Chapter 2.1
Solving Equations Graphically
Solution = a number that when substituted for the variable, produces a true statement
Ex. 2x + 1 =5 x = 2 is a solution 2(2) + 1 = 5
Solve = means to find all of its solutions
Equivalent = 2 equations are equivalent if they have the same solutions
** often there are no formulas that provide solutions to many other types of equations
 for such equations graphical approximation methods are practical alternatives
Complete Graph – if in a viewing window it shows all the important features of the
graph including all peaks, valleys and points where it touches an axis and suggests the
general shape of portions of the graph that are not in the window.
*best to use a window small enough to show as much detail as possible
A. INTERSECTION METHOD
1. Graph 2 equations on the same screen and find x-coordinate of the point where
the 2 graphs intersect.
** Coordinate can be approximated by zooming in and using the TRACE
CAUTION: if graphs don’t intersect then they have no common output value
B. X-INTERCEPT METHOD
1. Zero of a Function – is an input that produces an output of 0
** also called Solutions or Roots
2. A point where the graph of y=f(x) intersects the x-axis is of the form (a,0) because
every point on the x-axis has y coordinate = 0.
** the number a is called an x-intercept of the graph
Let f be a function. If r is a real # that satisfies any of the following statements then r
satisfies all the statements:
1. r is a zero of the function f – an input that produces an output of 0
2. r is an x-intercept of the graph f
3. x = r is a solution or root of the equation f(x) = 0
To Solve an Equation b the X-Intercept Method:
a)Write the equation in the equivalent form f(x) = 0
b) Graph y = f(x)
c) Find the x-intercepts of the graph – are the real solutions of the equation
ADVANTAGE: That solutions appear on the x-axis and no info.@ range of function is
needed
Do. Ex #2 p.84 Graphical Zero Finder (labeled Zero in the Calc menu)
Do Ex. #3 and Ex. #4 radical and rational examples –0when numerator =0 and deno. Is
A square root =0 only when 0
nonzero
Summary: To solve h(x) = g(x) use the following methods:
 Intersection method
a)Graph y1= h(x) and y2 = g(x)
b) Find the x-coordinate of each point of intersection
 The x-intercept Method
a) rewrite the equation as f(x) = 0 where f(x)=h(x) – g(x)
b) graph y = f(x)
c) Find the x-intercepts of the graph of f(x) – the x-intercepts are the
solutions of the equation
ADV: needs no info. About the range of the function
Do. P. 87#2-34 even
2.2 Solving Quadratic Equations Algebraically
** remember properties of equality
1. +/- the same quantity from both sides of the equation
2. x/÷ both sides of the equation by the same nonzero quantity
Quadratic Equation = a second degree equation is one written in the form ax²+ bx +c =0
For all real constants a, b, c with a ≠0
4 Techniques to Solve Quadratic Equations
1. Factoring – based on ZERO PRODUCT PROPERTY – if a product of real #’s is 0,
then at least one of the factors is 0, so if ab=0 then a=0 or b=0
ex. 2x²+11x=6
a) put f(x)=0 so 2x²+11x-6=0
b) then factor
(2x -1)(x+6) = 0
c) set each = 0
2x -1 = 0
and
x+6 =0
d) solve for x
x=½
x = -6
2. Taking the square root of both sides
solution x²=k for a real #k
x = ±√k
k<0 0 solutions
k=0 1 solution
o
k>0 2 solutions
+√k and -√k
ex. 4x²= 8
a) isolate x²
4x² = 8
x² = 2
b) take square root of each side √x² = √2
c) solve x = ±√2 so x = +1.414 and x = -1.414
ex. 5(x-3)² = 15
a) divide both sides by 5
so (x-3)² = 3
b) take the square root of both sides √(x-3)² = √3
c) Solve
x-3 = ±√3
x = 3+√3 = 4.73
x = 3-√3 = 1.27
3. Completing the Square – goes by the premise that an expression of the form x² + bx
can be changed into a perfect square by adding a suitable constant
*take ½ the middle term and square it
* then add to other side and you’ll get a perfect square trinomial
* must have a leading coefficient = 1
Ex. 8x² - 24x + 3 = 0
a) divide by 8 to get a 1 in front of x² - 3x + ⅜ = 0
b) move ⅜ to the other side by - = x² - 3x + ________ = -⅜
c) take ½ the middle term and square it ½(-3) = (-3/2)² = 9/4
x² - 3x + 9/4 = -⅜ + 9/4
d) x² -3x + 9/4 = 15/8
e) (x-3/2)(x-3/2) = 15/8
(x-3/2)² = 15/8
f) take square root of each side √(x-3/2)² = √15/8
g) x = 3/2 +√15/8
x=3/2 - √15/8
4. Quadratic Formula – in form ax² + bx + c = 0
x = -b ± √b² - 4ac
2a
ex. x²-10x-13 = 0
a=1 b = -10 c = -13
