Is is possible to solve this equation for x?

[fenixtx423]

y = x - x^-1

Thanks in advance for the help.
Scott

 

[Ishuda]

y = x - x^-1

Thanks in advance for the help.
Scott

Rewrite as
x² - x y -1 = 0; x\(\displaystyle \ne\)0
and solve as a quadratic equation. Note that the original function is double valued so you will get two answers for x; one positive and one negative.

 

[Ishuda]

Did you know that x^(-1) = 1/x ? So:
y = x - 1/x
x - y = 1/x

And when you use the quadratic formula on Ishuda's equation
(much to our surprise, he got that right ;-)) you will get
this discriminant: y^2 - 4 ; what does that tell you?

If it tells you nuttin' then you need classroom help.

Now Denis, it shouldn't come as a surprise since I'm right at least 50% of the time so the 'expected value of my answer being correct' is more often than not [although how much more often might be up for discussion]. BTW: I would look at that discriminate again.

EDIT: changed 'expected value of my corrected answer' to 'expected value of my answer being correct' BTW: Self corrections shouldn't be counted as an incorrect answer and correct answer as that might bring down my average.

 

[fenixtx423]

Thanks I was able to solve it with the quadratic!

 

[Ishuda]

Thanks I was able to solve it with the quadratic!

tanfastic!