Linear Inequalities (Quadratic Inequalites)

Solving Quadratic Inequalities

 

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		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 just like a solving equations ... we do most of the same things.

When solving equations we need try to find the 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: 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
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 in-between and test it:

At x=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 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

• 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 we 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:

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, +∞)

for more example I give a video