The side lengths of a triangle are $x,x+4,$ and $2x-4$. The area of the triangle is $32\sqrt{2}$. What is the perimeter of the triangle?
My attempt:
According to Heron's formula
$$A=\sqrt{s\left(s-a\right)\left(s-b\right)\left(s-c\right)}$$
where
$$s=\frac{a+b+c}{2}.$$
In this scenario, $$s=\frac{x+x+4+2x-4}{2}=\frac{4x}{2}=2x$$
We are left with the following situation:
$$32\sqrt{2}=\sqrt{2x \left(2x-x\right) \left(2x-\left(x+4\right)\right) \left(2x-\left(2x-4\right)\right) }$$
$$32\sqrt{2}=\sqrt{2x\left(2x-x\right)\left(2x-x-4\right)\left(2x-2x+4\right)}$$
$$32\sqrt{2}=\sqrt{2x\left(x\right)\left(x-4\right)\left(4\right)}$$
$$32\sqrt{2}=\sqrt{8x^3-32x^2}$$
This can be rewritten as
$$2^5\sqrt{2}=\sqrt{\left(2^{3}\right)\left(x^{3}\right)-\left(2^{5}\right)\left(x^{2}\right)}$$
Squaring both sides
$$2^{11}=\left(2^{3}\right)\left(x^{3}\right)-\left(2^{5}\right)\left(x^{2}\right)$$
Dividing both sides by $2^3$
$$2^{8}=x^{3}-2^{2}x^{2}$$
$$x^{3}-4x-256=0$$
How can I continue?
Actually, according to WolframAlpha, the real root of $x^{3}-4x-256=0$ is $$\frac{1}{3}\left(3456-24\sqrt{20733}\right)^{\frac{1}{3}}+\frac{2\left(144+\sqrt{20733}\right)^{\frac{1}{3}}}{3^{\frac{2}{3}}}$$
I don't think that's the right answer. What was my mistake in the approach? Or have I gone down the wrong path completely?
Your approach was okay. You just made an algebra mistake.
It should be $x^3-4x^{\color{red}2}-256=0$.
A real solution of this is $x=8$.