Quadratic Equations

We now turn to the task of finding an iterator G(x) whose fixed points are the solutions of ax² + bx + c = 0, ideally one with slope zero at those fixed points. Here we assume that b² - 4ac > 0, thereby assuring the existence of real solutions.

In the special case

a = 1, b = 0, and c = -k

we have found the desired iterator to be

A simple calculation

shows that the square roots of k are in fact fixed points of F(x), while the A-G Mean inequality can be used to confirm that its graph is flat at these fixed points.

Against this background, we begin our efforts to solve ax² + bx + c = 0 by setting b=0 and considering the quadratic equation

ax² + c = 0.

With a little algebra, this reduces to the previous case . For by writing this equation as x² - (-c/a) = 0 and setting k = -c/a, we are (for the case b = 0) led to the iterator

Exercise. By solving G(x) = x, confirm that x is a fixed point of G(x) whenever ax² + c = 0.

Having found the desired iterator for the case b = 0, we turn to the general quadratic equation: ax² + bx + c = 0. Our approach will be one of modifying the iterator G(x) as given above. Here it is readily confirmed that we could modify the numerator of G to arrive at the iterator

that has the "right fixed points" - i.e., if ax² + bx + c = 0, then G₁(x) = x. Similarly, we could modify the denominator of G to arrive at the iterator

Well, we have arrived at two iterators with the right fixed points (and there are other possibilities as well). To choose between them we could compare their effectiveness in approximating the roots of a particular quadratic equation, say x² - x - 1 = 0. Another possibility is to look at the graphs of both iterators and determine if one of these has slope zero at its fixed points.

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