Find Positive Values of x in the Quadratic Inequality y = 4x² + 8x

Q: Why do I need to factor the quadratic first?

Factoring $ 4x^2 + 8x = 4x(x + 2) $ makes finding the roots much easier! You can immediately see that $ x = 0 $ and $ x = -2 $ without using the quadratic formula.

Q: How do I know which intervals to test?

The roots divide the number line into regions. With roots at $ x = -2 $ and $ x = 0 $, you get three intervals: $ x , \( -2 , and \( x > 0 $. Test one value from each!

Q: Why is the answer 'or' instead of 'and'?

The function is positive in two separate regions: $ x OR \( x > 0 $. It's negative between the roots. Use 'and' only when the solution is one continuous interval.

Q: Can I use a sign chart instead of testing values?

Absolutely! A sign chart with the factored form $ 4x(x + 2) $ shows the same result. Mark where each factor is positive/negative, then combine the signs to find where the product is positive.

Quadratic Inequalities with Factoring Method

Look at the following function:

$y=4x^2+8x$

Determine for which values of $x$ the following holds:

$f(x) > 0$

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Step-by-step written solution

Follow each step carefully to understand the complete solution

Understand the problem

Look at the following function:

$y=4x^2+8x$

Determine for which values of $x$ the following holds:

$f(x) > 0$

Step-by-step solution

To solve this problem, we first need to find the roots of the quadratic function $y = 4x^2 + 8x$ .

Let's factor the quadratic equation:

$y = 4x^2 + 8x = 4x(x + 2)$

The roots of this equation are found by setting each factor to zero:

$4x = 0$ gives $x = 0$

$x + 2 = 0$ gives $x = -2$

Thus, the roots are $x = 0$ and $x = -2$ . These roots divide the number line into three intervals: $(-\infty, -2)$ , $(-2, 0)$ , and $(0, \infty)$ .

Next, we test a value from each interval to determine where the function is positive:

For the interval $(-\infty, -2)$ , test $x = -3$ :
$y = 4(-3)^2 + 8(-3) = 36 - 24 = 12$ (positive)
For the interval $(-2, 0)$ , test $x = -1$ :
$y = 4(-1)^2 + 8(-1) = 4 - 8 = -4$ (negative)
For the interval $(0, \infty)$ , test $x = 1$ :
$y = 4(1)^2 + 8(1) = 4 + 8 = 12$ (positive)

Based on these tests, $y = 4x^2 + 8x$ is positive in the intervals $x < -2$ and $x > 0$ .

Therefore, the solution to the problem is $x > 0$ or $x < -2$ .

Final Answer

$x > 0$ or $x < -2$

Key Points to Remember

Essential concepts to master this topic

Factoring Rule: Factor out common terms to find roots easily
Sign Analysis: Test values in each interval: $x = -3$ gives positive 12
Verification: Check endpoints and test points confirm $x > 0$ or $x < -2$ ✓

Common Mistakes

Avoid these frequent errors

Forgetting to test sign in each interval
Don't just find the roots $x = 0$ and $x = -2$ and guess the answer! Without testing values like $x = -1$ in the middle interval, you might incorrectly think the function is positive everywhere. Always test one value from each interval created by the roots.

Practice Quiz

Test your knowledge with interactive questions

The graph of the function below intersects the X-axis at points A and B.

The vertex of the parabola is marked at point C.

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

FAQ

Everything you need to know about this question

Why do I need to factor the quadratic first?

Factoring $4x^2 + 8x = 4x(x + 2)$ makes finding the roots much easier! You can immediately see that $x = 0$ and $x = -2$ without using the quadratic formula.

How do I know which intervals to test?

The roots divide the number line into regions. With roots at $x = -2$ and $x = 0$ , you get three intervals: $x < -2$ , $-2 < x < 0$ , and $x > 0$ . Test one value from each!

What if I get zero when testing a value?

If your test value gives zero, you've accidentally picked a root of the equation! Since we want $f(x) > 0$ (strictly greater than), roots don't count. Pick a different test value from that interval.

Why is the answer 'or' instead of 'and'?

The function is positive in two separate regions: $x < -2$ OR $x > 0$ . It's negative between the roots. Use 'and' only when the solution is one continuous interval.

Can I use a sign chart instead of testing values?

Absolutely! A sign chart with the factored form $4x(x + 2)$ shows the same result. Mark where each factor is positive/negative, then combine the signs to find where the product is positive.

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