Solve (5x-1)² < 0: Analyzing Negative Values of a Square Function

Question

Look at the function below:

$y=\left(5x-1\right)^2$

Then determine for which values of $x$ the following is true:

$f(x) < 0$

To find where $f(x) = (5x - 1)^2 < 0$ , we start by recognizing a fundamental property of squares:

The square of any real number is always non-negative. Therefore, $(5x - 1)^2 \geq 0$ for all real $x$ .

This implies that $(5x - 1)^2$ can never be less than zero for any real value of $x$ .

The inequality $(5x - 1)^2 < 0$ has no solution in the real numbers.

Therefore, there are no values of $x$ for which $f(x) < 0$ is true.

So the logical conclusion is: True for no values of $x$ .

True for no values of $x$