Thursday, June 9, 2016

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"
Graph of Inequality
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.
 x^2-x-6
 

What If It Doesn't Go Through Zero?

x^2-x-1Here 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:

Jump SketchThe 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:
Graph of Inequality
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

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