--10 ±√(-10²) – 4(1)(-13)
2(1)
x = 10±√152
2
= 11.2
x = 10-√152
2
= -1.2
Discriminant – expression b²-4ac in the quadratic formula and can be used to
Figure the # of solutions
b²-4ac>0
2 real solutions
b²-4ac=0
1 real solution
b²-4ac<0
0 real solutions (2 imaginary)
Ex. Solve 4x² = -x -2
a) 4x² + x + 2 = 0
a =4 b =1 c =2
b) b²-4ac (1)² - 4(4)(2) = 1-32 = -31
c) -31 < 0 so 2 imaginary solutions
**A Quadratic Equation w/ No x-intercepts has no real solutions **
-----------------------------------------------------------------------------------------------------------Polynomial Equations – of degree n is an equation that can be written in the form
anxn + an-1xn-1+…+a1x + a0 = 0 where the a’s are real numbers
ex. 4x6 – 3x5 + x4 + 7x3 – 8x2 + 4x + 9 = 0
= degree of 6
Have the following traits: 1. No variables in the denominators
2. No variables under radical signs
General Rule: polynomial equations of degree 3 are best solved by graphical methods
** some in quadratic form and can be solved algebraically
Ex. 3x4 – 5x² + 2 = 0 substitute u for x²
3(u)2 + 5u + 2 = 0
(3u – 2)(u – 1) =0
3u – 2 =0 u = ⅔
** now solve for x²
x² =⅔
** take the square root of each side x =±√⅔
u-1= 0 u = 1
x² = 1
x = ±√1
-----------------------------------------------------------------------------------------------------------2.3 Applications of Equations
Follow the guideline steps:
1. READ the problem carefully and determine what is asked for
2. LABEL the unknown quantities with variable
3. DRAW a picture if appropriate
4. TRANSLATE the verbal statements in the problem and relationship between
known and unknown
5. CONSOLIDATE the math info. Into an equation in 1 variable that can be solved
or an equation in 2 variables that can be graphed
6. SOLVE for at least 1 of the unknown variables
7. Find all remaining unknown quantities by using the relationship given in problem
8. CHECK and INTERPRET all quantities found in the original problem
Solution should: a)make sense
b) satisfy the given conditions
c) answer the original question
ex. The average of 2 real numbers is 39.625 and 1 number is 1500 times the reciprocal of
the other. Find the 2 #’s
READ: 2 #’s are asked for
LABEL: let the numbers be a and b
DRAW: A picture is not appropriate
TRANSLATE:
Eng. Lang.
Math Lang.
2 numbers
a and b
average is 39.625
a + b/2 = 39.625
1500 times the reciprocal
b = 1500 · 1/a
of the other
CONSOLIDATE:
a + b/2 =39.625
b = 1500 · 1/a
SOLVE: (substitute) a + 1500/a =39.625
a + 1500 = 79.25
2
a
Solve using x intercepts method (put x where a) x + 1500 -79.25 = 0
X
Graph in calculator – use ZOOM and TRACE to find solutions
a = 48
and
a =31.25
CHECK
48 + 31.25 = 39.625
2
1500 · 1/48 = 31.25
--------------------------------------------------------------------------------------------------EX. The width of a rectangle is three times its height. If it has an area of 60.75
square feet, what are its dimensions?
READ: Dimensions of rectangle
LABEL: w=width h=height
DRAW
TRANSLATE: width is 3 times height
w = 3h
Area is 60.75
wh = 60.75
CONSOLIDATE:
w=3h
wh=60.75
SOLVE:
3h(h) = 60.75 (substituted 3h in equation for w)
3h² = 60.75
h² = 20.25 take square root of each side
h = ±4.5 - since height is never negative, only positive root
applies in this situation so h = 4.5
--now find w by plugging in
W=3(4.5) = 13.5
--------------------------------------------------------------------------------------------Interest Applications
(P) Principal – when an amount is deposited or borrowed
(I) Interest – the fee paid for the use of the money and is calculated as a % of the
principal each year.
Simple Interest – when duration of a loan or a bank balance is less than a
year.
I = prt
r = annual interest rate t = time
Ex. A real estate investment yields a return of 10% per year and a CD pays 6%
interest per year. How much of $7500 should be put in the real estate investment
and how much should be put in the CD to get a return of 7% on the entire $7500.
(Return on s dollars at 10%) + (return on 7500-s dollars at 6%) =7% of 7500
READ/LABEL
TRANSLATE - .10s + .06(7500-s) = 7% of 7500
SOLVE:
.10s + .06(7500-s) = 525
.10s + 450 -.06s = 525
.04s = 75
s = 1875
CONCLUSION: $1875 in real estate
$7500 - $1875 = $5625 in CD’s
Distance Applications = involves distance and a constant rate of velocity
D=rxt
r = rate
t= time
d= distance
Ex. A canoeist paddled a total of 11 hours on a 48km round trip to Bear Lake.
