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

Given the function:

$y=4x^2+100$

Determine for which values of x the following holds:

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

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 $x$.