## College Algebra (6th Edition)

Domain: {$x|x$ belongs to all Real numbers}
This exercise has two types of functions: a square root function and a rational function. The square root of a negative number has no solution within the domain of Real numbers. That is to say, for all functions $f(x) = \sqrt x$, $x$ cannot be less than $0$. On the other hand, for rational functions, the denominator cannot be equal to zero. Knowing this, for we can calculate the domain of this exercise by using the denominator and solving for $4x^{2} - 5x + 2\gt0$. : $$x_{domain}: (4x^{2} - 5x + 2\gt0)$$ By using the quadratic formula $\frac{-b \frac{+}{} \sqrt {b^{2} - 4ac}}{2a}$ we can find the zeros of the function: $$4x^{2} - 5x + 2\gt0$$ $$x = \frac{-b \frac{+}{} \sqrt {b^{2} - 4ac}}{2a} = \frac{-(-5) \frac{+}{} \sqrt {(-5)^{2} - 4(4)(2)}}{2(4)} = \frac{5 \frac{+}{} \sqrt {25 - 32}}{8} = \frac{5 \frac{+}{} \sqrt {-7}}{8}$$ We've now hit a stumbling block where we have a negative square root. This means that, for the quadratic expression $4x^{2} - 5x +2$, there are no zeros within the domain of Real numbers. Because of this, we can infer that the inequality $4x^{2} - 5x +2\gt0$ is not bound, and its solution set includes all real numbers. The domain is therefore {$x|x$ belongs to all Real numbers}.