L06-Fri-16-Sep-2016-Sec-A-6-Equations-HW05-Moodle-Q05...

L06-Fri-16-Sep-2016-Sec-A-6-Equations-HW05-Moodle-Q05, page 39

L06-Fri-16-Sep-2016-Sec-A-6-Equations-HW05-Moodle-Q05


There are 3 general types of equations: Conditional, Contradiction, Identity

Conditional: These equations are true for some values of x but not others.

x + 2 = 3 is a conditional equation because 1 is a solution but other numbers are not.

Contradiction: An equation that is false for all values of x. They have NO SOLUTION.

x + 1 = x is a contradiction because there is no real number which, increased by 1, yields itself.

Example: How much snow would you need to pile up to form a glacier 4 miles high?

There is no amount because after a while the snow at the bottom turns to ice, and ice under great pressure behaves like a viscous liquid and starts to flow.

Identity: An equation that is true for all values of x. The solution is ALL REAL NUMBERS.

x + 1 = 1 + x is an identity. It is an example of the commutative property of addition.

Example: What color should internal steel support beams be painted to maximize their strength?

Any color will do. But, this is not true for the color of cereal boxes.


2 16 2 3 6 3 6 13 2

4 2 3 3 3 68 12 9 3 68 12 3 3

5 153

x x

x xx xx x

xx

We can also solve this graphically. Since we want the value of x that makes equal the expressions on the left and right sides of the equal sign, we could graph each and see where they intersect.

Let 2 2 33

f x x and 1 1

2g x x .

Find value of x on the graph where f x g x


5 4 3 2 5 45 20 3 2 10 4

6 23 6 1023 10

x x x xx x x x

x x

Simplifying this equation results in a contradiction because -23 is not equal to 10. That means the expression on the left can never equal the expression on the right no matter what number we put in for x. So the answer is no solution. That is, there is no value of x that makes the equation true.

Note how graphs of the two expressions are parallel lines.


Recall that the absolute value of a number is its distance from 0 on the number line. For example, 5 5 because +5 is 5 units from 0 on the number line. Likewise, 5 5 because -5 is 5 units

from 0 on the number line. Therefore, to solve this equation we will get the absolute value by itself on one side of the equal sign and then use its definition to form two equations.

5 3 2 3 75 3 2 10

3 2 2

xxx

The above equation will be true if the value inside the absolute value bars is either 2 or -2.

3 2 2 or 3 2 23 4 or 3 0

4 or 03

x xx x

x x

We can also solve this by graphing each expression:


This is called a degree 2 equation because the highest power of x is 2 (it is also called a quadratic equation). With a degree 2 equation we expect to get 2 solutions, although we could get no solution or one solution or an infinite number of solutions. One way to solve this type of equation is by factoring and using the Zero Product Property. This property says that if we have two or more factors and their product is 0 then either or both factors must equal 0 (because the product of any number and 0 is 0).

2

2

2

6 5 136 13 5 0

6 15 2 5 03 2 5 1 2 5 0

2 5 3 1 0

x xx x

x x xx x x

x x

Now, set each factor equal to 0 and solve each equation:

2 5 0 or 3 1 02 5 or 3 1

5 1or2 3

x xx x

x x

We can solve this graphically by seeing where the graphs of the left and right expressions cross.


Solutions are said to satisfy equations because they make both sides the same. This notion of satisfaction was put to music in 1964 by the Rolling Stones. Mick Jagger and Keith Richards wrote it as a lament of the fact that they could not do algebra in high school.


We could solve this like the last one by expanding the square of the binomial and then putting all terms on the left side and 0 on the right, and then factoring.

2

2

2 9

2 9 02 3 2 3 0

5 1 05 0 or 1 0

5 or 1

x

xx x

x xx x

x x

But, take a moment to see if there is a shorter way to solve this.

The left side is a perfect square so simply take the square root of both sides of the equation.

2

2

2 9

2 92 32 3 or 2 3

5 1

x

xxx x

x x

Here is the graphical solution.


We can try to solve this by factoring as we did before:

2

2

6 46 4 0

x xx x

The left side does not factor using integers because there are no two integers with product -4 and sum 6. However, if the left side were a perfect square we could simply take the square root of both sides. So, let’s add 9 to the left side to make the left side a perfect square. Of course, we also need to add 9 to the right side.

2

2

2

2

6 46 46 9 4 9

3 13

3 13

3 13

x xx xx x

x

xx

We can leave the answer in radical form.

Where did the 9 come from?

So, if we are given the first two terms we can find the third term by taking half of the coefficient of the second term and then squaring it.

For example, to make 2 6x x a perfect square we add 2

21 6 3 92

We get 2 6 9x x which can be written 23x , a perfect square.


Before completing the square, we must make the coefficient of the 2x term 1. So, divide by 2 first, and then complete the square.

2

2

2

2

2

2

32

3

2 3 13 12 2

12

Find what we need to complete the square

12

Now, add to both sides

3 12 2

3 174 16

1734 4

1734

916

916

9 916 1

4

2

6

x x

x x

x x

x x

x

x

x

We can leave this in radical form. Here is the graphical solution:


We can use completing the square to solve this equation to get the quadratic formula. We will derive this formula in section 3.3.

This is the same answer as we found using completing the square.


This one looks easy enough.

3

3

3 3 3

1 01

11

xxx

x

This is a degree 3 solution so where are the other two solutions? If we solve it differently, they will appear:

3

2

2

1 01 1 0

1 0 or 1 0

xx x x

x x x

The solution to the first equation is obvious. We can solve the second equation using the quadratic formula (we did this in a previous lecture). The solution is

312 2

x i

The value x = 1 is where the graph crosses the x-axis in a Cartesian graph. The other two solutions are imaginary and so do not appear on a Cartesian graph where both axes are real number lines. We will discuss imaginary numbers in sec A.7.


Set equal to 0 and see if we can factor this:

2

2 2

1 1

2 2 2

2

2

1

2

2 2 2

2 0

2 1 0

0 or 0 or 0 or 0

0 or

2 11 o

2

r

2

2

2

1 1

1 or 1

xx x x

x

x x

x x

x

x x x

x x

x

x

x x x x x

x x

x x

x

x

So, the solution is 1| 1, 0,2

x x x x

multiplhas ic 21 ityx

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