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

Given the function:

$y=4x^2+100$

Determine for which values of x the following holds:

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

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.