Solve the Quadratic Inequality: Determine x for which x² - 4 > 0

Q: Why do we factor the quadratic first?

Factoring $ x^2 - 4 = (x-2)(x+2) $ makes it much easier to determine where the expression is positive or negative. With two factors, you can analyze the sign of each factor separately!

Q: Why can't the answer include x = -2 or x = 2?

Because we need $ f(x) > 0 $, which means strictly greater than zero. At x = ±2, the function equals exactly zero, not greater than zero.

Q: What if I get confused about which intervals are positive?

Always substitute a test value from each interval back into the original expression $ x^2 - 4 $. If the result is positive, that interval is part of your solution!

Quadratic Inequalities with Factorization Method

Look at the function below:

$y=x^2-4$

Then determine for which values of $x$ the following is true:

$f(x) > 0$

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Step-by-step written solution

Follow each step carefully to understand the complete solution

Understand the problem

Look at the function below:

$y=x^2-4$

Then determine for which values of $x$ the following is true:

$f(x) > 0$

Step-by-step solution

To find the solution to when $y = x^2 - 4$ is greater than zero, we start by analyzing the inequality:

$x^2 - 4 > 0$

This expression can be factored as:

$(x - 2)(x + 2) > 0$

The roots of the equation $(x - 2)(x + 2) = 0$ are $x = 2$ and $x = -2$ . These points divide the real number line into intervals. We need to determine on which intervals the expression is positive. Let's analyze the intervals:

$x < -2$ : Choose $x = -3$ , then both factors are negative: $(x - 2) < 0$ and $(x + 2) < 0$ , making their product $(x - 2)(x + 2) > 0$ .
$-2 < x < 2$ : Choose $x = 0$ , then one factor is positive and the other is negative: $(x - 2) < 0$ and $(x + 2) > 0$ , making their product negative.
$x > 2$ : Choose $x = 3$ , then both factors are positive: $(x - 2) > 0$ and $(x + 2) > 0$ , making their product positive.

Therefore, the function $x^2 - 4$ is positive for $x > 2$ or $x < -2$ .

Thus, the solution to the inequality $y > 0$ is:

$x > 2$ or $x < -2$ .

Final Answer

$x > 2$ or $x < -2$

Key Points to Remember

Essential concepts to master this topic

Factoring: Factor $x^2 - 4$ as $(x-2)(x+2)$ difference of squares
Test Intervals: Use critical points -2 and 2 to create three intervals
Sign Check: Test x = -3, 0, 3 in factored form: (+)(+), (-)(+), (+)(+) ✓

Common Mistakes

Avoid these frequent errors

Solving as equation instead of inequality
Don't just find where $x^2 - 4 = 0$ and stop = missing the intervals! This only gives boundary points x = ±2. Always test intervals between critical points to find where the expression is positive or negative.

Practice Quiz

Test your knowledge with interactive questions

The graph of the function below does not intersect the $$ x $$ -axis.

The parabola's vertex is marked A.

Find all values of $$ x $$ where
$f\left(x\right) > 0$ .

FAQ

Everything you need to know about this question

Why do we factor the quadratic first?

Factoring $x^2 - 4 = (x-2)(x+2)$ makes it much easier to determine where the expression is positive or negative. With two factors, you can analyze the sign of each factor separately!

How do I know which intervals to test?

The critical points (where the expression equals zero) divide the number line into intervals. For $x = -2$ and $x = 2$ , test any point in each interval: x < -2, -2 < x < 2, and x > 2.

Why can't the answer include x = -2 or x = 2?

Because we need $f(x) > 0$ , which means strictly greater than zero. At x = ±2, the function equals exactly zero, not greater than zero.

How do I remember the sign pattern?

When both factors have the same sign (both + or both -), the product is positive. When factors have opposite signs, the product is negative. This creates the pattern: +, -, + across the intervals.

What if I get confused about which intervals are positive?

Always substitute a test value from each interval back into the original expression $x^2 - 4$ . If the result is positive, that interval is part of your solution!

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