Solve for v.
v^2=–11
You asked:
Solve the equation \[{v}^{2} = -11\] for the unknown \(v\).
MathBot Answer:
\[\begin{aligned}v &= i \sqrt{11} \approx 3.3166248 i\\v &= - i \sqrt{11} \approx - 3.3166248 i\end{aligned}\]
Take the square root of both sides of the equation $$\begin{aligned}v^{2} &= -11 \\ \sqrt{v^{2}} &= \sqrt{-11} \end{aligned}$$
Remember that both positive and negative numbers will result in a positive number when squared. This means that the square root of a negative number is undefined over the real numbers. To resolve this, the imaginary unit, \(i\) is defined as \(\sqrt{-1}\). A number containing the imaginary unit is called an imaginary number. Square roots of negative numbers can thus be expressed in terms of \(i\). $$\begin{aligned}v &= \sqrt{11} \cdot \sqrt{-1}\\v &= \pm \sqrt{11} i \\ v = \sqrt{11} i &,\, v=-\sqrt{11} i\end{aligned}$$
\(i\) is the imaginary unit, defined as \(i^2 = -1\).