To solve the inequality $x^2 \geq 3$, we'll proceed with the following steps:

1. Rewrite the Inequality:

$x^2 \geq 3$ This can be rewritten as: $x^2 - 3 \geq 0$

2. Find the Roots:

First, solve the equation $x^2 - 3 = 0$ to find the critical points (roots). $x^2 = 3$ Taking the square root of both sides: $x = \pm \sqrt{3}$ So, the critical points are $x = \sqrt{3}$ and $x = -\sqrt{3}$.

3. Determine the Sign of the Expression:

To determine where the expression $x^2 - 3$ is positive or negative, consider the intervals defined by the critical points:

• $(-\infty, -\sqrt{3})$
• $(-\sqrt{3}, \sqrt{3})$
• $(\sqrt{3}, \infty)$

Test the sign of the expression in each interval:

• For $x \in (-\infty, -\sqrt{3})$: Choose $x = -2$ $(-2)^2 - 3 = 4 - 3 = 1 \geq 0$ The expression is positive.
• For $x \in (-\sqrt{3}, \sqrt{3})$: Choose $x = 0$ $0^2 - 3 = -3 < 0$ The expression is negative.
• For $x \in (\sqrt{3}, \infty)$: Choose $x = 2$ $2^2 - 3 = 4 - 3 = 1 \geq 0$ The expression is positive.

4. Write the Solution in Interval Notation:

The inequality $x^2 \geq 3$ is satisfied when $x \in (-\infty, -\sqrt{3}]$ or $x \in [\sqrt{3}, \infty)$.

Thus, the solution in interval notation is: $(-\infty, -\sqrt{3}] \cup [\sqrt{3}, \infty)$

Would you like further details or have any questions about this solution?

Here are some related questions:

1. How do you solve $x^2 > 4$ using interval notation?
2. What is the process for solving $x^2 < 2$?
3. How do you graph the solution set for $x^2 \geq 5$?
4. What are the key steps to find intervals for quadratic inequalities?
5. Can you explain why the inequality flips when dividing by a negative number?

Tip: When solving inequalities, always consider testing points within each interval to determine the sign of the expression.