Let's solve the problems presented in the image one by one.

1. Solve the equation:

$\sqrt[3]{5x + x^2 - 8} = -2$

Solution:

• Start by cubing both sides to eliminate the cube root: $5x + x^2 - 8 = (-2)^3$ Simplifying the right side: $5x + x^2 - 8 = -8$

• Add 8 to both sides to cancel the constant: $5x + x^2 = 0$

• Factor the equation: $x(5 + x) = 0$

• This gives two solutions: $x = 0 \quad \text{or} \quad x = -5$

Solution set: $x = 0 \quad \text{or} \quad x = -5$

2. Find the domain and zeros of the function:

$y = \frac{\sqrt{|x| - 1}}{x^2 - 3}$

Solution:

• Domain:

  • The expression inside the square root must be non-negative: $|x| - 1 \geq 0 \quad \Rightarrow \quad |x| \geq 1$ This gives two intervals: $x \leq -1 \quad \text{or} \quad x \geq 1$
  • The denominator cannot be zero: $x^2 - 3 \neq 0 \quad \Rightarrow \quad x \neq \pm\sqrt{3}$
  • Therefore, the domain of the function is: $(-\infty, -\sqrt{3}) \cup (-\sqrt{3}, -1] \cup [1, \sqrt{3}) \cup (\sqrt{3}, \infty)$

• Zeros:

  • For the function to be zero, the numerator must be zero: $\sqrt{|x| - 1} = 0 \quad \Rightarrow \quad |x| - 1 = 0 \quad \Rightarrow \quad |x| = 1$ Thus, $x = \pm 1$.

Zeros: $x = -1 \quad \text{and} \quad x = 1$

3. Solve the inequality using the interval method:

$\frac{x^7(x^2 + 4x + 4)}{x^2 - 3} \leq 0$

Solution:

• Factor the quadratic in the numerator: $x^2 + 4x + 4 = (x + 2)^2$ So the inequality becomes: $\frac{x^7(x + 2)^2}{x^2 - 3} \leq 0$

• Critical points occur where the expression equals zero or is undefined:

  • $x^7 = 0 \Rightarrow x = 0$
  • $(x + 2)^2 = 0 \Rightarrow x = -2$
  • $x^2 - 3 = 0 \Rightarrow x = \pm\sqrt{3}$

• These critical points divide the real line into intervals. We need to test each interval to see where the expression is negative or zero: $(-\infty, -\sqrt{3}), (-\sqrt{3}, -2), (-2, 0), (0, \sqrt{3}), (\sqrt{3}, \infty)$

  • Analyze the sign of the expression in each interval and consider the zeros.

• Final solution: The inequality is satisfied in the following intervals: $(-\infty, -\sqrt{3}) \cup (-2, 0] \cup (0, \sqrt{3})$

Would you like more details on any of these solutions?

Here are 5 related questions for practice:

1. How would the equation change if the cube root in problem 1 were replaced with a square root?
2. What is the range of the function in problem 2?
3. How would the solution change in problem 3 if the inequality were strict $< 0$?
4. Can you solve $\sqrt[3]{5x + x^2 - 8} = 3$?
5. How would you find the maximum or minimum of the function in problem 2?

Tip: Always check the domain of a function before solving, especially if there are square roots or denominators.