Solve the Quadratic Inequality: x²-9<0 Step by Step

Q: Why do I need to factor x² - 9 first?

Factoring $ x^2 - 9 = (x-3)(x+3) $ reveals the critical points where the expression changes sign. Without factoring, it's much harder to see where the parabola is above or below the x-axis!

Q: How do I know which intervals to test?

The critical points x = -3 and x = 3 divide the number line into three regions: $ (-\infty, -3) $, $ (-3, 3) $, and $ (3, \infty) $. Pick any number from each interval and test!

Q: Why isn't x = -3 or x = 3 included in the answer?

Because the inequality is strictly less than zero ($ (x-3)(x+3) = 0 $, not negative. If it were ≤, then we'd include these points.

Q: What if I test x = 0 and get a positive result?

If you get a positive result when testing x = 0, double-check your calculation! For this problem: $ (0-3)(0+3) = (-3)(3) = -9 $, which is negative and satisfies our inequality.

Q: Can I graph this instead of using intervals?

Absolutely! The graph of $ y = x^2 - 9 $ is a parabola opening upward. You want where it's below the x-axis, which is between the x-intercepts at x = -3 and x = 3.

Quadratic Inequalities with Factoring Method

Solve the following equation:

$x^2-9<0$

❤️ Continue Your Math Journey!

We have hundreds of course questions with personalized recommendations + Account 100% premium

Step-by-step video solution

Watch the teacher solve the problem with clear explanations

Step-by-step written solution

Follow each step carefully to understand the complete solution

Understand the problem

Solve the following equation:

$x^2-9<0$

Step-by-step solution

To solve the inequality $x^2 - 9 < 0$ , we will perform the following steps:

Step 1: Factor the inequality $x^2 - 9 = (x - 3)(x + 3)$ .
Step 2: Identify the critical values from the factored expression, which occur at $x = 3$ and $x = -3$ .
Step 3: Use these critical points to divide the number line into intervals: $(-\infty, -3)$ , $(-3, 3)$ , and $(3, \infty)$ .
Step 4: Test each interval to determine where the inequality holds:
- For $x = 0$ in the interval $(-3, 3)$ , $(0 - 3)(0 + 3) = -9$ , which satisfies $< 0$ .
- For $x = -4$ in $(-\infty, -3)$ , $(-4 - 3)(-4 + 3) = 7$ , which does not satisfy $< 0$ .
- For $x = 4$ in $(3, \infty)$ , $(4 - 3)(4 + 3) = 7$ , which does not satisfy $< 0$ .

Therefore, the inequality $x^2 - 9 < 0$ holds in the interval $-3 < x < 3$ . This means any $x$ that falls between these values will satisfy the inequality.

The correct answer is $\mathbf{-3 < x < 3}$ .

Final Answer

$-3 < x < 3$

Key Points to Remember

Essential concepts to master this topic

Factor First: Rewrite $x^2 - 9$ as $(x-3)(x+3) < 0$
Critical Points: Set each factor to zero: x = 3 and x = -3
Test Intervals: Check x = 0 in middle interval: $(0-3)(0+3) = -9 < 0$ ✓

Common Mistakes

Avoid these frequent errors

Solving x² - 9 = 0 instead of the inequality
Don't find where x² - 9 equals zero and stop there = only getting x = ±3! This gives critical points but ignores the inequality sign. Always test intervals between critical points to find where the expression is actually negative.

Practice Quiz

Test your knowledge with interactive questions

Solve the following equation:

$$ x^2+4>0 $$

FAQ

Everything you need to know about this question

Why do I need to factor x² - 9 first?

Factoring $x^2 - 9 = (x-3)(x+3)$ reveals the critical points where the expression changes sign. Without factoring, it's much harder to see where the parabola is above or below the x-axis!

How do I know which intervals to test?

The critical points x = -3 and x = 3 divide the number line into three regions: $(-\infty, -3)$ , $(-3, 3)$ , and $(3, \infty)$ . Pick any number from each interval and test!

Why isn't x = -3 or x = 3 included in the answer?

Because the inequality is strictly less than zero (<). At x = ±3, we get $(x-3)(x+3) = 0$ , not negative. If it were ≤, then we'd include these points.

What if I test x = 0 and get a positive result?

If you get a positive result when testing x = 0, double-check your calculation! For this problem: $(0-3)(0+3) = (-3)(3) = -9$ , which is negative and satisfies our inequality.

Can I graph this instead of using intervals?

Absolutely! The graph of $y = x^2 - 9$ is a parabola opening upward. You want where it's below the x-axis, which is between the x-intercepts at x = -3 and x = 3.

🌟 Unlock Your Math Potential

Get unlimited access to all 18 Equations and Systems of Quadratic Equations questions, detailed video solutions, and personalized progress tracking.

📹

Unlimited Video Solutions

Step-by-step explanations for every problem

📊

Progress Analytics

Track your mastery across all topics

🚫

Ad-Free Learning

Focus on math without distractions

No credit card required • Cancel anytime