Question

solve the equation $-x^2 - 3x = 5 - 3x$ and re-express the answer as a complex number with the imaginary unit. (1 point) ± \square

Explanation:

Step1: Simplify the equation

Start with the given equation \(-x^{2}-3x = 5 - 3x\). Add \(3x\) to both sides to eliminate the \(-3x\) terms.

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Step2: Solve for \(x^{2}\)

Multiply both sides by \(- 1\) to get \(x^{2}=- 5\).

Step3: Find \(x\)

Take the square root of both sides. Remember that for a negative number \(a=-b\) (\(b>0\)), \(\sqrt{a}=\pm\sqrt{b}i\). So, \(x = \pm\sqrt{-5}=\pm\sqrt{5}i\) (since \(\sqrt{-5}=\sqrt{5\times(- 1)}=\sqrt{5}\times\sqrt{-1}=\sqrt{5}i\) and we have the \(\pm\) from the square root operation).

Answer:

\(\sqrt{5}i\) (and also \(-\sqrt{5}i\), but the box is for the part after \(\pm\), so the value inside the box is \(\sqrt{5}i\) but actually the number we put in the box is \(\sqrt{5}i\) but since the problem has \(\pm\) before the box, the value in the box is \(\sqrt{5}i\) (the imaginary part magnitude with \(i\)). Wait, actually when we solve \(x^{2}=-5\), \(x = \pm\sqrt{5}i\), so the number in the box is \(\sqrt{5}i\) (but the standard form for the square root of -5 is \(\pm\sqrt{5}i\), so the value inside the box is \(\sqrt{5}i\) (the non - sign part with \(i\)). So the answer in the box is \(\sqrt{5}i\) (but more precisely, the number is \(\pm\sqrt{5}i\), so the part after \(\pm\) is \(\sqrt{5}i\)).