Let's solve these equations step by step:

a. $x^3 = -i$

To solve this, we'll find the cube roots of the complex number $-i$.

1. Express $-i$ in polar form: $-i = 1 \cdot e^{-i\frac{\pi}{2}}$.

2. Find the cube roots using the formula $x_k = r^{1/3} \cdot e^{i(\theta + 2k\pi)/3}$, where $k = 0, 1, 2$.

   Thus, we get three roots: $x_0 = e^{-i\frac{\pi}{6}}, \quad x_1 = e^{i\frac{5\pi}{6}}, \quad x_2 = e^{i\frac{13\pi}{6}}.$ These can be converted back to rectangular form.

b. $2x^3 + 14x^2 + 32x + 20 = 0$

Use the Rational Root Theorem to find possible rational roots, and then perform synthetic division or factorization to simplify the polynomial.

• Checking possible roots like $x = -2$, we find $x + 2$ is a factor.
• Factorize to get the other roots.

c. $x^2 = -3$

Solve for $x$: $x = \pm \sqrt{-3} = \pm i\sqrt{3}.$

d. $x^2 - 5x - 24 = 0$

Use the quadratic formula: $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a},$ where $a = 1$, $b = -5$, and $c = -24$. This gives the solutions.

e. $3x^2 + 22 = 0$

Rewrite it as: $x^2 = -\frac{22}{3},$ giving the roots: $x = \pm i\sqrt{\frac{22}{3}}.$

Would you like further details on any of these solutions or a different approach?

Relative Questions:

1. How do we convert a