Find the x for the minimum area of triangle

Question

Triangle area = f(x) = $$\frac{{(4+a^2)}^2}{-4a}$$ So, to find min area, i take the derivative and equals to 0 $$f'(x) = 16 -8a^2-3a^4$$ $$0 = 16 -8a^2-3a^4$$ How to factorize it? Thanks

Answer 1

Hint: $3a^4+8a^2-16$ is in quadratic form, and you can also factor it. The factoring would be $(a^2+4)(3a^2-4)$=0, or $(a+2i)(a-2i)(\sqrt{3}+2)(\sqrt{3}-2)$.

Answer 2

At the risk of being pedantic, even if you arrive to the correct equation, you must know that your derivative if wrong.

$$f(x)=-\frac{\left(a^2+4\right)^2}{4 a}\implies f'(x)=-a^2-4+\frac{\left(a^2+4\right)^2}{4 a^2}=-\frac{3 a^2}{4}+\frac{4}{a^2}-2$$ What you wrote, as $f'(x)$ is in fact $4a^2 f'(x)$ assuming that $a\neq 0$.

Now, to get the solution of $f'(x)=0$, you need to solve $$-\frac{3 a^2}{4}+\frac{4}{a^2}-2=0$$ Let $x=a^2$ to make the equation $$-\frac{3 x}{4}+\frac{4}{x}-2=0\implies 3 x^2+8 x-16=0$$ which is a quadratic in $x$; its solutions are $x_1=-4$ and $x_2=\frac 43$. $x_1$ must be discarded since $x=a^2>0$ and the only root is then $a^2=x_2=\frac 43$.