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Quadratics and Polynomial Roots

As noted above, the purpose of using algebra in physics is so that we can take known expressions that e.g. describe laws of nature and a particular problem and transform these ``truths'' into a ``true'' statement of the answer by isolating the symbol for that answer on one side of an equation.

For linear problems that is usually either straightforward or impossible. For ``simple'' linear problems (a single linear equation) it is always possible and usually easy. For sets of simultaneous linear equations in a small number of variables (like the ones represented in the course) one can ``always'' use a mix of composition (substitution) and elimination to find the answer desired16.

What about solving polynomials of higher degree to find values of their variables that represent answers to physics (or other) questions? In general one tries to arrange the polynomial into a standard form like the one above, and then finds the roots of the polynomial. How easy or difficult this may be depends on many things. In the case of a quadratic (second degree polynomial involving at most the square) one can - and we will, below - derive an algebraic expression for the roots of an arbitrary quadratic.

For third and higher degrees, our ability to solve for the roots is not trivially general. Sometimes we will be able to ``see'' how to go about it. Other times we won't. There exist computational methodologies that work for most relatively low degree polynomials but for very high degree general polynomials the problem of factorization (finding the roots) is hard. We will therefore work through quadratic forms in detail below and then make a couple of observations that will help us factor a few e.g. cubic or quartic polynomials should we encounter ones with one of the ``easy'' forms.

In physics, quadratic forms are quite common. Motion in one dimension with constant acceleration (for example) quite often requires the solution of a quadratic in time. For the purposes of deriving the quadratic formula, we begin with the ``standard form'' of a quadratic equation:

$\displaystyle a x^2 + b x + c = 0$ (60)

(where you should note well that $ c = a_0$ , $ b = a_1$ , $ a = a_2$ in the general polynomial formula given above).

We wish to find the (two) values of $ x$ such that this equation is true, given $ a, b, c$ . To do so we must rearrange this equation and complete the square.

$\displaystyle a x^2 + b x + c$ $\displaystyle =$ 0  
$\displaystyle a x^2 + b x$ $\displaystyle =$ $\displaystyle -c$  
$\displaystyle x^2 + \frac{b}{a} x$ $\displaystyle =$ $\displaystyle -\frac{c}{a}$  
$\displaystyle x^2 + \frac{b}{a} x + \frac{b^2}{4a^2}$ $\displaystyle =$ $\displaystyle \frac{b^2}{4a^2} -
$\displaystyle (x + \frac{b}{2a})^2$ $\displaystyle =$ $\displaystyle \frac{b^2}{4a^2} -
$\displaystyle (x + \frac{b}{2a})$ $\displaystyle =$ $\displaystyle \pm \sqrt{\frac{b^2}{4a^2} - \frac{c}{a}}$  
$\displaystyle x$ $\displaystyle =$ $\displaystyle - \frac{b}{2a} \pm \sqrt{\frac{b^2}{4a^2} - \frac{c}{a}}$  
$\displaystyle x_{\pm}$ $\displaystyle =$ $\displaystyle \frac{- b \pm \sqrt{b^2 - 4ac}}{2a}$ (61)

This last result is the well-known quadratic formula and its general solutions are complex numbers (because the argument of the square root can easily be negative if $ 4ac > b^2$ ). In some cases the complex solution is desired as it leads one to e.g. a complex exponential solution and hence a trigonometric oscillatory function as we shall see in the next section. In other cases we insist on the solution being real, because if it isn't there is no real solution to the problem posed! Experience solving problems of both types is needed so that a student can learn to recognize both situations and use complex numbers to their advantage.

Before we move on, let us note two cases where we can ``easily'' solve cubic or quartic polynomials (or higher order polynomials) for their roots algebraically. One is when we take the quadratic formula and multiply it by any power of $ x$ , so that it can be factored, e.g.

$\displaystyle a x^3 + b x^2 + cx$ $\displaystyle =$ 0  
$\displaystyle (a x^2 + b x + c)x$ $\displaystyle =$ 0 (62)

This equation clearly has the two quadratic roots given above plus one (or more, if the power of $ x$ is higher) root $ x = 0$ . In some cases one can factor a solvable term of the form $ (x + d)$ by inspection, but this is generally not easy if it is possible at all without solving for the roots some other way first.

The other "tricky" case follows from the observation that:

$\displaystyle x^2 - a^2 = (x+a)(x-a)$ (63)

so that the two roots $ x = \pm a$ are solutions. We can generalize this and solve e.g.

$\displaystyle x^4 - a^4 = (x^2 - a^2)(x^2 + a^2) = (x-a)(x+a)(x-ia)(x+ia)$ (64)

and find the four roots $ x = \pm a, \pm ia$ . One can imagine doing this for still higher powers on occasion.

In this course we will almost never have a problem that cannot be solved using ``just'' the quadratic formula, perhaps augmented by one or the other of these two tricks, although naturally a diligent and motivated student contemplating a math or physics major will prepare for the more difficult future by reviewing the various factorization tricks for ``fortunate'' integer coefficient polynomials, such as synthetic division. However, such a student should also be aware that the general problem of finding all the roots of a polynomial of arbitrary degree is difficult17 . So difficult, in fact, that it is known that no simple solution involving only arithmetical operations and square roots exists for degree 5 or greater. However it is generally fairly easy to factor arbitrary polynomials to a high degree of accuracy numerically using well-known algorithms and a computer.

Now that we understand both inverse functions and Taylor series expansions and quadratics and roots, let us return to the question asked earlier. What happens if we extend the domain of an inverse function outside of the range of the original function? In general we find that the inverse function has no real solutions. Or, we can find as noted above when factoring polynomials that like as not there are no real solutions. But that does not mean that solutions do not exist!

next up previous contents
Next: Complex Numbers and Harmonic Up: Functions Previous: The Taylor Series and   Contents
Robert G. Brown 2011-04-19