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Slide 1
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P.4
Solving Equations Algebraically
and Graphically
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Slide 2
• 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.
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Slide 3
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Diophantis
• Father of Algebra
• First to use abbreviated word forms in
English
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Slide 4
• ... his boyhood lasted 1/6th of his life; he
married after 1/7th more; his beard grew after
1/ th more, and his son was born 5 years
12
later; the son lived to half his father's age,
and the father died 4 years after the son.
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Slide 5
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
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Slide 6
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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.
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Slide 7
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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.)
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Slide 8
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Solve:
x 3x

2
3 4
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Slide 9
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Solve:
1
3
6x


x  2 x  2 x2  4
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Slide 10
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Intercepts and Solutions
A point at which the graph of an
equation meets the x-axis is called an xintercept. 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 yintercept. We find it be replacing x with
0 and solving for y.
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Slide 11
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Find the x- and y- intercepts of the
graph of each equation.
2x +3y = 6
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Slide 12
• y = x2 + x - 6
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Slide 13
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.
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Slide 14
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Use a graphing utility to approximate
the solutions of x3 + 4x + 1 = 0
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Slide 15
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.
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Slide 16
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Find the points of intersection of the
graphs of:
 x - 2y = 1 and 3x - y = 7
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Slide 17
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Find the points of intersection of the
graphs of:
y = x2 + 2x - 8 and y = x3 + x2 -6x + 2
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Slide 18
Solving Polynomial
Equations Algebraically
Quadratic Equations
Factoring
Square root principle
Completing the square
Quadratic Formula
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Slide 19
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Solve by factoring:
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x2 + 7x +12 = 0
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Slide 20
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The Illegal Move
ax 2  bx  c  0
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The illegal move is used to factor a quadratic equation
when the leading coefficient (a) is not 1.
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Slide 21
6x2  7x  3  0
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 2  7 x  18  0
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Slide 22
x 2  7 x  18  0
Step 2: Now, factor this new trinomial.
x  9x  2  0
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Slide 23
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

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Slide 24
9 
2

 x   x    0
6 
6

Step 4: Reduce the fractions.
3 
1

 x   x    0
2 
3

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Slide 25
3 
1

 x   x    0
2 
3

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Step 5: Clear the fractions by moving the
denominator in front of the x.
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2x  33x 1  0
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Slide 26
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Solve by factoring:
2x2
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+ 3x = -1
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Slide 27
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The Square Root Principle
• Get “squared” part by itself.
• Take the square root of both sides.
• Solve both equations for x.
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Slide 28
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Solve using the square root principle:
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16x2 = 25
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Slide 29
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Solve using the square root principle:
(x -
4)2
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+ 3 = 12
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Slide 30
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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.  b  2
b2
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2
 x    c 
2
4

• Step 4: Complete using the Square
Root Principle.
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Slide 31
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Solve by completing the square:
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x2 + 4x = 5
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Slide 32
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Solve using the quadratic formula:
3x2
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-x-5=0
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Slide 33
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Solving Quadratic Type Equations
x 4  5x 2  6  0
•Write in quadratic form.
x 
2 2
•Factor.
x
2
 
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 5 x2  6  0


 2 x2  3  0
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•Solve.
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Slide 34
x 4  8 x 2  15  0
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Slide 35
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Please solve
x  9x
3
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Slide 36
x3  x 2  4 x  4  0
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Slide 37
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Other types of equations
WARNING!!!!!!!
The following methods can produce
extraneous roots. Therefore all alleged
solutions should be checked in the
original equation.
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Slide 38
Solving An Equation With Rational Exponents
3
2
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x  27  0
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Slide 39
Solving an Equation Involving
Fractions
x
6
 2
x 1 x
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Slide 40
Solving An Equation Involving
Absolute Value
x2  6  x
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