How to re-write this equation?

topsquark

Math Team
May 2013
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There are two things I don't understand about this.

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
 
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Feb 2020
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20200314_153558.jpg

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
Jul 2011
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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) ?
 
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skipjack

Forum Staff
Dec 2006
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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.