Solve the Quadratic Inequality: When is -x² + 9 Greater Than Zero?

Q: Why do I need to find the roots first?

The roots divide the number line into intervals where the function keeps the same sign (positive or negative). These critical points at $ x = ±3 $ tell us exactly where the parabola crosses the x-axis!

Q: How do I know which intervals to test?

The roots create three intervals: $ (-∞, -3) $, $ (-3, 3) $, and $ (3, ∞) $. Pick any easy number from each interval and substitute it into the original function.

Q: What if I get y = 0 at my test point?

That means you accidentally picked a root as your test point! Choose a different number that's clearly inside the interval, not on the boundary.

Q: Why is the parabola positive in the middle interval?

Since $ y = -x^2 + 9 $ opens downward (negative coefficient of x²), it's positive between its roots and negative outside them. This upside-down U-shape is key!

Q: Do I include the endpoints x = -3 and x = 3?

No! Since we want $ y > 0 $ (strictly greater), and $ y = 0 $ at the endpoints, we use open intervals: -3

Q: Can I solve this by factoring instead?

Yes! Factor as $ -(x-3)(x+3) > 0 $, which means \( (x-3)(x+3) . This product is negative when the factors have opposite signs, giving the same answer!

Quadratic Inequalities with Critical Points

Look at the following function:

$y=-x^2+9$

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:

$y=-x^2+9$

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

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

Step-by-step solution

We begin with the function $y = -x^2 + 9$ and want to find for which values of $x$ , $y > 0$ .

Step 1: Solve for the roots of the related equation $-x^2 + 9 = 0$ .
Rearranging gives: $x^2 = 9$ .

Step 2: Solve for $x$ .
Taking the square root of both sides gives: $x = \pm 3$ .

Step 3: Determine intervals for the inequality $-x^2 + 9 > 0$ .
Consider test points in the intervals determined by $x = -3$ and $x = 3$ .

For $x$ in the interval: $(-\infty, -3)$ , select $x = -4$ .
Calculate: $y = -(-4)^2 + 9 = -16 + 9 = -7$ . Thus, $y < 0$ .
For $x$ in the interval: $(-3, 3)$ , select $x = 0$ .
Calculate: $y = -(0)^2 + 9 = 9$ . Thus, $y > 0$ .
For $x$ in the interval: $(3, \infty)$ , select $x = 4$ .
Calculate: $y = -(4)^2 + 9 = -16 + 9 = -7$ . Thus, $y < 0$ .

Therefore, $y = -x^2 + 9$ is positive for $x$ values within the interval $-3 < x < 3$ .

Therefore, the solution is $-3 < x < 3$ .

Final Answer

$-3 < x < 3$

Key Points to Remember

Essential concepts to master this topic

Method: Find roots first, then test intervals between them
Technique: Test x = 0 in (-3,3): -0² + 9 = 9 > 0
Check: Verify boundaries: at x = ±3, y = 0 (not included) ✓

Common Mistakes

Avoid these frequent errors

Including boundary points in strict inequality
Don't write -3 ≤ x ≤ 3 when solving y > 0! At x = ±3, we get y = 0, which doesn't satisfy y > 0. Always use strict inequalities -3 < x < 3 when the original inequality is strict.

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 I need to find the roots first?

The roots divide the number line into intervals where the function keeps the same sign (positive or negative). These critical points at $x = ±3$ tell us exactly where the parabola crosses the x-axis!

How do I know which intervals to test?

The roots create three intervals: $(-∞, -3)$ , $(-3, 3)$ , and $(3, ∞)$ . Pick any easy number from each interval and substitute it into the original function.

What if I get y = 0 at my test point?

That means you accidentally picked a root as your test point! Choose a different number that's clearly inside the interval, not on the boundary.

Why is the parabola positive in the middle interval?

Since $y = -x^2 + 9$ opens downward (negative coefficient of x²), it's positive between its roots and negative outside them. This upside-down U-shape is key!

Do I include the endpoints x = -3 and x = 3?

No! Since we want $y > 0$ (strictly greater), and $y = 0$ at the endpoints, we use open intervals: -3 < x < 3.

Can I solve this by factoring instead?

Yes! Factor as $-(x-3)(x+3) > 0$ , which means $(x-3)(x+3) < 0$ . This product is negative when the factors have opposite signs, giving the same answer!

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