Is this solution for a equation above 2nd degree correct?

Question

I have a simple equation above 2nd degree that I solved, with that I mean that my solution matches the one on the book, but I wonder if my method is correct.

$2x^5-32x=0 $

I factor out 2x

$2x(x^4-16)=0 $

Now by the zero-product property I say $2x=0$ (basically it would be $0/2$ which is $0$. Therefore $0$ is one of the solutions.

Then I have $x^4-16=0 $. What I did was:

$x^4-16=0 \rightarrow x^4=+16 \rightarrow \sqrt{x^4=+16} \rightarrow x^2=+4 \rightarrow \sqrt{x^2=+4} \rightarrow x=+-2 $

$+2$ and $-2$ are solutions of the equation.

Is this fine?

Answer 1

Here is an approach

$$2x(x^4-16)=0$$ $$\iff x(x^2+4)(x^2-4)=0$$ $$\iff x(x^2+4)(x+2)(x-2)=0$$

$$\iff x\in\{0,-2,2\}$$

since for all real $x$, $x^2+2\ne 0$.

Answer 2

Firstly $\sqrt{x^2 = +4}$ is not a valid mathematical expression. You can't square root an equation like that.

Now $x^4-16$ is a degree 4 equation so you'd expect 4 roots (counting with multiplicity). You took the square root of both sides and you got $x^2=4$. However any solution to $x^2=-4$ would also be a solution. This is because the solutions to $x^2=a$ is not only $\sqrt{a}$, but also $-\sqrt{a}$.

Therefore by taking square roots you lose some solutions. The correct set of solutions is $\{0, -2, 2, -2i, 2i\}$