Solving Quadratic Inequalities
... and more ...
Quadratic
A Quadratic Equation (in Standard Form) looks like:
A Quadratic Equation in Standard Form
(a, b, and c can have any value, except that a can't be 0.)
That is an equation (=) but sometimes we need to solve inequalities like these:
Symbol 
Words 
Example 


> 
greater than 
x^{2} + 3x > 2 

< 
less than 
7x^{2} < 28 

≥ 
greater than or equal to 
5 ≥ x^{2} − x 

≤ 
less than or equal to 
2y^{2} + 1 ≤ 7y 

Solving
Solving inequalities is very like solving equations ... we do most of the same things.
When solving equations we try to find points, such as the ones marked "=0" 
But when we solve inequalities
we try to find interval(s), such as the one marked "<0" 
So this is what we do:
 find the "=0" points
 in between the "=0" points, are intervals that are either
 greater than zero (>0), or
 less than zero (<0)
 then pick a test value to find out which it is (>0 or <0)
Here is an example:
Example: x^{2} − x − 6 < 0
x^{2} − x − 6 has these simple factors (because I wanted to make it easy!):
(x+2)(x−3) < 0
Firstly, let us find where it is equal to zero:
(x+2)(x−3) = 0
It is equal to zero when x = −2 or x = +3
because when x = −2, then (x+2) is zero
and
when x = +3, then (x−3) is zero
So between −2 and +3, the function will either be
 always greater than zero, or
 always less than zero
We don't know which ... yet!
Let's pick a value inbetween and test it:
At x=0: x^{2} − x − 6
= 0 − 0 − 6
= −6
So between −2 and +3, the function is less than zero.
And that is the region we want, so...
x^{2} − x − 6 < 0 in the interval (−2, 3)
Note: x^{2} − x − 6 > 0 on the interval (−∞,−2) and (3, +∞)
And here is the plot of x^{2} − x − 6:


What If It Doesn't Go Through Zero?
Here is the plot of x^{2} − x + 1 There are no "=0" points! But that makes things easier! 

Because the line does not cross through y=0, it must be either:
So all we have to do is test one value (say x=0) to see if it is above or below. 
A "Real World" Example
A stuntman will jump off a 20 m building.
A highspeed camera is ready to film him between 15 m and 10 m above the ground.
When should the camera film him?
We can use this formula for distance and time:
d = 20 − 5t^{2}
 d = distance above ground (m), and
 t = time from jump (seconds)
(Note: if you are curious about the formula, it is simplified from d = d_{0} + v_{0}t + ½a_{0}t^{2} , where d_{0}=20, v_{0}=0, and a_{0}=−9.81, the acceleration due to gravity.)
OK, let's go.
First, let us sketch the question:
The distance we want is from 10 m to 15 m: 10 < d < 15 And we know the formula for d: 10 < 20 − 5t^{2} < 15 
Now let's solve it!
First, let's subtract 20 from both sides:
−10 < −5t^{2} <−5
Now multiply both sides by −(1/5). But because we are multiplying by a negative number, the inequalities will change direction ... read Solving Inequalities to see why.
2 > t^{2} > 1
To be neat, the smaller number should be on the left, and the larger on the right. So let's swap them over (and make sure the inequalities still point correctly):
1 < t^{2} < 2
Lastly, we can safely take square roots, since all values are greater then zero:
√1 < t < √2
We can tell the film crew:
"Film from 1.0 to 1.4 seconds after jumping"
Higher Than Quadratic
The same ideas can help us solve more complicated inequalities:
Example: x^{3 }+ 4 ≥ 3x^{2} + x
First, let's put it in standard form:
x^{3} − 3x^{2 }− x + 4 ≥ 0
This is a cubic equation (the highest exponent is a cube, i.e. x^{3}), and is hard to solve, so let us graph it instead:
The zero points are approximately:
 −1.1
 1.3
 2.9
And from the graph we can see the intervals where it is greater than (or equal to) zero:
 From −1.1 to 1.3, and
 From 2.9 on
In interval notation we can write:
Approximately: [−1.1, 1.3] U [2.9, +∞)