Solving a Simple Cubic Equation

Task

1. Find all the values of $x$ for which the equation $9x=x^3$ is true.
2. Use graphing technology to graph $f(x)=x^3-9x$. Explain where you can see the answers from part (a) in this graph, and why.
3. Someone attempts to solve $9x=x^3$ by dividing both sides by $x$, yielding $9=x^2$, and going from there. Does this approach work? Why or why not? 

IM Commentary

The purpose of this task is twofold. First, it prompts students to notice and explain a connection between the factored form of a polynomial and the location of its zeroes when graphed. Second, it highlights a complication that results from a seemingly innocent move that students might be tempted to make: "dividing both sides by x." When students solve equations in earlier grades they tend to be limited to linear equations in one variable, where "doing the same thing to both sides" is encouraged and never causes problems. This is because moves like "add one to both sides" are invertible. Once students try to solve more complicated equations, though, "doing the same thing to both sides" isn't necessarily always okay. In this case, the problem comes from dividing both sides by something that could take the value zero. Once students understand the Zero Product Property, and are able to rewrite expressions in factored form, they have another strategy for solving equations besides "performing operations on both sides." This task suggests that checking the graph of an associated function for locations of x-intercepts is a way to avoid inadvertently "losing" solutions.

A technology note: the default graphing window for many graphing calculators is from -10 to 10 both vertically and horizontally. Strictly speaking, that's fine for this task, because all of the zeroes of the polynomial will be visible. However, the maximum and minimum values will be outside the viewing window, and it's a good habit for students to consider all the important features of a graph when analyzing it. Some students default to "zooming out" to change the window, which is often too coarse a move to be effective, which it is in this case. This could be an opportunity to help students understand the benefit of strategically selecting a viewing window in a graphing calculator's settings. A suggested viewing window is x:[-10,10] and y:[-15,15].

Solution

1. The values of $x$ making the equation true are 0, -3, and 3. A student could find these by reasoning about equality and operations. Or, one could write the equivalent equation $x^3-9x=0$, rewrite the left side as $x(x^2-9)=0$, use the Zero Product Property to reason that if that is true, then $x=0$ or $x^2-9=0$. So one solution is $x=0$. We can deal with $x^2-9=0$ by either factoring one more time to get $(x+3)(x-3)=0$ (hence $x=3$ or $x=-3$) or reasoning directly that if $x^2=9$ then $x$ could equal $3$ or $-3$.

2. This graph intercepts the $x$-axis when $x$ is $-3$, $0$, and $3$, which matches the solutions to the equation given in part (a). The equation was equivalent to $0=x^3-9x$. In the $xy$-plane we graphed $y=x^3-9x$. The $x$-intercepts are the values of $x$ when $y$ or $x^3 - 9x$ is $0$. From part (a), we know that this is true when $x$ is $0$, $-3$, or $3$.

3. If you first divide both sides by $x$ and then solve $x^2=9$, you get only two solutions: $x = -3$ and $x = 3$. So something is obviously wrong with this approach, because there were three solutions to the original equation: $x = 0$,$x = -3$, and $x = 3$. When the person divided by $x$, they neglected the possibility that $x$ could be zero. So solving $x^2 = 9$ gives the same non-zero solutions as the original equation $x^3 = 9x$ but the solution $x = 0$ is lost because, in this case, we performed the illegal move of dividing both sides of the equation $x^3 = 9x$ by $0$.

   The operation ''dividing by $x$'' can be undone (namely by ''multiplying by $x$'') as long as $x$ is non-zero so we do not lose the non-zero solutions to the equation $x^3 = 9x$ when we divide both sides by $x$. Dividing both sides by $x$ will not produce spurious solutions but it does remove a valid solution. In a similar way, multiplying both sides by $x-1$, for example, will preserve all of the solutions to the equation but will introduce an additional solution, namely $x = 1$.