Find the point of intersection between $y=\sqrt{x}$ and $y=\dfrac{x^2}{8}$.
I know that I have to set them equal to each other and solve for zero. The problem I have is that these terms are nice and will not factor nicely. This is as far as I got $$\sqrt{x}=\dfrac{x^2}{8}$$ $$\sqrt{x}-\dfrac{1}{8}x^2=0$$
What technique can I use to factor this? Could someone shine some light on this problem for me? Thank you!
It's easier to begin by squaring:
$$\sqrt{x} = \frac{x^2}{8} \implies x = \frac{x^4}{64}$$
Rearranging, this is equivalent to
$$x^4 - 64x = 0$$
which, in turn, we can factor as
$$x(x^3 - 64) = 0$$
Solving this:
There are hence $4$ solutions: $x = 0$, $x = \sqrt[3]{64}$, and two complex numbers (in particular, they are $\sqrt[3]{64} \omega$ and $\sqrt[3]{64} \omega^2$ where $\omega = \frac{1}{2} + \frac{\sqrt{3}}{2} i$ is a primitive third root of unity). But I'm guessing that you're only looking for the real solutions, so we'll just go with the first two.
It is necessary to verify that they are indeed solutions to the given equation, since we could have introduced extraneous solutions by squaring both sides. I'll leave it to you to check that these indeed work.
If you rewrite the equation using $y=\sqrt x$, the equation becomes $$\begin{align} y -\frac{1}{8}y^4 =& 0\\ -\frac{1}{8}y\left(y^3-8\right)=& 0\\ \end{align}$$
and you can continue to factorise this to find some more roots. If you do this, you have to check those real $y$'s you obtained can be written as $\sqrt x$ for some $x$. That is, you should only accept non-negative real $y$'s, since $\sqrt x \ge 0$.
