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.)

Sometimes we need to solve inequalities like these:

Symbol	Words	Example

>	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

Solving inequalities is very like solving equations ... you 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 to show you:

Example: x² - x - 6 < 0

x² - x - 6 has these simple factors (because I wanted to make it easy for you!):

(x+2)(x-3) < 0

Firstly, let us find where it is equal to zero:

(x+2)(x-3) = 0

It will be equal to zero when x = -2 or x = +3

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 in-between and test it!

At 0: x² - 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...

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.

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 you 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?

You can use this formula for distance and time:

d = 20 - 5t²

d = distance above ground (m), and
t = time from jump (seconds)

(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:

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 you 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:

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 would write:

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