Determine Where 4x² + 100 is Positive: An Inequality Challenge

Q: Why doesn't this quadratic have any x-intercepts?

The discriminant $ b^2 - 4ac = 0^2 - 4(4)(100) = -1600 $ is negative, meaning no real solutions exist. The parabola never touches the x-axis!

Q: How can I tell if a quadratic is always positive?

Check two things: positive leading coefficient (opens upward) and negative discriminant (no x-intercepts). If both are true, the function is always positive.

Q: What's the minimum value of this function?

Since $ 4x^2 \geq 0 $ for all x, the minimum occurs when $ x = 0 $. The minimum value is $ 4(0)^2 + 100 = 100 $.

Q: Could this function ever equal zero or be negative?

Never! Since $ 4x^2 \geq 0 $ and we're adding 100, the smallest possible value is 100. The function is always at least 100 units above the x-axis.

Quadratic Inequalities with Always-Positive Functions

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

Understand the problem

Given the function:

$y=4x^2+100$

Determine for which values of x the following holds:

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

Step-by-step solution

To determine for which values of $x$ the function $y = 4x^2 + 100$ is greater than 0, we analyze the structure of the quadratic equation:

The function is $y = 4x^2 + 100$ , where $4x^2$ is always non-negative for any real number $x$ because squaring any real number gives a non-negative result, and multiplying by a positive constant (4) remains non-negative.

The constant term in the function is $100$ , which is positive. Therefore, the smallest value $4x^2$ can take is 0 (when $x = 0$ ), making the minimum value of the function $y = 0^2*4 + 100 = 100$ .

Since $y = 4x^2 + 100$ is always greater than $0$ , for all values of $x$ , this quadratic function never reaches $0$ or becomes negative.

In conclusion, the function $y = 4x^2 + 100$ is positive for all values of $x$ .

Therefore, the solution to the problem is All $x$ .

Final Answer

All x

Key Points to Remember

Essential concepts to master this topic

Key Insight: $4x^2$ is always non-negative for any real x
Technique: Find minimum value: when x = 0, y = 100
Check: Verify minimum is positive: 100 > 0, so all values positive ✓

Common Mistakes

Avoid these frequent errors

Trying to solve 4x² + 100 = 0 for roots
Don't set the function equal to zero and solve for x = this has no real solutions! Students waste time trying to find where it crosses the x-axis when it never does. Always recognize when a quadratic has no real roots and stays entirely above 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 have any x-intercepts?

The discriminant $b^2 - 4ac = 0^2 - 4(4)(100) = -1600$ is negative, meaning no real solutions exist. The parabola never touches the x-axis!

How can I tell if a quadratic is always positive?

Check two things: positive leading coefficient (opens upward) and negative discriminant (no x-intercepts). If both are true, the function is always positive.

What's the minimum value of this function?

Since $4x^2 \geq 0$ for all x, the minimum occurs when $x = 0$ . The minimum value is $4(0)^2 + 100 = 100$ .

Could this function ever equal zero or be negative?

Never! Since $4x^2 \geq 0$ and we're adding 100, the smallest possible value is 100. The function is always at least 100 units above the x-axis.

How is this different from other quadratic inequalities?

Most quadratics cross the x-axis and change signs. This one never crosses the x-axis, so it has the same sign everywhere - always positive in this case.

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