Solve the Quadratic Inequality: When Does -4x² - 12 Fall Below Zero?

Q: Why doesn't this quadratic equation have any solutions?

The discriminant $ \Delta = b^2 - 4ac = -192 $ is negative, which means there are no real x-intercepts. The parabola never touches the x-axis!

Q: How do I know the function is always negative?

Since the parabola opens downward (a = -4 \( y = -4(0)^2 - 12 = -12 .

Q: What if the coefficient of x² was positive instead?

If a > 0, the parabola would open upward. With a negative discriminant, it would stay above the x-axis, so f(x) > 0 for all x values.

Q: Can I just plug in numbers to check?

Yes! Try any x-value: x = 1 gives $ y = -4(1)^2 - 12 = -16 $, x = -2 gives $ y = -4(-2)^2 - 12 = -28 $. All results are negative!

Q: Why is the vertex at (0, -12)?

The vertex x-coordinate is $ x = -\frac{b}{2a} = -\frac{0}{2(-4)} = 0 $. Substituting: $ y = -4(0)^2 - 12 = -12 $. So the vertex is (0, -12).

Quadratic Inequalities with Negative Discriminants

Look at the function below:

$y=-4x^2-12$

Then 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 function below:

$y=-4x^2-12$

Then determine for which values of $x$ the following is true:

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

Step-by-step solution

Let's solve this step-by-step:

The function given is $y = -4x^2 - 12$ .
This quadratic function is of the form $y = ax^2 + bx + c$ , where $a = -4$ , $b = 0$ , and $c = -12$ .
The parabola opens downwards because $a = -4$ is negative.
The vertex form of the quadratic equation is used to find the maximum point, which in this configuration is the vertex. For this standard form, $x = -\frac{b}{2a} = 0$ .
The vertex is located at $(0, -12)$ .
Calculate the discriminant to identify the x-intercepts (if any):
The discriminant $\Delta = b^2 - 4ac = 0^2 - 4(-4)(-12) = -192$ .
Since the discriminant is negative, there are no real roots for the quadratic equation. This means the function does not intersect the x-axis and is always below it.

Given that the parabola opens downwards and never meets the x-axis, the function $y = -4x^2 - 12$ is always less than zero for all real $x$ .

Therefore, the solution to the problem is that the function is negative for all values of $x$ .

Final Answer

All values of $x$

Key Points to Remember

Essential concepts to master this topic

Parabola Direction: When a < 0, the parabola opens downward
Discriminant Test: Calculate b² - 4ac = 0² - 4(-4)(-12) = -192
Verification: Test any x-value: y = -4(0)² - 12 = -12 < 0 ✓

Common Mistakes

Avoid these frequent errors

Trying to solve the quadratic equation instead of analyzing the inequality
Don't set -4x² - 12 = 0 and try to find roots when the discriminant is negative! This wastes time and leads nowhere since there are no x-intercepts. Always check the discriminant first and analyze whether the parabola stays above or below the x-axis.

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 doesn't this quadratic equation have any solutions?

The discriminant $\Delta = b^2 - 4ac = -192$ is negative, which means there are no real x-intercepts. The parabola never touches the x-axis!

How do I know the function is always negative?

Since the parabola opens downward (a = -4 < 0) and never crosses the x-axis, it stays completely below the x-axis. Test any point like x = 0: $y = -4(0)^2 - 12 = -12 < 0$ .

What if the coefficient of x² was positive instead?

If a > 0, the parabola would open upward. With a negative discriminant, it would stay above the x-axis, so f(x) > 0 for all x values.

Can I just plug in numbers to check?

Yes! Try any x-value: x = 1 gives $y = -4(1)^2 - 12 = -16$ , x = -2 gives $y = -4(-2)^2 - 12 = -28$ . All results are negative!

Why is the vertex at (0, -12)?

The vertex x-coordinate is $x = -\frac{b}{2a} = -\frac{0}{2(-4)} = 0$ . Substituting: $y = -4(0)^2 - 12 = -12$ . So the vertex is (0, -12).

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