1. ## Solving quadratic inequality by wavy curve method

I was solving some problems on inequalities but got stuck on this one

(x-1)2 (x2-5)/(x2+x-1)>=0

3. ## Re: Solving quadratic inequality by wavy curve method

I don't believe I've ever heard of the "wavy curve method" but I suspect it is this: all rational functions are "continuous" except where the denominator is 0, which, here, means that they cannot go from a positive value to a negative value with "passing through 0". That means that the points where the numerator equals 0 (so the function is 0) or where the denominator is 0 (so the function is not continuous) are the only places where value can change from "< 0" to "> 0" or vice-versa. (The only places where the graph, a "wavy curve", can cross the x-axis.)

Where is the numerator $(x- 1)(x^2- 5)^2= 0$? Where is the denominator $x^2+ x- 1= 0$?
Mark those points on a number line and check the value of function at one point in each interval to see whether the function is '>0' or '<0' at every point in that interval.

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# wavy curve method with examples

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