### Instruction: Find all values of $k$ so that the domain of
\[b(x) = \frac{kx^2 + 2x - 5}{-5x^2 + 2x + k}\]is the set of all real numbers.### Response: I know that the domain of a rational function is the set of all real numbers except those that make the denominator zero.
So I want to find the values of $k$ that prevent the denominator from having any real roots.
To do that, I can use the discriminant of the quadratic equation $-5x^2 + 2x + k = 0$.
The discriminant is $b^2 - 4ac$, where $a = -5$, $b = 2$, and $c = k$.
If the discriminant is negative, then the quadratic has no real roots, and the domain of $b(x)$ is all real numbers.
So I need to solve the inequality $b^2 - 4ac < 0$.
Plugging in the values of $a$, $b$, and $c$, I get $(2)^2 - 4(-5)k < 0$.
Simplifying, I get $4 + 20k < 0$.
Dividing by 4, I get $1 + 5k < 0$.
Subtracting 1, I get $5k < -1$.
Dividing by 5, I get $k < -\frac{1}{5}$.
This is the final answer.