Respuesta :
Answer:
Explanation:
Given:
[tex]\begin{gathered} 2x^2+32=0 \\ \text{The answer is x=}+-4i \end{gathered}[/tex]To fully understand how we get the given answer, we simplify the equation first:
[tex]\begin{gathered} 2x^2+32=0 \\ \text{Simplify and rearrange} \\ 2x^2=-32 \\ x^2=-\frac{32}{2} \\ x^2=-16 \\ \end{gathered}[/tex]Next, we apply the rule:
[tex]\begin{gathered} \text{For x}^2=f(a),\text{ the solutions are } \\ x=\sqrt[]{f(a)} \\ x=-\sqrt[]{f(a)} \end{gathered}[/tex]So,
[tex]\begin{gathered} x^2=-16 \\ x=\sqrt[]{-16},x=-\sqrt[]{-16} \end{gathered}[/tex]Then, we also apply the radical rule:
[tex]\begin{gathered} \sqrt[]{-a}=\sqrt[]{-1}\sqrt[]{a} \\ So, \\ x=\sqrt[]{-16} \\ =\sqrt[]{-1}\sqrt[]{16} \\ \text{Then, apply the imaginary number rule:} \\ \sqrt[]{-1}=i \\ \text{Hence,} \\ x=4i \end{gathered}[/tex]For
[tex]\begin{gathered} x=-\sqrt[]{-16} \\ Use\text{ the same steps} \\ x=-4i \end{gathered}[/tex]Therefore the x-values are: x=4i, x=-4i. The i on the answer means imaginary number. It is a number that, when squared, has a negative result.
