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P.4
Solving Equations Algebraically and
Graphically
• An ancient Egyptian papyrus, discovered in
1858, contains one of the earliest examples
of mathematical writing in existence.
• The papyrus itself dates back to around
1650 B.C., but it is actually a copy of
writings form two centuries earlier.
• The algebraic expressions on the papytrus
were written in words.
Diophantis
• Father of Algebra
• First to use abbreviated word forms in
English
• ... his boyhood lasted 1/6th of his life; he married
after 1/7th more; his beard grew after 1/12th more,
and his son was born 5 years later; the son lived
to half his father's age, and the father died 4 years
after the son.
• So he married at the age of 26 and had a son who
died at the age of 42, four years before Diophantus
himself died aged 84. Based on this information
we have given him a life span of 84 years.
Equations and Solutions of
Equations
An equation in x is a statement that
variable expressions are equal.
A solution of an equation is a number r,
such that when x is replaced by r, the
resulting equation is a true statement
The solution set of an equation is the set of
all solutions of the equations.
To solve an equation means to find its
solution set.
Types of Equations
Identity - every real number in the domain of
the variable is a solution.
Conditional - only some of the numbers in the
domain of the variable are solutions. (These are
the types we have to solve.)
Solve:
x 3x

2
3 4
24
x
13
Solve:
1
3
6x

 2
x2 x2 x 4
No solution
Intercepts and Solutions
A point at which the graph of an equation
meets the x-axis is called an x-intercept. We
find it be replacing y with 0 and solving for
x.
A point at which the graph of an equation
meets the y-axis is called a y-intercept. We
find it be replacing x with 0 and solving for
y.
Find the x- and y- intercepts of the graph of
each equation.
2x +3y = 6
0,2 and 3,0
• y = x2 + x - 6
x - intercepts :  3,0 and 2,0 
y - intercept : 0,-6
Finding Solutions Graphically
 Write the equation in general form, f(x) = 0
 Use a graphing utility to graph the function y =
f(x). Be sure the viewing window shows all the
relevant features of the graph.
 Use the zero or root feature or the zoom and
trace features of the graphing utility to
approximate the x-intercept of the graph of f.
Use a graphing utility to approximate the
solutions of x3 + 4x + 1 = 0
Points of Intersection of Two
Graphs
Points at which two graphs meet are called
points of intersection. Their corresponding
ordered pairs are solutions to both of the
equations.
Find the points of intersection of the graphs
of:
 x - 2y = 1 and 3x - y = 7
3,2
Find the points of intersection of the graphs
of:
y = x2 + 2x - 8 and y = x3 + x2 -6x + 2
(-3.31863, -3.62396).
Solving Polynomial Equations
Algebraically
Quadratic Equations
Factoring
Square root principle
Completing the square
Quadratic Formula
Solve by factoring:
x2 + 7x +12 = 0
x  3,4
The Illegal Move
ax  bx  c  0
2
The illegal move is used to factor a quadratic equation when
the leading coefficient (a) is not 1.
6x  7x  3  0
2
Step 1: Multiply the leading coefficient (a) by
c and form a new trinomial where a is now
1, b is the same, and c is now ac.
x  7 x  18  0
2
x  7 x  18  0
2
Step 2: Now, factor this new trinomial.
x  9x  2  0
x  9x  2  0
Step 3: Undo the “Illegal Move” by dividing
the original leading coefficient (a).
9 
2

 x   x    0
6 
6

9 
2

 x   x    0
6 
6

Step 4: Reduce the fractions.
3 
1

 x   x    0
2 
3

3 
1

 x   x    0
2 
3

Step 5: Clear the fractions by moving the
denominator in front of the x.
2x  33x 1  0
Solve by factoring:
2x2 + 3x = -1
1
x  1,
2
The Square Root Principle
• Get “squared” part by itself.
• Take the square root of both sides.
• Solve both equations for x.
Solve using the square root principle:
16x2 = 25
5
x
4
Solve using the square root principle:
(x - 4)2 + 3 = 12
x  7,1
Completing the Square
• Step 1: Move c to the other side.
• Step 2: Add  b2 to both sides.
• Step 3: Factor your perfect square trinomial.
2
2
b
b2

 x    c 
2  using the
4 Square Root

• Step 4: Complete
Principle.
Solve by completing the square:
x2 + 4x = 5
x  5,1
Solve using the quadratic formula:
3x2 - x - 5 = 0
1 61
x
6
Solving Quadratic Type Equations
x  5x  6  0
4
2
•Write in quadratic form.
x   5x  6  0
2 2
2
•Factor.
x
2


2 x 3  0
•Solve.
No solution
2
x  8 x  15  0
4
2
x   3 , 5
Please solve
x  9x
3
x  3,3
x  x  4x  4  0
3
x  2,1,2
2
Other types of equations
WARNING!!!!!!!
The following methods can produce
extraneous roots. Therefore all alleged
solutions should be checked in the original
equation.
Solving An Equation With Rational Exponents
3
2
x  27  0
x 9
Solving an Equation Involving
Fractions
x
6
 2
x 1 x
x  2  10
Solving An Equation Involving
Absolute Value
x2  6  x
x  2,3