Solve the Quadratic Inequality: x² - 16 > 0

Q: Why do I need to test intervals instead of just finding where x² - 16 = 0?

Finding where $ x^2 - 16 = 0 $ gives you the boundary points (x = -4 and x = 4), but the inequality asks where the expression is greater than zero. Testing intervals tells you which regions actually satisfy the inequality!

Q: How do I know which test points to choose in each interval?

Pick any convenient number in each interval! For $ x , try x = -5. For \( -4 , try x = 0. For \( x > 4 $, try x = 5. The specific numbers don't matter as long as they're in the right interval.

Q: What does the notation x < -4, 4 < x mean?

This means x is less than -4 OR x is greater than 4. It's the same as writing $ (-\infty, -4) \cup (4, \infty) $ in interval notation. The expression is positive on both ends, but negative in the middle.

Q: Why isn't x = -4 or x = 4 included in the solution?

Because the inequality uses $ > 0 $, not $ \geq 0 $! At x = ±4, the expression equals zero, which doesn't satisfy greater than zero. If the problem asked for $ x^2 - 16 \geq 0 $, then we'd include those points.

Q: How can I double-check my interval testing?

After testing, verify your answer makes sense: quadratic expressions open upward when the coefficient of x² is positive. So $ x^2 - 16 $ should be positive for very large and very small x values, which matches our solution!

Step-by-step video solution

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Step-by-step written solution

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1

Understand the problem

Solve the following equation:

$x^2-16>0$

2

Step-by-step solution

To solve the inequality $x^2 - 16 > 0$ , we'll first factor the quadratic expression.

Step 1: Recognize that $x^2 - 16$ is a difference of squares, and factor it as $(x - 4)(x + 4) > 0$ .
Step 2: Identify the critical points where the expression equals zero: $x - 4 = 0$ gives $x = 4$ , and $x + 4 = 0$ gives $x = -4$ .
Step 3: Test intervals determined by these critical points. We consider the intervals:
a) $x < -4$
b) $-4 < x < 4$
c) $x > 4$
Step 4: Determine the sign of the expression in each interval:
- For $x < -4$ , choose $x = -5$ . Then $(x - 4)(x + 4) = (-5 - 4)(-5 + 4) = (-9)(-1) = 9 > 0$ . So, this interval satisfies the inequality.
- For $-4 < x < 4$ , choose $x = 0$ . Then $(x - 4)(x + 4) = (0 - 4)(0 + 4) = (-4)(4) = -16 < 0$ . This interval does not satisfy the inequality.
- For $x > 4$ , choose $x = 5$ . Then $(x - 4)(x + 4) = (5 - 4)(5 + 4) = (1)(9) = 9 > 0$ . So, this interval satisfies the inequality.
Step 5: Combine these results to state the solution as $x < -4$ or $x > 4$ .

Hence, the solution to the inequality is: $x < -4, \, 4 < x$ .

3

Final Answer

$x < -4,4 < x$

Key Points to Remember

Essential concepts to master this topic

Factoring Rule: Recognize difference of squares pattern $x^2 - 16 = (x-4)(x+4)$
Sign Testing: Check intervals using test points like x = -5, 0, 5
Verification: Confirm critical points make expression equal zero: (-4)² - 16 = 0 ✓

Common Mistakes

Avoid these frequent errors

Solving the inequality like an equation
Don't just solve x² - 16 = 0 and think x = ±4 is the answer! This only finds where the expression equals zero, not where it's positive. Always test intervals between critical points to determine where the inequality is satisfied.

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 test intervals instead of just finding where x² - 16 = 0?

+

Finding where $x^2 - 16 = 0$ gives you the boundary points (x = -4 and x = 4), but the inequality asks where the expression is greater than zero. Testing intervals tells you which regions actually satisfy the inequality!

How do I know which test points to choose in each interval?

+

Pick any convenient number in each interval! For $x < -4$ , try x = -5. For $-4 < x < 4$ , try x = 0. For $x > 4$ , try x = 5. The specific numbers don't matter as long as they're in the right interval.

What does the notation x < -4, 4 < x mean?

+

This means x is less than -4 OR x is greater than 4. It's the same as writing $(-\infty, -4) \cup (4, \infty)$ in interval notation. The expression is positive on both ends, but negative in the middle.

Why isn't x = -4 or x = 4 included in the solution?

+

Because the inequality uses $> 0$ , not $\geq 0$ ! At x = ±4, the expression equals zero, which doesn't satisfy greater than zero. If the problem asked for $x^2 - 16 \geq 0$ , then we'd include those points.

How can I double-check my interval testing?

+

After testing, verify your answer makes sense: quadratic expressions open upward when the coefficient of x² is positive. So $x^2 - 16$ should be positive for very large and very small x values, which matches our solution!

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Solve the Quadratic Inequality: x² - 16 > 0

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