2}}= \frac{6(2- \sqrt{2})}{(2+ \sqrt{2})(2- \sqrt{2}}= \frac{12- 6\sqrt{2}}{4- 2}= 6- 3\sqrt{2}$
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November 30th, 2017, 04:55 PM   #4
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Quote:
Originally Posted by EvanJ View Post
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)
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November 30th, 2017, 05:47 PM   #5
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Quote:
Originally Posted by Denis View Post
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.
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November 30th, 2017, 05:54 PM   #6
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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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