Answer
$$(-∞,-\frac{1}{2})∪(\frac{1}{3},+∞,)$$
Work Step by Step
Let $6x^2 + x \gt 1 $ be a function $f$.
Express the inequality in the form $f(x) \gt 0$. Begin by the rewriting the inequality so that $0$ is on the right side.
$6x^2 + x \gt 1 $
$6x^2 + x -1\gt 1-1 $
$6x^2 + x -1\gt 0 $
Find the $x$-intercepts by solving $6x^2 + x -1=0$
Factor: $6x^2 + x -1 = (3x-1)(2x+1)$
$3x-1=0$ or $2x+1=0$
$x = \frac{1}{3}$ or $x = -\frac{1}{2}$
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:
$(-∞, -\frac{1}{2})(-\frac{1}{2},\frac{1}{3})(\frac{1}{3}, +∞)$
Choose one test value within each interval and evaluate $f$ at that number.
$(-∞, -\frac{1}{2})$
Test Value = $-1$
$f=(3x-1)(2x+1)$
$f=(3(-1)-1)(2(-1)+1)$
$f=4$
Conclusion: $f (x) > 0$ for all $x$ in $(-∞, -\frac{1}{2})$.
$(-\frac{1}{2}, \frac{1}{3})$
Test Value = $0$
$f=(3x-1)(2x+1)$
$f=(3(0)-1)(2(0)+1)$
$f=-1$
Conclusion: $f (x) < 0$ for all $x$ in $(-\frac{1}{2}, \frac{1}{3})$.
$(\frac{1}{3},+∞)$
Test Value = $1$
$f=(3x-1)(2x+1)$
$f=(3(1)-1)(2(1)+1)$
$f=6$
Conclusion: $f (x) > 0$ for all $x$ in $(\frac{1}{3},+∞)$.
Write the solution set, selecting the interval or intervals that satisfy the given inequality. We are interested in solving $f (x) \gt 0$, where $f (x) = 6x^2+x-1$. Based on the solution above, $f(x)\gt0$ for all $x$ in $(-∞,-\frac{1}{2})$ or $(\frac{1}{3},+∞,)$.
Thus, the solution set of the given inequality is:
$$(-∞,-\frac{1}{2})∪(\frac{1}{3},+∞,)$$
