Solve the Quadratic Inequality: x² - 25 < 0

Q: Why do I need to test points in each interval?

Testing points tells you the sign of the expression in each region! Since $ (x-5)(x+5) $ can only change sign at x = -5 and x = 5, each interval has a consistent sign.

Q: What happens at the boundary points x = -5 and x = 5?

At these points, the expression equals zero. Since our inequality is strictly less than zero (

Q: How do I remember which interval to choose?

Always pick the interval where your test calculation gives a negative result! For example, testing x = 0 gives \( (0-5)(0+5) = -25 , so (-5, 5) is correct.

Q: What if the inequality was ≤ instead of <?

If it were $ x^2 - 25 ≤ 0 $, you would include the boundary points! The solution would be $ -5 ≤ x ≤ 5 $ instead of \( -5 .

Q: Can I solve this without factoring?

You could use the quadratic formula, but factoring is much faster for difference of squares like $ x^2 - 25 $. Always look for special patterns first!

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:

$x^2-25<0$

2

Step-by-step solution

To solve the inequality $x^2 - 25 < 0$ , follow these steps:

Step 1: Factor the quadratic expression $x^2 - 25$ . This is a difference of squares, which factors as $(x - 5)(x + 5)$ .
Step 2: Identify the critical points where the expression equals zero, i.e., where $(x-5)(x+5) = 0$ . The solutions are $x = 5$ and $x = -5$ .
Step 3: Determine the intervals defined by these critical points: $(-∞, -5)$ , $(-5, 5)$ , and $(5, ∞)$ .
Step 4: Test each interval to see where the inequality $(x-5)(x+5) < 0$ holds:

Choose a test point from $(-∞, -5)$ , such as $x = -6$ . Then, $(-6-5)(-6+5) = (-11)(-1) > 0$ . This interval does not satisfy the inequality.
Choose a test point from $(-5, 5)$ , such as $x = 0$ . Then, $(0-5)(0+5) = (-5)(5) = -25 < 0$ . This interval satisfies the inequality.
Choose a test point from $(5, ∞)$ , such as $x = 6$ . Then, $(6-5)(6+5) = (1)(11) > 0$ . This interval does not satisfy the inequality.

Step 5: Conclude the solution. The inequality holds true in the interval $(-5, 5)$ .

Therefore, the solution to the inequality $x^2 - 25 < 0$ is $-5 < x < 5$ .

3

Final Answer

$-5 < x < 5$

Key Points to Remember

Essential concepts to master this topic

Factoring: Recognize difference of squares: $x^2 - 25 = (x-5)(x+5)$
Critical Points: Set factors equal to zero: x = -5 and x = 5
Test Intervals: Check sign in each region: (-∞,-5), (-5,5), (5,∞) ✓

Common Mistakes

Avoid these frequent errors

Solving the equation instead of the inequality
Don't just solve $x^2 - 25 = 0$ and stop at x = ±5! This only gives you the boundary points, not the solution region. The inequality asks where the expression is negative, so you need to test intervals between these critical points. Always determine which intervals satisfy the original inequality sign.

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 points in each interval?

+

Testing points tells you the sign of the expression in each region! Since $(x-5)(x+5)$ can only change sign at x = -5 and x = 5, each interval has a consistent sign.

What happens at the boundary points x = -5 and x = 5?

+

At these points, the expression equals zero. Since our inequality is strictly less than zero (< 0), we don't include these endpoints in our solution.

How do I remember which interval to choose?

+

Always pick the interval where your test calculation gives a negative result! For example, testing x = 0 gives $(0-5)(0+5) = -25 < 0$ , so (-5, 5) is correct.

What if the inequality was ≤ instead of +

If it were $x^2 - 25 ≤ 0$ , you would include the boundary points! The solution would be $-5 ≤ x ≤ 5$ instead of $-5 < x < 5$ .

Can I solve this without factoring?

+

You could use the quadratic formula, but factoring is much faster for difference of squares like $x^2 - 25$ . Always look for special patterns first!

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Solve the Quadratic Inequality: x² - 25 < 0

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