Solve for X: When Is y = -7x² Less Than Zero?

To solve the problem of finding for which values of $x$ the function $y = -7x^2$ is negative, we perform the following analysis:

Step 1: We are given a quadratic function $y = -7x^2$. The function is quadratic because it is of the form $ax^2 + bx + c$ where $a = -7$, $b = 0$, and $c = 0$.

Step 2: Observe the form, $y = -7x^2$, which implies that $y$ depends solely on $x^2$. Since $x^2 \geq 0$ for all real numbers $x$, multiplying by -7 (a negative constant) ensures that $y$ will always be less than or equal to zero.

Step 3: Specifically, $f(x) = -7x^2$ is negative ($f(x) < 0$) wherever $x^2 > 0$. The only time $x^2 = 0$ occurs is when $x = 0$. Therefore, $y = 0$ when $x = 0$.

Step 4: We conclude that the only condition under which $f(x) \geq 0$ is precisely when $x = 0$, meaning for all other real numbers $x \neq 0$, $f(x) < 0$.

Thus, the function $y = -7x^2$ is negative for all real numbers except for when $x = 0$.

Therefore, the solution is $x \ne 0$.