Determine When the Quadratic Function 4x² + 8x Is Less Than Zero

Q: Why do we factor first instead of using the quadratic formula?

Factoring reveals the critical points where the function equals zero, which divide the number line into intervals. This makes it easy to test the sign in each interval!

Q: What's the difference between < 0 and ≤ 0?

With , you exclude points where f(x) = 0 (use open intervals). With ≤ 0, you include those points (use closed intervals). Check your original inequality carefully!

Q: Can I solve this by graphing instead?

Yes! Graph $ y = 4x^2 + 8x $ and look where the parabola is below the x-axis. You'll see it dips negative between x = -2 and x = 0.

Q: What if my quadratic doesn't factor nicely?

Use the quadratic formula to find the critical points first, then proceed with interval testing. The process is the same once you have those boundary values!

Q: Why is my parabola opening upward?

Because the coefficient of $ x^2 $ is positive (4 > 0). This means the function starts positive, dips negative between the roots, then becomes positive again!

Quadratic Inequalities with Factoring Method

Look at the following function:

$y=4x^2+8x$

Determine for which values of $x$ the following is true:

$f\left(x\right) < 0$

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Step-by-step written solution

Follow each step carefully to understand the complete solution

Understand the problem

Look at the following function:

$y=4x^2+8x$

Determine for which values of $x$ the following is true:

$f\left(x\right) < 0$

Step-by-step solution

To solve for which values of $x$ the function $y = 4x^2 + 8x$ is negative:

Step 1: Start by factoring the quadratic expression:
$y = 4x^2 + 8x = 4x(x + 2)$ .
Step 2: Set each factor to zero to find the roots:
$4x = 0$ or $x + 2 = 0$ , giving roots $x = 0$ and $x = -2$ .
Step 3: Analyze the intervals determined by these roots:
- For $x < -2$ , pick a test point like $x = -3$ ; plug into $4x(x + 2)$ :
$4(-3)((-3) + 2) = 4(-3)(-1) = 12 > 0$
- For $-2 < x < 0$ , pick a test point like $x = -1$ ; plug into $4x(x + 2)$ :
$4(-1)((-1) + 2) = 4(-1)(1) = -4 < 0$
- For $x > 0$ , pick a test point like $x = 1$ ; plug into $4x(x + 2)$ :
$4(1)(1 + 2) = 4(1)(3) = 12 > 0$ .

Hence, the quadratic is negative in the interval $-2 < x < 0$ .

The correct answer is therefore $-2 < x < 0$ .

Final Answer

$-2 < x < 0$

Key Points to Remember

Essential concepts to master this topic

Rule: Factor quadratic completely before finding critical points
Technique: Test intervals using sign analysis: $4(-1)(1) = -4 < 0$
Check: Verify boundary points aren't included in solution set ✓

Common Mistakes

Avoid these frequent errors

Including boundary points in inequality solution
Don't include x = 0 and x = -2 in the final answer when solving f(x) < 0 = wrong solution set! These points make f(x) = 0, not f(x) < 0. Always use open intervals like (-2, 0) for strict inequalities.

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 first instead of using the quadratic formula?

Factoring reveals the critical points where the function equals zero, which divide the number line into intervals. This makes it easy to test the sign in each interval!

How do I know which interval to test?

Pick any number inside each interval created by your critical points. For $-2 < x < 0$ , try x = -1. The sign of your result tells you if that entire interval is positive or negative.

What's the difference between < 0 and ≤ 0?

With < 0, you exclude points where f(x) = 0 (use open intervals). With ≤ 0, you include those points (use closed intervals). Check your original inequality carefully!

Can I solve this by graphing instead?

Yes! Graph $y = 4x^2 + 8x$ and look where the parabola is below the x-axis. You'll see it dips negative between x = -2 and x = 0.

What if my quadratic doesn't factor nicely?

Use the quadratic formula to find the critical points first, then proceed with interval testing. The process is the same once you have those boundary values!

Why is my parabola opening upward?

Because the coefficient of $x^2$ is positive (4 > 0). This means the function starts positive, dips negative between the roots, then becomes positive again!

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