Question

5/26 name fredl 2. $x^{2}+8=-1$

Explanation:

Step1: Isolate \(x^2\) term

Subtract 8 from both sides of the equation \(x^2 + 8=-1\).
\(x^2=-1 - 8\)

Step2: Simplify the right - hand side

Calculate \(-1-8\), we get \(x^2=-9\)

Step3: Analyze the solution

For real numbers, the square of any real number \(a\) (i.e., \(a^2\)) is non - negative (\(a^2\geq0\)). Here, we have \(x^2=-9\), and since \(-9<0\), there is no real solution for \(x\). If we consider complex numbers, we can use the definition of the imaginary unit \(i = \sqrt{- 1}\). Then \(x=\pm\sqrt{-9}=\pm3i\)

Answer:

If we are in the real number system, there is no solution. If we are in the complex number system, the solutions are \(x = 3i\) and \(x=-3i\)