# Find perimeter of triangle with side lengths $x,x+4$, and $2x-4$

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?

• it should be $x^3-4x^{\color{red}2}-256=0$ Commented Mar 25 at 2:51

It should be $$x^3-4x^{\color{red}2}-256=0$$.
A real solution of this is $$x=8$$.
• Then the perimeter is $2s=4x$ Commented Mar 25 at 3:09