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: $-4.5657x + 265.58 = 300x^{-1}$
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.
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 $$x\approx 1.15244$$ $$x\approx 57.0161$$