To solve the equation \( x^4 - 2 = -10 \), we start by moving \(-10\) to the other side:
\[ x^4 - 2 + 10 = 0 \]
This simplifies to:
\[ x^4 + 8 = 0 \]
Next, we isolate \( x^4 \):
\[ x^4 = -8 \]
Since \( x^4 \) cannot be negative for real numbers, we conclude that there are no real solutions. However, we can find complex solutions.
To solve \( x^4 = -8 \), we can express \(-8\) using its polar form:
\[ -8 = 8 \text{cis}(180^\circ) \quad \text{or} \quad 8 \text{cis}(\pi) \]
where \(\text{cis}(\theta) = \cos(\theta) + i\sin(\theta)\).
To find the fourth roots, we divide the angle by 4:
\[ \text{Angle} = \frac{\pi + 2k\pi}{4} \quad \text{for } k = 0, 1, 2, 3 \]
This gives us the four roots:
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For \( k = 0 \): \[ x_0 = \sqrt[4]{8}\text{cis}\left(\frac{\pi}{4}\right) = \sqrt[4]{8}\left(\cos\frac{\pi}{4} + i\sin\frac{\pi}{4}\right) = 2\sqrt{2}\left(\frac{\sqrt{2}}{2} + i\frac{\sqrt{2}}{2}\right) = 2 + 2i \]
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For \( k = 1 \): \[ x_1 = \sqrt[4]{8}\text{cis}\left(\frac{3\pi}{4}\right) = 2\sqrt{2}\left(-\frac{\sqrt{2}}{2} + i\frac{\sqrt{2}}{2}\right) = -2 + 2i \]
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For \( k = 2 \): \[ x_2 = \sqrt[4]{8}\text{cis}\left(\frac{5\pi}{4}\right) = 2\sqrt{2}\left(-\frac{\sqrt{2}}{2} - i\frac{\sqrt{2}}{2}\right) = -2 - 2i \]
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For \( k = 3 \): \[ x_3 = \sqrt[4]{8}\text{cis}\left(\frac{7\pi}{4}\right) = 2\sqrt{2}\left(\frac{\sqrt{2}}{2} - i\frac{\sqrt{2}}{2}\right) = 2 - 2i \]
Thus, the four complex solutions to the equation \( x^4 - 2 = -10 \) are:
\[ x = 2 + 2i, -2 + 2i, -2 - 2i, 2 - 2i \]