Food for thought

# How not to solve a cubic... Add to your resource collection Remove from your resource collection Add notes to this resource View your notes for this resource

## Things you might have noticed

• Can you find a cubic equation $ax^3+bx+c=0$ where (at least) one of the roots of the corresponding quadratic $ax^2+bx+c=0$ is also a root of the cubic?

Before diving into this question we might begin by taking a step back to think about what it is asking us to find.

• Can we represent the problem graphically?

• Do we expect there to be a unique solution?

We are essentially looking for a quadratic and cubic curve that both pass through a given point (or pair of points) on the $x$-axis. It feels as though there may be lots of pairs of curves like this, and indeed we may have already found an example of a cubic and quadratic curve that each pass through $(0,3)$ in the Warm-up problem.

But are both equations in the form $ax^3+bx+c=0$ and $ax^2+bx+c=0$ as required?

This constraint narrows down the problem. We might need to consider algebraic representations of the problem to help us make progress.

Let’s label the quadratic and cubic equations $f(x)$ and $g(x)$ so that \begin{align*} f(x)&=ax^2+bx+c\\ g(x)&=ax^3+bx+c. \end{align*}

Now if $f(k)=0$ and $g(k)=0$, so that $k$ is a root of both equations, we can write $ak^2 + bk + c = 0\quad\text{and}\quad ak^3 + bk + c =0.$ We can now equate the left hand sides to get $ak^2 + bk + c = ak^3 + bk + c$. Subtracting $bk+c$ gives $ak^2=ak^3$, and this is only true for $k=1$ or $k=0$. (It is also true if $a=0$, but then we don’t have a cubic equation, so we can ignore this possibility.)

Did you remember to consider the possibility that $k=0$?

If $k=0$, there is a common root at $x=0$, so $c=0$ and we can rewrite the equations as \begin{align*} f(x)&=ax^2+bx=x(ax+b)\\ g(x)&=ax^3+bx=x(ax^2+b). \end{align*}

How did we know that $c=0$ in this case?

You might like to use the GeoGebra applet below to convince yourself of our findings so far. You can adjust the sliders to see how the functions change for different values of $a$ and $b$ (but with our common root staying fixed at $x=0$).

Can you adjust the sliders to find a case where the functions share two roots?

Is this case unique or can you find another?

Where do you expect the second shared root to be?

We began this line of thinking by considering the case where $k=0$. How might we now consider the other possibility that $k=1$, so there is a common root at $x=1$?

By the factor theorem we can say that if $f(k)=0$ and $g(k)=0$, so that $k$ is a root of both equations, we also have $(x-k)$ as a factor of both $f(x)$ and $g(x)$. So in this case, $x-1$ is a factor of both $f(x)$ and $g(x)$.

Can you suggest a pair of functions for which $x=1$ is a root but $x=0$ is not?

Looking at the graph, it is also interesting to think about the behaviour of the graphs at the points where they meet: how do the behaviours at $x=0$ and $x=1$ compare?

#### On reflection…

When we reflected on our approach, we realised that there is an easier way to think about the problem.

Where do the cubic and quadratic graphs intersect?

#### An alternative algebraic approach

One way to approach this problem using algebra is to write the cubic and quadratic in terms of their roots. Let us assume that $\alpha$ is a common root, let the other roots of the cubic be $\beta$ and $\gamma$, and the other root of the quadratic be $\delta$. Then we can write the cubic as $a(x-\alpha)(x-\beta)(x-\gamma)=0$ and the quadratic as $a(x-\alpha)(x-\delta)=0$.

We could now expand the left hand sides to get $ax^3-a(\alpha+\beta+\gamma)x^2+a(\alpha\beta+\beta\gamma+\gamma\alpha)x-a\alpha\beta\gamma=ax^3+bx+c$ and $ax^2-a(\alpha+\delta)x+a\alpha\delta=ax^2+bx+c.$

Equating coefficients (and noting $a\ne0$ throughout, as the first equation is a cubic), we obtain: \begin{align} \alpha+\beta+\gamma &= 0 \label{eq:1}\\ \alpha\beta+\beta\gamma+\gamma\alpha &= -(\alpha+\delta) \label{eq:2}\\ \alpha\beta\gamma &= -\alpha\delta. \label{eq:3} \end{align}

Now all that is left to do is to solve these equations!