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Real-life scenario, not theory, homework, etc. My math is so rusty it's basically powder at this point. No need to beat me up; I am already beating myself up for not understanding this very simple problem.

I want to find where a power function line and a linear line intersect. I know the answer is about 1.15243535589047 but I want to know how to do this with simple pen and paper (so-to-speak).

Solving for x: $-4.5657x + 265.58 = 300x^{-1}$

xy scatter

My thought process:

The place where the two lines intersect is where the $x$ in the linear equation equals the $x$ in the power equation. Or, the linear equation minus the power equation should equal zero.

My attempt:

\begin{align} -4.5657x + 265.58 = \frac{300}{x^1} &\implies \frac{x^1}{1} (-4.5657x + 265.58) = \frac{300}{x^1} \cdot \frac{x^1}{1} \\ &\implies x^1(-4.5657x + 265.58) = 300 \\ &\implies -4.5657x^2 + 265.58x = 300 \\ &\implies -4.5657x^2 + 265.58x -300 = 0. \end{align}

Now to test:

If I plug in $1$ for $x$, I get $-38.9857$. If I plug in $1.15243535589047$, I get $0.0000425483196977439$. Since this number is very close to zero, I take that to mean $1.15$ is the $x$ value where both lines intersect (roughly). But I had to use solver or an optimizer function to figure this out.

My question:

How do I take these two equations and set them up so the output will tell me that $x = 1.15$ ? I guess I'm saying I want to find where the two equations equal the same number.

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    $\begingroup$ The last equation you have is quadratic in $x$, which means that you are in luck! There is a well-known closed-form solution to such equations, provided by the quadratic formula. $\endgroup$ – Xander Henderson Mar 2 '18 at 20:13
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solve the quadratic equation by the formula $$x_{1,2}=-\frac{b}{2a}\pm\frac{1}{2a}\sqrt{b^2-4ac}$$ with your values i get $$x\approx 1.15244$$ $$x\approx 57.0161$$

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    $\begingroup$ I can't believe I was asking about the quadratic formula without even knowing it. I guess a perfect example of why they teach us this in school but I didn't realize when to apply it. Thank you! $\endgroup$ – Jarad Mar 2 '18 at 20:14
  • $\begingroup$ ok nice that your Problem is solved now, have a nice evening! $\endgroup$ – Dr. Sonnhard Graubner Mar 2 '18 at 20:16
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$x$ is definitely non-zero, this means your steps in setting up the quadratic is correct, which is multiplying both sides by $x$. You don't really have to plug in values to test and you can instead use the quadratic formula for solving questions in the form \begin{align}ax^2+bx+c=0\end{align} The formula is given as: \begin{align}\frac{-b\pm\sqrt{b^2-4ac}}{2a}\end{align} Plugging in the values for a, b, c and you should get: $x=57.016082234325$ and $x=1.1524351890714$

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