# Solving Quadratic Inequalities

... and more ...

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 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, +∞)

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