Solve the Quadratic Inequality: When is 3x² + 21 Less Than Zero?

Given the function:

$y=3x^2+21$

Determine for which values of x is $f\left(x\right) < 0$ true

To solve the given problem, follow these steps:

Step 1: Determine the direction of the parabola by examining the coefficient of $x^2$ . Since $a = 3$ , the parabola opens upwards.
Step 2: Analyze the quadratic to find where $y = 3x^2 + 21 < 0$ .
Step 3: Find the vertex to determine the minimum point of the parabola. In this case, since there is no $x$ term (i.e., $b = 0$ ), the vertex lies on the y-axis, specifically at $x = 0$ .
Step 4: Evaluate the function at the vertex, $y(0) = 3(0)^2 + 21 = 21$ , confirming it is always positive and does not drop below zero.
Step 5: As the discriminant $\Delta = b^2 - 4ac = 0^2 - 4 \times 3 \times 21 = -252$ is negative, there are no real roots, meaning the parabola does not cross the x-axis.

Given the function is always positive and never reaches zero or becomes negative, there are no values of $x$ for which $f(x) < 0$ .

Thus, the solution to this problem is No x.

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