To solve the equation \( x^4 - 2 = -10 \), we first isolate \( x^4 \):
\[ x^4 - 2 + 2 = -10 + 2 \]
This simplifies to:
\[ x^4 = -8 \]
Next, we observe that \( x^4 = -8 \) implies that we are looking for real or complex solutions. Since the left side \( x^4 \) is always non-negative for real \( x \), there are no real solutions. However, we can find complex solutions.
Taking the fourth root, we have:
\[ x = \sqrt[4]{-8} \]
We can express \(-8\) in polar form. The modulus of \(-8\) is 8, and the argument (angle) is \( \pi \) (since \( -8 \) is on the negative real axis):
\[ -8 = 8 \text{ (cis } \pi) = 8 (\cos \pi + i \sin \pi) \]
Thus, the fourth roots of \(-8\) can be calculated using:
\[ \sqrt[4]{r} \text{ (cis } \frac{\theta + 2k\pi}{n} \text{) for } k = 0, 1, 2, 3 \]
Where \( r = 8 \) and \( \theta = \pi \), and \( n = 4 \).
Calculating the modulus:
\[ \sqrt[4]{8} = 8^{1/4} = 2^{3/4} = 2 \sqrt{2} \]
For the angle, we have:
\[ \frac{\pi + 2k\pi}{4} = \frac{\pi(1 + 2k)}{4} \]
Calculating for \( k = 0, 1, 2, 3 \):
-
For \( k = 0 \): \[ x_0 = 2\sqrt{2} \left( \cos\left(\frac{\pi}{4}\right) + i \sin\left(\frac{\pi}{4}\right) \right) = 2\sqrt{2} \left( \frac{\sqrt{2}}{2} + i \frac{\sqrt{2}}{2} \right) = 2 + 2i \]
-
For \( k = 1 \): \[ x_1 = 2\sqrt{2} \left( \cos\left(\frac{3\pi}{4}\right) + i \sin\left(\frac{3\pi}{4}\right) \right) = 2\sqrt{2} \left( -\frac{\sqrt{2}}{2} + i \frac{\sqrt{2}}{2} \right) = -2 + 2i \]
-
For \( k = 2 \): \[ x_2 = 2\sqrt{2} \left( \cos\left(\frac{5\pi}{4}\right) + i \sin\left(\frac{5\pi}{4}\right) \right) = 2\sqrt{2} \left( -\frac{\sqrt{2}}{2} - i \frac{\sqrt{2}}{2} \right) = -2 - 2i \]
-
For \( k = 3 \): \[ x_3 = 2\sqrt{2} \left( \cos\left(\frac{7\pi}{4}\right) + i \sin\left(\frac{7\pi}{4}\right) \right) = 2\sqrt{2} \left( \frac{\sqrt{2}}{2} - i \frac{\sqrt{2}}{2} \right) = 2 - 2i \]
Thus, the four complex solutions are:
\[ \begin{align*} x_0 &= 2 + 2i, \ x_1 &= -2 + 2i, \ x_2 &= -2 - 2i, \ x_3 &= 2 - 2i. \end{align*} \]
1 answer