Answer
$$(-∞,\frac{5-\sqrt{33}}{4}]∪[\frac{5+\sqrt{33}}{4},+∞)$$
Work Step by Step
Let $2x^2 -5x \geq 1 $ be a function $f$.
Express the inequality in the form $f(x) \geq 0$. Begin by the rewriting the inequality so that $0$ is on the right side.
$2x^2 -5x \geq 1$
$2x^2 -5x -1\geq 1-1$
$2x^2 -5x -1\geq 0 $
Find the $x$-intercepts by solving $2x^2 -5x -1=0$
Use the quadratic formula: $x = \frac{-b±\sqrt{b^2-4ac}}{2a}$ where $a=2$, $b=-5$, $c=-1$
$x = \frac{-(-5)±\sqrt{-5^2-4⋅2⋅-1}}{2⋅2}$
$x = \frac{5+\sqrt{33}}{4}$ or $\frac{5-\sqrt{33}}{4}$
These $x$-intercepts serve as boundary points that separate the number line into intervals.
The boundary points of this equation therefore divide the number line into three intervals:
$(-∞, \frac{5-\sqrt{33}}{4})(\frac{5-\sqrt{33}}{4},\frac{5+\sqrt{33}}{4})(\frac{5+\sqrt{33}}{4}, +∞)$
Choose one test value within each interval and evaluate $f$ at that number.
$(-∞, \frac{5-\sqrt{33}}{4})$
Test Value = $-1$
$f=2x^2 -5x -1$
$f=2(-1)^2 -5(-1) -1$
$f=6$
Conclusion: $f (x) > 0$ for all $x$ in $(-∞,\frac{5-\sqrt{33}}{4})$.
$(\frac{5-\sqrt{33}}{4}, \frac{5+\sqrt{33}}{4})$
Test Value = $0$
$f=2x^2 -5x -1$
$f=2(0)^2 -5(0) -1$
$f=-1$
Conclusion: $f (x) < 0$ for all $x$ in $(\frac{5-\sqrt{33}}{4}, \frac{5+\sqrt{33}}{4})$.
$(\frac{5+\sqrt{33}}{4},+∞)$
Test Value = $3$
$f=2x^2 -5x -1$
$f=2(3)^2 -5(3) -1$
$f=2$
Conclusion: $f (x) > 0$ for all $x$ in $(\frac{5+\sqrt{33}}{4},+∞)$.
Write the solution set, selecting the interval or intervals that satisfy the given inequality. We are interested in solving $f (x) \gt 0$, where $f (x) = 2x^2 -5x -1$. Based on the solution above, $f(x)\gt0$ for all $x$ in $(\frac{5-\sqrt{33}}{4},-∞)$ or $(\frac{5+\sqrt{33}}{4},+∞)$. However, because the inequality involves $\geq$ (greater than or equal to), we must also include the solutions of $2x^2-5x-1$, namely $\frac{5-\sqrt{33}}{4}$ and $\frac{5+\sqrt{33}}{4}$, in the solution set.
Thus, the solution set of the given inequality is:
$$(-∞,\frac{5-\sqrt{33}}{4}]∪[\frac{5+\sqrt{33}}{4},+∞)$$
