The inequality $-x^2-6<0$ has □ .
The problem asks to find all real numbers \(x\) that satisfy the inequality \(-x^2 - 6 < 0\). We are dealing with a quadratic expression \(f(x) = -x^2 - 6\). Our goal is to determine the values of \(x\) for which this expression is negative.
To make the inequality easier to analyze, we can multiply both sides by \(-1\). Remember that multiplying or dividing an inequality by a negative number reverses the direction of the inequality sign. Multiplying \(-x^2 - 6 < 0\) by \(-1\) yields:
\[x^2 + 6 > 0\]
This modified form asks for which real numbers \(x\) the expression \(x^2 + 6\) is strictly greater than zero.
We know that for any real number \(x\), the square of \(x\) is always non-negative:
\[x^2 \ge 0\]
Adding \(6\) to both sides of this inequality gives us:
\[x^2 + 6 \ge 0 + 6 = 6\]
Since \(x^2 + 6\) is always at least \(6\), it follows that \(x^2 + 6\) is always strictly greater than \(0\) for any real number \(x\). Therefore, the condition \(x^2 + 6 > 0\) is satisfied for every real number \(x\).
Because the inequality holds for all real numbers, the solution set is the set of all real numbers, denoted as \((-\infty, \infty)\). Thus, the original inequality \(-x^2 - 6 < 0\) is true for all real numbers. The final answer is \(\boxed{(-\infty, \infty)}\).
This problem demonstrates that not all quadratic expressions behave in the same way. In real-world engineering or physics, such inequalities might arise when determining the stability of a system or the range of possible energy levels. If you have an energy function \(E(x) = -x^2 - 6\), knowing that this value is always negative helps you understand that the system's total energy is always below a certain threshold, ensuring safety or stability under all operating conditions defined by the variable \(x\).