How do you solve 6x^2+x>1?

Sep 2, 2016

$\textcolor{g r e e n}{x < - \frac{1}{2}}$ or $\textcolor{g r e e n}{x > \frac{1}{3}}$

Explanation:

$6 {x}^{2} + x > 1$ is equivalent to $6 {x}^{2} + x - 1 > 0$

Factoring $6 {x}^{2} + x - 1$
we have $\left(3 x - 1\right) \left(2 x + 1\right)$
which will be equal to zero when $x = \frac{1}{3}$ or $x = - \frac{1}{2}$

We not interested in when $6 {x}^{2} + x - 1$ is equal to zero.
Rather we are interested in the ranges set up by these "boundaries"

That is we are interested in the ranges:
$\textcolor{w h i t e}{\text{XXX}} x < - \frac{1}{2}$
$\textcolor{w h i t e}{\text{XXX}} x \in \left(- \frac{1}{2} , \frac{1}{3}\right)$
$\textcolor{w h i t e}{\text{XXX}} x > \frac{1}{3}$

We can test each of these ranges by picking any easily evaluated sample value of $x$ within each range
and then seeing if that value satisfies the required inequality.

For example:
$\textcolor{w h i t e}{\text{XXX}}$in the range $x < - \frac{1}{2}$ we could use $x = - 1$
$\textcolor{w h i t e}{\text{XXX}}$in the range $x \in \left(- \frac{1}{2} , \frac{1}{3}\right)$ we could use $x = 0$
$\textcolor{w h i t e}{\text{XXX}}$in the range $x > \frac{1}{3}$ we could use $x = + 1$

{:(,"|","in range " x < -1/2,"in range "x in (-1/2,1/3),"in range "x > 1/3), (,"|",underline("sample: " x=-1),underline("sample: "x=0),underline("sample: "x=+1)), (6x^2+x,"|",color(white)("XXX")5,color(white)("XXX")-1,color(white)("XXX")7), (6x^2+x > 1?,"|","True","False","True") :}