Re: Math Conundrum
No, if you use the quadratic equation to solve for X, you get an imaginary number, not a real number.
(x+2/x)^2=6
x + 2/x = sqrt(6)
x^2 - sqrt(6)x + 2 = 0
plug into the quadratic equation [-b +/- sqrt(b^2 - 4ac)]/2a
notice b^2 - 4ac = 6 - 8 = -2, so you end up with an imaginary number.
[edit] and I see this point has already been made [img]/forums/images/icons/smile.gif[/img] [/edit]
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