Thursday, September 6, 2007

Precalculus Prerequisite 4a - Solving Equations Algebraically and Graphically

Prerequisite 4a - Solving Equations Algebraically and Graphically

I. Vocabulary:

A. An Equation - is a statement that two algebraic expressions are equal.
Example: 2x + 4 = 10

B. To Solve an equation in "x" means to find all values of "x" for which the statement is true. These values are solutions.
Example: 2x + 4 = 10
2x = 6
x = 3

C. An identity equation- an equation that is true for every real number in the domain of the variable.
Example: x2 - 6x + 9 = (x - 3)2

D. Conditional Equation - an equation that is true for just some (or even none of the real numbers in the domain of the variable.
Example 1: x2 - 6x + 9 = 0
(x - 3)2 = 0
x = 3 only

Example 2: 3x + 2 = 4x - 5
x = 7 only

Example 3: x2 + 3x + 4 = 4x - 5
no solutions

E. Notations:
1. all real numbers - R
2. no solutions - { } empty set or Æ is the null set

F. Solving an Equation Involving Fractions:
Example 1:
x/4 + (2x)/3 = 6

multiply by the LCD
4, 3 are the denominators so the LCD = 12, so multiply each term by 12:

12(x/4) + (12)(2x)/3 = (12)(6)
3x + 8x = 72
11x = 72
x = 72/11

Example 2: 5/x + (3x)/2 = 7

LCD = 2x

(2x)(5/x) + (2x)(3x)/2 = (2x)(7)
10 + 3x2 = 14x
3x2 - 14 x + 10 = 0

use the quadratic equation

x = (-b ± Ö(b2 - 4ac))/(2a)

x = (14 ± Ö(142 - (4)(3)(10)))/((2)(3))

x = (14 ± Ö76)/3

x = (7 + Ö 19)/3 and x = (7 - Ö19)/3

or

x = 3.786299648 and x = .8803670188 (check to make sure they both work - they do!)

You can also check by using your graphing calculator -
put in
y1 = 5/x + (3x)/2 and
y2 = 7

calc ® intersect, shows the same answers!

G. Extraneous Solutions - an answer or solution that does not satisfy the original equation.
Example:

6/x - 2/(x+3) = (3(x+5))/(x(x + 3))

LCD = x(x+3)

6(x + 3) - 2x = 3(x + 5)
6x + 18 - 2x = 3x + 15
4x + 18 = 3x + 15
x = -3

substituting x=-3 back into the original equation, we have
-2 - 2/0 = 6/0
This is impossible so there is not a solution

H. To find the x-intercepts (a, 0), let y = 0 and solve for x.
To find the y-intercepts (0, b), let x = 0 and solve for y.

Example:
2x2 - 5x + 2 = y

x-intercepts, let y = 0
2x2 - 5x + 2 = 0
(2x -1)(x - 2) = 0
2x - 1 = 0 and x - 2 = 0
x = 1/2 and x = 2
(.5, 0) and (2, 0)

y-intercepts, let x = 0
2(0)2 - 5(0) + 2 = y
2 = y
(0,2)

I. Finding Solutions Graphically:
24x3 - 36x + 17 = 0

graph this on your calculator and you see there is only one solution:
window:
x-values: -10 to 10
y-values: -5 to 40
You can see it is hard to tell how many times the equation crosses the x-axis
so change the y-values to -3 to 3 and you see it crosses y=0 only once

x = -1.414486, y=0

J. Can find answers more than one way, here are a couple using the calculator:
1. y1 = 24x3 - 36x + 17
2nd trace ® zero
enter left-bound before x=-2 and then right-bound after x=-1, then guess about -1.5, gives the answer.
2. y1 = 24x3 - 36x + 17
y2 = 0
2nd trace, #5 intersect, enter 3 times about where the two equations intersect.

K. Remember with points of intersection, always look for all solutions:
Example:
24x3 - 36x + 17 = 2x + 5 has how many solutions?

(-1.393598, 2.212803)
(.3407855, 5.681571)
(1.052813, 7.105626)

remember there can be no points of intersection (therefore no solutions), one solution or many solutions.

L. Solving Polynomial Equations Algebraically:

1. First degree equation - linear equation -
example: 2x + 4 = 7

2. Second degree equation - quadratic equation -
example: 2x2 + 4x + 6 = 0

3. Third degree equation - Cubic equation -
example: 2x3 + 2x2 + 5x + 2 = 0

4. Fourth degree equation - Quartic equation -
example: 2x4 + 3x3 - 2x2 + 3x + 2 = 0

5. Fifth degree equation - Quintic equation -
example: 4x5 + 2x4 - 3x3+ 2x2 + x - 4 = 0