# 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 $$e$$,

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

We can also express $$p$$ in terms of $$b$$,

$$p + 2p + 3p + 4p = 10p = -b$$

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

$$125 e = \frac {3}{10} b^4$$