Solve the Quadratic Inequality: y = -1/7x^2 - 23/7x > 0

Q: How do I know which interval makes the function positive?

Look at the coefficient of x²! Since it's $ -\frac{1}{7} $ (negative), the parabola opens downward. This means the function is positive between the roots, not outside them.

Q: Why do I need to factor instead of using the quadratic formula?

Factoring is often faster and clearer for inequalities! When you can factor easily like $ -\frac{1}{7}x(x + 17) $, you immediately see the roots and can analyze the sign changes.

Q: What if I can't factor the quadratic?

Then use the quadratic formula to find the roots first. Once you have the roots, the process is the same: determine the parabola direction and find where it's positive or negative.

Q: How do I test which intervals are positive?

Pick a test point in each interval created by the roots. Substitute it into the original function. If the result is positive, that entire interval satisfies $ f(x) > 0 $!

Q: Do I include the roots in my final answer?

Not for strict inequalities! Since we want $ f(x) > 0 $ (not ≥), the roots where $ f(x) = 0 $ are excluded. Use open intervals like (-17, 0).

Quadratic Inequalities with Factoring Method

Look at the following function:

$y=-\frac{1}{7}x^2-2\frac{3}{7}x$

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 following function:

$y=-\frac{1}{7}x^2-2\frac{3}{7}x$

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

$f(x) > 0$

Step-by-step solution

First, we need to find the roots of the quadratic equation:

The quadratic is given by:
$y = -\frac{1}{7}x^2 - \frac{17}{7}x$

Setting $y = 0$ to find the $x$ -intercepts (roots):
$-\frac{1}{7}x^2 - \frac{17}{7}x = 0$

Factor out the common factor, $-\frac{1}{7}x$ :
$-\frac{1}{7}x(x + 17) = 0$

This gives the roots:
$x = 0$ and $x + 17 = 0 \rightarrow x = -17$

These roots divide the number line into intervals. We need to determine where $y > 0$ . Because the coefficient of $x^2$ is negative, the parabola opens downward. The function will be positive between the roots.

Thus, we test the interval:
$(-17, 0)$

Since the parabola opens downward, the function $y > 0$ is true in the interval $-17 < x < 0$ .

Therefore, the solution to the problem is $-17 < x < 0$ , which corresponds to choice 3.

Final Answer

$-17 < x < 0$

Key Points to Remember

Essential concepts to master this topic

Rule: Factor out common terms to find the roots easily
Technique: Factor $-\frac{1}{7}x(x + 17) = 0$ gives roots x = 0, x = -17
Check: Test x = -10: $-\frac{1}{7}(-10)(-7) = 10 > 0$ ✓

Common Mistakes

Avoid these frequent errors

Forgetting parabola direction determines solution interval
Don't assume the function is positive outside the roots = wrong interval! With negative coefficient of x², the parabola opens downward, so it's positive between the roots. Always check the coefficient of x² to determine parabola direction.

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

How do I know which interval makes the function positive?

Look at the coefficient of x²! Since it's $-\frac{1}{7}$ (negative), the parabola opens downward. This means the function is positive between the roots, not outside them.

Why do I need to factor instead of using the quadratic formula?

Factoring is often faster and clearer for inequalities! When you can factor easily like $-\frac{1}{7}x(x + 17)$ , you immediately see the roots and can analyze the sign changes.

What if I can't factor the quadratic?

Then use the quadratic formula to find the roots first. Once you have the roots, the process is the same: determine the parabola direction and find where it's positive or negative.

How do I test which intervals are positive?

Pick a test point in each interval created by the roots. Substitute it into the original function. If the result is positive, that entire interval satisfies $f(x) > 0$ !

Do I include the roots in my final answer?

Not for strict inequalities! Since we want $f(x) > 0$ (not ≥), the roots where $f(x) = 0$ are excluded. Use open intervals like (-17, 0).

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