Solve for q.

q^2=–28

asked by guest
on Nov 23, 2024 at 3:57 pm



You asked:

Solve the equation \[{q}^{2} = -28\] for the unknown \(q\).

MathBot Answer:

\[\begin{aligned}q &= 2 i \sqrt{7} \approx 5.2915026 i\\q &= - 2 i \sqrt{7} \approx - 5.2915026 i\end{aligned}\]

Take the square root of both sides of the equation $$\begin{aligned}q^{2} &= -28 \\ \sqrt{q^{2}} &= \sqrt{-28} \end{aligned}$$

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 \(i\). $$\begin{aligned}q &= \sqrt{28} \cdot \sqrt{-1}\\q &= \pm 2 \sqrt{7} i \\ q = 2 \sqrt{7} i &,\, q=-2 \sqrt{7} i\end{aligned}$$

\(i\) is the imaginary unit, defined as \(i^2 = -1\).

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