Question

given $2x^2 = -8$, $x$ is.

Explanation:

Step1: Divide both sides by 2

To isolate \(x^2\), we divide both sides of the equation \(2x^2 = -8\) by 2. This gives us \(x^2=\frac{-8}{2}=-4\).

Step2: Analyze the square of a real number

For any real number \(x\), the square of \(x\) (i.e., \(x^2\)) is always non - negative (greater than or equal to 0). In our case, we have \(x^2=-4\), and since - 4 is negative, there is no real number \(x\) that satisfies this equation. If we consider complex numbers, we can solve it as follows:
From \(x^2=-4\), we can write \(x^2 = 4i^2\) (since \(i^2=-1\)). Then \(x=\pm\sqrt{4i^2}=\pm2i\), where \(i\) is the imaginary unit with the property \(i^2 = - 1\). But if we are restricted to real numbers, there is no solution.

Answer:

If we are in the set of real numbers, there is no real solution. If we are in the set of complex numbers, \(x = \pm2i\) (where \(i\) is the imaginary unit and \(i^2=-1\)).