# How to re-write this equation?

#### topsquark

Math Team

First, what exactly is your question? Your f(x) has no bearing on this problem.

Second, the given solution, p = -1, q = -20, is not a possible solution because $$\displaystyle x^3 - x^2 - 20 = 0$$ also has no bearing on this problem.

The only function that can be written as $$\displaystyle f(x) = x^2$$ is, well, $$\displaystyle x^2$$ itself.

Is there more to this problem?

-Dan

idontknow

#### Cyan

Yes, my bad. The question is a part of this graph, I managed to plot it but I'm not sure how it's linked to p= -1 q = -20

#### skeeter

Math Team
in what looks to be part (a), $f(x) = \dfrac{20}{x}+x$

your initial post seems to be part (f) (i) , $f(x) = x^2$

are there parts (b), (c), (d), and (e) that maybe lead up to part (f) ?

topsquark

#### skipjack

Forum Staff
The only function that can be written as . . .
The question uses the word "equation", not "function". It turns out that $\text{f}(x)$ was previouly defined.

The question starts with $$\displaystyle \text{f}(x) = \frac{20}{x} + x,\ x \neq 0$$.
That implies the equation $$\displaystyle \text{f}(x) = x^2$$ means $$\displaystyle x^2 = \frac{20}{x} + x$$,
which can be written as $$\displaystyle x^3 - x^2 - 20 = 0$$, so $p = -1$ and $q = -20$ if $p$ and $q$ are constants.