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I was trying to rewrite this equation in terms of $s$:

$$ p = 4s \frac{(s - 1)}{2} + s (2r + 1) $$

After failing at that, I tried with Wolfram Alpha, and got the answer I wanted. But, how did it get there? What are the techniques I should know to work with equations like this?

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  • $\begingroup$ The equation is quadratic in $s$. There is a formula to get the 2 solutions for $s$: en.wikipedia.org/wiki/… $\endgroup$
    – Nameless
    Mar 24, 2014 at 23:46

4 Answers 4

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The techniques are the same as if there are only one or two variables, it can just get more confusing. You want to isolate $s$, so let's see:

$$\begin{align} p &= 4s\frac{s-1}{2}+s(2r+1) \\ &= 2s^2-2s+2rs+s \\ &=2s^2+2rs-s\end{align}$$

At this point, we can see that the equation is "quadratic in $s$", i.e., as far as that variable is concerned, it's a quadratic equation. Let's collect terms according to their degree in $s$:

$$2s^2+(2r-1)s-p = 0$$

Now you can apply the quadratic formula, using $2$ for $a$, $(2r-1)$ for $b$, and $-p$ for $c$.

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  • $\begingroup$ Thanks! "Let's collect terms according to their degree in s" makes things clear to me. $\endgroup$ Mar 24, 2014 at 23:37
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This is actually a quadratic in $s$, rewrite this as:

$2s^2-2s+s(2r+1)-p=0 \Rightarrow 2s^2+(2r-1)s-p=0$ now use the quadratic formula with $a=2,b=2r-1,c=-p$, to obtain:

$s=\frac{1-2r +-\sqrt{(2r-1)^2+16p}}{4}$

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  • $\begingroup$ I think you want $2s^2$, not $4s^2$. $\endgroup$ Mar 24, 2014 at 23:31
  • $\begingroup$ Yes you're right, thanks $\endgroup$
    – Ellya
    Mar 24, 2014 at 23:32
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We have $$p = 4s \frac{(s - 1)}{2} + s (2r + 1)\\ \implies p = 2s^2-2s + 2rs +s=2s^2-s(-2+2r+1)\\ \implies 2s^2+s(2-2r-1)-p=0\\ \implies\text{By the quadratic formula, }s=-(2-2r-1)\pm\sqrt{(2-2r-1)^2+8p}\\ \implies \boxed{s=2r-1\pm\sqrt{4r^2-4r+1+8p}}$$

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Note that $p$ is a quadratic function of $s$. The quadratic formula will yield s as a function of $p$ and $r$.

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