Finding Values Where x² Exceeds 16: Inequality Solution

Q: Why do we factor instead of just solving x² = 16?

Great question! Solving $ x^2 = 16 $ only gives you the boundary points where the expression equals zero. But we need to know where $ x^2 - 16 > 0 $, which requires testing the sign in different intervals.

Q: How do I remember which intervals are positive?

Use the sign chart method! Draw a number line with your critical points (-4 and 4), then test one point in each interval. The signs follow a pattern: positive, negative, positive for this type of factored expression.

Q: What's the difference between > and ≥ in the final answer?

The symbol matters! Since we have $ x^2 - 16 > 0 $ (strict inequality), we cannot include the points where the expression equals zero. So we write $ x or \( x > 4 $, not ≤ or ≥.

Q: Can I solve this by taking square roots?

Not directly! If you try $ x^2 > 16 $ then $ |x| > 4 $, you still need to consider both positive and negative values. The factoring method is clearer and shows exactly why we get two separate intervals.

Q: What if I get confused about which intervals to include?

Remember: we want the intervals where the product is positive. After factoring to $ (x-4)(x+4) > 0 $, look for where both factors have the same sign (both positive OR both negative).

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

The objective is to find the values of $x$ such that the inequality $x^2 - 16 > 0$ is satisfied.

Step 1: Factor the inequality expression.

The expression $x^2 - 16$ can be factored using the difference of squares formula:

$x^2 - 16 = (x - 4)(x + 4)$ .

Step 2: Determine the critical points.

Set the factors equal to zero to find the critical points:

$x - 4 = 0$ gives $x = 4$ .
$x + 4 = 0$ gives $x = -4$ .

Step 3: Analyze the sign changes on the number line.

We test the intervals defined by the critical points $-4$ and $4$ on a number line: $(-∞, -4)$ , $(-4, 4)$ , $(4, ∞)$ .

Choose a test point from each interval and substitute into the factored expression to check the sign.

For $x = -5$ (interval $(-∞, -4)$ ): $(x - 4)(x + 4) = (-5 - 4)(-5 + 4) = (-9)(-1) > 0$ .
For $x = 0$ (interval $(-4, 4)$ ): $(x - 4)(x + 4) = (0 - 4)(0 + 4) = (-4)(4) < 0$ .
For $x = 5$ (interval $(4, ∞)$ ): $(x - 4)(x + 4) = (5 - 4)(5 + 4) = (1)(9) > 0$ .

Step 4: Extract the solution.

The inequality $x^2 - 16 > 0$ holds true in the intervals where the product is positive, which are $(-∞, -4) \cup (4, ∞)$ .

Therefore, the solution to the inequality is $x < -4$ or $x > 4$ .

The correct choice is $x < -4, 4 < x$ .

3

Final Answer

$x < -4,4 < x$

Key Points to Remember

Essential concepts to master this topic

Factoring Rule: Use difference of squares $a^2 - b^2 = (a-b)(a+b)$
Sign Analysis: Test intervals: $(-5-4)(-5+4) = (+)(+) > 0$
Verification: Check boundary points aren't included since inequality is strict > ✓

Common Mistakes

Avoid these frequent errors

Including the critical points in the solution
Don't write $x ≤ -4$ or $x ≥ 4$ for strict inequality = wrong answer! The original inequality uses >, not ≥, so the expression equals zero (not greater than zero) at x = ±4. Always use open intervals for strict inequalities.

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 we factor instead of just solving x² = 16?

+

Great question! Solving $x^2 = 16$ only gives you the boundary points where the expression equals zero. But we need to know where $x^2 - 16 > 0$ , which requires testing the sign in different intervals.

How do I remember which intervals are positive?

+

Use the sign chart method! Draw a number line with your critical points (-4 and 4), then test one point in each interval. The signs follow a pattern: positive, negative, positive for this type of factored expression.

What's the difference between > and ≥ in the final answer?

+

The symbol matters! Since we have $x^2 - 16 > 0$ (strict inequality), we cannot include the points where the expression equals zero. So we write $x < -4$ or $x > 4$ , not ≤ or ≥.

Can I solve this by taking square roots?

+

Not directly! If you try $x^2 > 16$ then $|x| > 4$ , you still need to consider both positive and negative values. The factoring method is clearer and shows exactly why we get two separate intervals.

What if I get confused about which intervals to include?

+

Remember: we want the intervals where the product is positive. After factoring to $(x-4)(x+4) > 0$ , look for where both factors have the same sign (both positive OR both negative).

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Finding Values Where x² Exceeds 16: Inequality Solution

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