Solving the Quadratic Inequality: Where Does 4x² + 100 Become Negative?

Quadratic Functions with Positivity Analysis

Given the function:

$y=4x^2+100$

Determine for which values of x the following holds:

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

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Step-by-step written solution

Follow each step carefully to understand the complete solution

1

Understand the problem

Given the function:

$y=4x^2+100$

Determine for which values of x the following holds:

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

2

Step-by-step solution

To solve for the values of $x$ where $y = 4x^2 + 100 < 0$ , follow these steps:

Step 1: Analyze the given function $y = 4x^2 + 100$ .
Step 2: Recognize that the quadratic function is in the standard form $y = ax^2 + bx + c$ with $a = 4$ , $b = 0$ , and $c = 100$ .
Step 3: Note that since the coefficient of $x^2$ (i.e., $a = 4$ ) is positive, the parabola opens upwards.
Step 4: The vertex of the parabola, which occurs at the minimum point $x = -\frac{b}{2a} = 0$ , gives the minimum function value $y = 4(0)^2 + 100 = 100$ .

Since the minimum value of the function is $y = 100$ , which is greater than zero, the function never reaches below zero. Hence, there are no $x$ values for which the function $f(x) < 0$ .

Therefore, the solution is No x.

3

Final Answer

No x

Key Points to Remember

Essential concepts to master this topic

Rule: Upward parabolas have minimum values, never negative below minimum
Technique: Find vertex using $x = -\frac{b}{2a}$ to get minimum value
Check: Substitute vertex x-value: $y = 4(0)^2 + 100 = 100 > 0$ ✓

Common Mistakes

Avoid these frequent errors

Setting the quadratic equal to zero instead of analyzing its range
Don't solve $4x^2 + 100 = 0$ to find where it's negative = wrong approach! This only finds zeros, not negative regions. Always analyze the parabola's shape and minimum value to determine where it can be negative.

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 can't this quadratic ever be negative?

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Since a = 4 > 0, the parabola opens upward like a U-shape. The minimum value occurs at the vertex, which is $y = 100$ . Since the lowest point is positive, the function never dips below zero!

What if the constant term was smaller, like 4x² + 1?

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Same result! Any quadratic $ax^2 + c$ where a > 0 and c > 0 has a minimum value of c, which is positive. It can never be negative.

How do I know when a quadratic CAN be negative?

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A quadratic $ax^2 + bx + c$ with a > 0 can be negative only if its minimum value is negative. Calculate the y-coordinate of the vertex - if it's negative, then the quadratic has negative regions!

What's the difference between f(x) < 0 and f(x) = 0?

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f(x) = 0 finds where the graph crosses the x-axis (roots). f(x) < 0 finds where the graph is below the x-axis. For this problem, the graph never touches or crosses the x-axis!

Could I graph this to double-check my answer?

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Absolutely! Graphing $y = 4x^2 + 100$ shows a U-shaped curve with its lowest point at (0, 100). Since the entire graph sits above the x-axis, you can visually confirm there are no negative y-values.

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