Course:MATH110/Archive/2010-2011/003/Groups/Group 10/Basic skills project

Here is what we have so far from our Group 10 Page. We will be adding more. Can you help with formatting like you have done for your section?

Solving Quadratic Inequalities

To solve a quadratic inequality, follow these steps:

1. Solve the inequality as though it were an equation. The real solutions to the equation become boundary points for the solution to the inequality.

2. Make the boundary points solid circles if the original inequality includes equality; otherwise, make the boundary points open circles.

3. Select points from each of the regions created by the boundary points. Replace these “test points” in the original inequality.

4. If a test point satisfies the original inequality, then the region that contains that test point is part of the solutions.

5. Represent the solution in graphic form and in solution test form.

Example 1: Solve (x-3)(x+2)>0

By the zero product property, x-3=0 or x+2=0, x=3 and x=-2

Make the boundary points.

Here, the boundary points are open circles because the original inequality does not include equality.

Select points from different regions created.

Three regions are created:

X=-3

X=0

X=4

See if the test points satisfy the original inequality

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

(-3-3)(-3+2)>0

6>0 therefore, it works

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

(0-3)(0+2)>0

-6>0 no, it does not work

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

(4-3)(-3+2)>0

6>0 therefore, it works

One of the first steps to solving inequalities is the symbol and meanings of the inequalities.

> means greater than

< means less than

≥ means greater than or equal to

≤ means less than or equal to

Because you want to get x alone, you can rather:

• Add or subtract a number from both sides

• Multiply or divide both sides by a positive number

• Simplify a side

However, doing the following things will change the direction of the inequality:

• Multiplying or dividing both sides by a negative number

• Swapping left and right hand sides

Have a look at: http://www.mathsisfun.com/algebra/inequality-solving.html for more information.

If you’re having trouble in the book there is a good section starting on page 1061 which is a review of algebra and sets of real numbers. It gives number lines and shows inequalities to match.

If you want to check your answer on how to solve an inequality try: http://webmath.com/solverineq.html

What is an inequality?

It is exactly what it sounds, it's definition is that two numbers are not equal to each other. Such as "y" is not equal to "x".

1. It may be that "x" is greater than "y" which can be written as

  x>y.

2. x is greater than or equal to y

  x $\geq$ y

3. x is less than y

  x < y

4. x is less than or equal to y

 x $\leq$ y

5. x is not equal to y

x $\neq$ y

How to represent the solutions of inequalities?

They can be represented in an interval notation (such as the notation used for defining a domain), as it specifies what group of number that fits under this rule.

How to solve linear inequalities?

Solve it like a linear equation.

Goal: to isolate the variable so that you can determine the interval of "x"

Addition/Subtraction

The equation can be seen as a normal equation, instead of the greater than, less than, not equal to sign, just let the = sign replace them for now. Then solve the equation algebraically, afterwards substitute the equations' original inequality sign.

EX

x + 8 > 10 $\longrightarrow$ substitute the = sign

x + 8 = 10 $\longrightarrow$ isolate the variable

x = 2 $\longrightarrow$ replace the equal sign with the original inequality sign

x > 2

In this example "x" can be any number grater than two, and thus can be written like this (2,$\infty$)

Multiplication/Division

is similar to solving addition/subtraction equations

if 2x $\geq$ 5 $\longrightarrow$ then to isolate x, divide both sides by 2

x $\geq$ $\{5 \over 2}\$

How to solve quadratic inequalities?

Let us look at an example of a quadratic inequality:

-x² + 9 < 0

Our first step is to associate this with and equation. Therefore:

y= -x² + 9

The next step is to find out where the equation cuts the graph on the x-axis. This means equating it to 0

-x² + 9 = 0

Now we can solve:

-x² + 9 = 0

x² + 9 = 0

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

x=-3 x=3 Now we know that our quadratic equation crosses the x-axis at these values

We now get 3 different intervals at the points where the x-axis is cut

1.) x < -3

2.) -3 < x < 3

3.) x > 3

The next step is to find out on which intervals is the graph below the x-axis. In order to do this we must look at our original equation:

y = -x² + 9 since it is a negative quadratic we can say that the graph would be facing down. Thus, in order to solve our original inequality of

-x² + 9 < 0 we must look at where our y values are less than 0

From this we can see that the solution is: x < -3 or x > 3

Online References/extension

[1] Written step-by-step explanation

[2] Video Explanation

http://www.youtube.com/watch?v=VgDe_D8ojxw