A calculus of the absurd

4.11.2 Some further examples
  • Example 4.11.3 The equation \(x^3 + bx^3 + cx^2 + dx + e\) has roots \(p\), \(2p\), \(3p\) and \(4p\). Show that \(125e = \frac {3}{10} b^4\).

As always, the first thing to do is to try to orient oneself. What is the question asking for? A relationship between \(e\) and \(b\) is one way to think about this. A logical next question is whether we can find a simple way to relate \(e\) and \(b\) directly to one another (I can’t). Perhaps another question which is not unreasonable to ask after this is whether it is possible to express \(e\) in terms of \(b\) by first expressing \(e\) in terms of some third quantity, and then expressing this quantity in terms of \(b\). Considering the question gives us the answer; we can express \(e\) in terms of \(p\), and \(b\) in terms of \(p\) (which also means we can express \(p\) in terms of \(b\)).

Using Vieta’s formulae we can express \(p\) in terms of \(e\),

\begin{equation} (p)(2p)(3p)(4p) = 24 p^4 = e \label {P and E} \end{equation}

We can also express \(p\) in terms of \(b\),

\begin{equation} p + 2p + 3p + 4p = 10p = -b \end{equation}

Therefore, \(p = -\frac {b}{10}\). We can then just substitute this into Equation 4.145,

\begin{align} e &= 24 p^4 \\ &= 24 \left (-\frac {b}{10}\right )^4 \\ &= \frac {24}{10,000} b^4 \\ &= \frac {3}{1250} b^4 \end{align}

Therefore, after multiplying by \(125\), the end result is that

\begin{equation} 125 e = \frac {3}{10} b^4 \end{equation}