In Precalculus, it's essential that you can easily and efficiently solve sentences like ‘$\,\frac{3x}{2}1 \ge \frac 15  7x\,$’ and ‘[beautiful math coming... please be patient]$\,x^2 \ge 3\,$’ .
The first sentence is an example of a linear inequality in one variable; the prior lesson covers this type of sentence.
For linear inequalities, the variable appears in the simplest possible way—all you have are numbers, times $\,x\,$ to the first power
(i.e., terms of the form $\,kx\,$, where $\,k\,$ is a
real number).
The second sentence is an example of a nonlinear inequality in one variable, and is covered in this lesson and the next two.
In nonlinear sentences, the variable appears in a more complicated way—perhaps
you have an $\,x^2\,$ (or higher power), or $\,x\,$, or $\,\sin x\,$.
For solving nonlinear inequalities, more advanced tools are needed.
To get started, review the essential concepts in these two lessons, being sure to clickclickclick to practice the concepts from each:
There are two basic methods for solving nonlinear inequalities in one variable.
Both are called ‘test point methods’, because they involve identifying important intervals, and then ‘testing’ a number from each of these intervals.
Below, the sentence ‘$\,x^2 \ge 3\,$’ is solved using both methods, so you can get a sense of which appeals to you more.
The solutions below are written extremely compactly—this is pretty much the bare minimum that a teacher would want to see.
For all the underlying concepts and details, study the next two lessons:
YOU WRITE THIS DOWN  COMMENTS 
$x^2 \ge 3$  original sentence 
$x^2  3\ge 0$  rewrite the inequality with zero on the righthand side; since the inequality symbol is $\,\ge\,$, we need to determine where the graph of $\,x^2  3\,$ lies on ($\,=\,$) or above ($\,\gt\,$) the $x$axis 
$f(x) := x^2  3$  if desired, name the function on the lefthand side $\,f\,$, so it can be easily referred to in later steps; recall that ‘$:=$’ means ‘equals, by definition’ 
$x^2  3 = 0$ $x^2 = 3$ $x = \pm\sqrt 3$ (no breaks in graph of $\,f\,$) 
identify the candidates for sign changes for $\,f\,$:

SIGN OF $\,f(x)\,$ 

solution set: $(\infty,\sqrt 3] \cup [\sqrt 3,\infty)$ sentence form of solution: $x\le \sqrt 3\ \ \text{ or }\ \ x\ge \sqrt 3$ 
read off the solution set, using correct interval notation; or, give the sentence form of the solution 
When a graph is easy to obtain (as in this example), then you may not need to ‘officially’
use the test point method.
Just graph $\,f(x) = x^2  3\,$ (see below);
the solutions of ‘$\,x^2  3 \ge 0\,$’ are the values of $\,x\,$ for which
the graph of $\,f\,$ lies on or above the $x$axis.
$x^2 \ge 3$  original sentence 
$f(x) := x^2$, $g(x) := 3$  if desired, define functions $\,f\,$ (the lefthand side) and $\,g\,$ (the righthand side), so they can be easily referred to in later steps 
$x^2 = 3$ $x = \pm\sqrt 3$ (no breaks in either graph) 
find where $\,f(x)\,$ and $\,g(x)\,$ are equal (intersection points); find any breaks in the graphs of $\,f\,$ and $\,g\,$ 
TRUTH OF ‘$\,x^2 \ge 3\,$’ 

solution set: $(\infty,\sqrt 3] \cup [\sqrt 3,\infty)$ sentence form of solution: $x\le \sqrt 3\ \ \text{ or }\ \ x \ge\sqrt 3$ 
read off the solution set, using correct interval notation; or, give the sentence form of the solution 
When graphs are easy to obtain (as in this example), then you may not need to ‘officially’
use the test point method.
Just graph $\,f(x) = x^2\,$ (in red below) and
$\,g(x) = 3\,$ (in green below):
then, the solutions of ‘$\,x^2 \ge 3\,$’ are the values of $\,x\,$ for which
the graph of $\,f\,$ lies on or above the graph of $\,g\,$.
Just for fun, jump up to wolframalpha.com
and key in: x^2 >= 3
Voila!
On this exercise, you will not key in your answer. However, you can check to see if your answer is correct. 
PROBLEM TYPES:
