Thread: Maths Help
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Old 03-15-2008, 01:29 AM   #7
portoskins

Join Date
Oct 2005
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386
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Whoops I think the roots are supposed to be x = (1 +/- i*sqrt(3))/2

I had an extra (-) there

edit: BTW, your answers are incorrect in the sense that you did not have common roots that satisfy both equations.

edit: and yet...I found an error in my post above. Sigh. Will try to work it out.

For a general quadratic equation, ax^2 + bx^2 + c = 0:
roots, x = (-b +/- sqrt(b^2 - 4ac))/2a

So, solving for the two equations, the common root would be
(-k +/- sqrt(k^2-4))/2 = (-1 +/- sqrt(1-4k))/2 [ERROR]

It turns out
k = 1 satisfies the equation

And so the common roots x = (1 +/- i*sqrt(3))/2

...I think. Unfortunately k = 1 is *not* the solution. I mostly did it by inspection (and I made and error)...so now I'm hand-calculating this and I've reached a point where I don't feel like working out the algebra. Perhaps I'm misinterpreting the term "common roots".

I don't feel like solving the following:

(-k +/- sqrt(k^2 - 4))/2 = (1 +/- sqrt(1-4k))/2
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