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October 1, 2014

October 1, 2014

Posted by **Kay** on Monday, September 16, 2013 at 1:38pm.

F(x)=x^2+(1/(500x^2-1000))

- Math -
**Steve**, Monday, September 16, 2013 at 2:13pmx^2 + 1/(500(x^2-2)) = 0

If f(x) were x^2, then there would be a double root near 0. But, the graph is shifted down just a smidgen, so there will now be two roots near zero. Also, since there are asymptotes at ±√2, the graph will cross the x-axis near ±√2, making two more roots.

Or, we can adjust things a bit to work with a straight polynomial.

x^2 (x^2-2) + 1/500 = 0

If f(x) were x^2 (x^2-2) then there would be a double root at x=0 and roots at x = ±√2.

But our f(x) is translated up by 1/500, meaning we still have two roots near ±√2, but also now two roots near x=0. So, 4 distinct roots in all.

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