Solve for x in terms of y

$\displaystyle y=x+\frac{1}{x}$

It was so interesting, but after 30 minutes I gave up.

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- Sep 22nd 2009, 06:07 PMchengbinSolving for x
Solve for x in terms of y

$\displaystyle y=x+\frac{1}{x}$

It was so interesting, but after 30 minutes I gave up. - Sep 22nd 2009, 06:47 PMpacmany = x + 1/x
http://www.mathhelpforum.com/math-he...bff81749-1.gif

y = x + 1/x

xy = x^2 + 1

x^2 - xy + 1 = 0, quadratic in x,

x = [-(-y) +/- ((-y)^2 - 4(1)(1))^1/2]/2

x = [y (+/-) (y^2 - 4)^(1/2)]/2

positive root:

x = y/2 +[sqrt (y^ - 4)]/2

negative root;

x = y/2 -[sqrt (y^ - 4)]/2

plot below of y = x + 1/x - Sep 23rd 2009, 07:59 AMwoodsmith
The above answer missed "2", it should be

positive root:

x = y/2 +[sqrt (y^2 - 4)]/2

negative root;

x = y/2 -[sqrt (y^ 2- 4)]/2 - Sep 24th 2009, 04:28 AMpacman
**thanks woodsmith**, you have an eye of the eagle