Solving Quadratic Inequalities

... and more ...


A Quadratic Equation (in Standard Form) looks like:

Quadratic Equation
A Quadratic Equation in Standard Form
(a, b, and c can have any value, except that a can't be 0.)


The above is an equation (=) but sometimes we need to solve inequalities like these:

greater than
x2 + 3x > 2
less than
7x2 < 28
greater than or equal to
5 ≥ x2 − x
less than or equal to
2y2 + 1 ≤ 7y


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"
Graph of Inequality
But when we solve inequalities we try to find interval(s),
such as the ones marked ">0" or "<0"

So this is what we do:

Here is an example:

Example: x2 − x − 6 < 0

x2 − 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
or when x = +3, then (x−3) is zero


So between −2 and +3, the function will either be

We don't know which ... yet!

Let's pick a value in-between and test it:

At x=0:x2 − x − 6  
 =  0 − 0 − 6  

So between −2 and +3, the function is less than zero.

And that is the region we want, so...

x2 − x − 6 < 0 in the interval (−2, 3)


Note: x2 − x − 6 > 0 on the interval (−∞,−2) and (3, +∞)


And here is the plot of x2 − x − 6:

  • The equation equals zero at −2 and 3
  • The inequality "<0" is true between −2 and 3.

Also try the Inequality Grapher.

What If It Doesn't Go Through Zero?


Here is the plot of x2 − 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:

  • always > 0, or
  • always < 0

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 high-speed 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 − 5t2

(Note: if you are curious about the formula, it is simplified from d = d0 + v0t + ½a0t2 , where d0=20, v0=0, and a0=−9.81, the acceleration due to gravity.)

OK, let's go.


First, let us sketch the question:

Jump Sketch

The distance we want is from 10 m to 15 m:

10 < d < 15

And we know the formula for d:

10 < 20 − 5t2 < 15


Now let's solve it!

First, let's subtract 20 from both sides:

−10 < −5t2 <−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 > t2 > 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 < t2 < 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: x3 + 4 ≥ 3x2 + x

First, let's put it in standard form:

x3 − 3x2 − x + 4 ≥ 0

This is a cubic equation (the highest exponent is a cube, i.e. x3), and is hard to solve, so let us graph it instead:

Graph of Inequality

The zero points are approximately:

And from the graph we can see the intervals where it is greater than (or equal to) zero:

In interval notation we can write:

Approximately: [−1.1, 1.3] U [2.9, +∞)