Misunderstood Relationship between Roots and Zeros

Misunderstood Relationship between Roots and Zeros (MRRZ)

Much of one's mathematical experience prior to the calculus is spent in solving equations. There are the kind of equations, known as identities, where every number in the domain is a solution. The equation

$\begin{displaymath}(x - 2)(2x + 3) = 2x^2 - x - 6, \end{displaymath}$

is such an identity. It is not this, but the other type of equation, known as a conditional equality, that one learns to solve, precisely because solutions, also known as roots, of conditional equalities are not everywhere to be found. Often there are very few numbers, perhaps even none at all, which make a conditional equality true.

Another fact about conditional equalities is that comparatively few of them may be solved exactly. Leaps in technology have made it commonplace for students, with the purchase of a handheld calculator, to have at their fingertips powerful graphing capability and numerical methods for finding approximate solutions to many, perhaps even most conditional equalities. This does not mean that one should forego learning the algebraic techniques which lead to exact solutions, thinking that deftness in pushing the right pair of buttons is an appropriate substitute for the thinking processes such algebraic methods introduce. Still, there is added value in the knowledge one gets by investigating graphical methods. By these methods one comes to think of the solutions of, say,

$\begin{displaymath}3x^2 - 2x = 2x^3 + x^2 - 1 \end{displaymath}$

as the

-values of points of intersection between the graphs of the two functions

$\begin{displaymath}y = 3x^2 - 2x \qquad\mbox{and}\qquad y = 2x^3 + x^2 - 1. \end{displaymath}$

As another example, solutions of the equation

$\begin{displaymath}x^2 + 5x = -6 \end{displaymath}$

would be found at points of intersection between the graphs of

$\begin{displaymath}y = x^2 + 5x \quad\mbox{(a parabola)}\qquad\mbox{and}\qquad y = -6 \quad\mbox{(a horizontal line)}. \end{displaymath}$

It is in solving equations like this latter one that students become confused. What some students do is the following:

$\begin{eqnarray*} x^2 + 5x = -6 & \quad\Rightarrow\quad & x(x+5) = -6 \quad\mb... ...or}\;\; x = -5. \mbox{\hspace{0.6in} \epsfig{0.6in}{mrrz.eps}} \end{eqnarray*}$

In mathematical terms, the student who does these steps has found the zeros of

; that is, the values for

which make the output of

be zero. There are several ways to see that this work is wrong. One way to see it is that, in the equation, we want values of

whose output value is

, not zero. Another angle which reveals the errors is the one that notes that, while there are a lot of pairs of numbers which may be multiplied to give (-6) -- (-1) and 6, 12 and (-1/2), 55 and (-6/55) are three such pairs -- one thing which we can say for certain is that neither of the numbers in the pair is zero, which is quite counter to the idea of setting the factors equal to zero. (Of course, neither is it enough to set the factors equal to (-6), as in

$\begin{eqnarray*} x(x+5) = -6 & \qquad\Rightarrow\qquad & x = -6 \;\;\mbox{or} ... ... x = -11, \mbox{\hspace{0.6in} \epsfig{0.6in}{mrrzPrime.eps}}, \end{eqnarray*}$

since it is not enough for either one of these conditions to hold by itself; that is, if

then we would need the other factor

to be equal to 1 in order for their product to be

, and clearly these things cannot occur at the same time.)

In summary, the error occurs in finding the zeros of a function and taking to be the roots of the equation, when the two concepts do not coincide. There is a simple way to make them coincide. We simply make one side zero (using a valid algebraic step, of course).

$\begin{eqnarray*} x^2 + 5x = -6 & \quad\Rightarrow\quad & x^2 + 5x + 6 = 0 \\ ... ...x + 3 = 0 \\ & \Rightarrow & x = -2 \;\;\mbox{or}\;\; x = -3. \end{eqnarray*}$

The zeros of

are the numbers which make

equal zero, and that is exactly what we want in a solution of the equation

, so the two concepts coincide. Why do students mess this up? The most likely answer is that many are looking to do as little work as possible, and bringing the

over makes factoring a more difficult job (it is harder to factor

than to factor

); of course, the quadratic formula is an option for this case. What may help to avoid this confusion is remembering this graphical interpretation of what one is doing (still applied to the example above):

Solutions of an equation like correspond to points of intersection between the two sides, considered as functions, of the equation (i.e., the function and the function, in this case a constant one, ).
If the two functions are combined into one function on one side of the equation, there is still a second function, the zero function, that remains on the other side. Now we have in place of the old problem a new one (but entirely equivalent) of finding the solutions that correspond to points of intersection between the new left-hand side (in this case ) and the new right-hand side (here zero).
When our combining of terms has left one side of the equation zero (which, when considered as a function, has the -axis as its graph), one may solve the equation by finding the zeros of the nonzero side of the equation.

Top Algebra Errors Made by Calculus Students (full document)
Full List of Grading Codes

Thomas L. Scofield 2003-09-04