The upstream trip followed a period of heavy rain and the current he paddled
against was 3km/h. For the downstream return trip, he paddled with the current
that slowed to 2 km/hr. At what speed would he paddle the canoe in still water?
D
R
T (don’t know in this problem)
Upstream
24
r-3
24/r-3
Downstream 24
r+2
24/r+2
Time upstream + Time Downstream = 11
24/r-3
+
24/r+2
= 11
(r-3)(r+2) 24/r-3 + 24/r+2 (r-3)(r+2) = 11 (r-3)(r+2)
(r+2)24
+ 24( r-3)
= 11(r-3)(r+2)
24r + 48 + 24r – 72 = 11r² -11r – 66
48r – 24 = 11r² - 11r -66
11r² - 59r – 42 = 0
(11r+7)(r-6) = 0
r = -7/11 and r = 6
** negative solution doesn’t apply so r = 6
** Check 6 in original equation and see if works so upstream he paddles 3km/h
and downstream he paddles 8 km/hr
----------------------------------------------------------------------------------ex. A walkway of uniform width is to be built around a square swimming pool
with 18 foot edger. How wide should the walkway be to make the area of the
walkway’s surface 400 square feet?
READ : Let x = width of walkway in fee
(Area of outer square) - (Area of pool) = Area of Walkway
(18 + 2x)(18+2x)
- (18)²
=
400
4x²+72x+324 - 324 = 400
4x²+72x-400 = 0
(Divide everything by 4 to get leading coeff. Of 1)
x²+18x-100 = 0
a = 1 b = 18 c= -100
Use Quad formula: -b ± √b²-4ac/2a
Get x = 4.45 or -22.45 can’t be negative so x = 4.45
-----------------------------------------------------------------------------------
ex. A chemist has 30mL of a 40% acid solution in a test tube. How much of the
solution should be poured off and replaced with pure acid so the new mixture is
70% acid?
Solution: Let x = # of solution to be replaced by pure acid
When x mL are drained there are 30 –x mL left in solution, 40% of
Which is acid
Amount of acid solution
After pouring off x mL
+
x mL of pure acid =
Amount of pure
acid in final solution
40% of (30-x)
+
x
= 70% of 30
.4(30-x) + x = .7(30)
12 - .4x + x = 21
.6x = 9
x = 15
Chemist should pour off 15mL of the solution and replace with pure acid.
HW p. 105 #1-4, 9, 12, 13, 15, 16, 19
--------------------------------------------------------------------------------------------------2.4
Other Types of Equations
Absolute Value – of a # c, is denoted │c│ and defined as follows:
If c≥0 then │c│= c
If c<0 then │c│ = -c
** also indicates # of spaces from 0
Ex. │-5-11│= │-16│=16
Absolute Value and Distance: if c and d are real #’s then │c-d│is the
distance between c and d on the number line.
Geometric Definition of Absolute Value: if c is a real #, then │c│ is the
distance from c to 0 on the number line.
Ex. │-3.5│= distance from -3.5 to 0 on the number line
Properties of Absolute Value: Let c and d represent real #’s
1. │c│≥0 and │c│>0 when c≠0
2. │c│= │-c│ ex. c = 3 then c=│3│=│-3│
3. │cd│ = │c│•│d│ ex. let c=6 d=-2 │6(-2)│= │6│•│-2│
4. │c│ where d≠0
│d│
Triangle Inequality: for any real #’s c and d
│c+d│≤│c│ + │d│ ex. │-3+5│≤│-3│+│5│
Square Root of Square: for every real #c, √c² = │c│
Solving Absolute Value Equations
Ex. solve │x-4│=8
2 choices
x-4 = 8
and x-4 = -8
x = 12
x = -4
check : │12-4│= 8 and │-4-4│= 8
Extraneous Solutions – equations that are not true when checked by substitution
(extraneous roots)
Ex. Solve │x+4│=5x-2
a) x+4 = 5x -2
3/2 = x
or
b) x+4 = -(5x-2)
-1/3 = x
Check: only solution is 3/2 because -1/3 is extraneous
Ex. │x²+4x-3│ = 2
a) x²+4x-3=2
or
b) x²+4x -3 = -2
x²+4x-5 = 0
x²+4x-1 use quad formula
(x+5)(x-1) = 0
x = -5 x=1
x = -4±√20
** check by graphing
Radical Equation: equations that contain expressions with a variable under a radical
symbol.
Follow the fact: If A and B are algebraic expressions and A=B then An=Bn for
Every positive integer n --**may introduce extraneous roots
Power Principle : If both sides of an equation are raised to the same positive integer
power, then every solution of the original equation is a solution of the
new equation. However, the new equation may have solutions that are
not solutions of the original one.
Solving a Radical Equation:
Ex. 1 + √4x + 33 = 2x
Steps: 1. Get radical on side by itself: √4x + 33 = 2x -1
2. Square both sides: (√4x + 33)² = (2x – 1)²
4x+33 = 4x²-4x+1
0 = 4x² -8x-32
(4x+8)(x-4)=0 x=-2 x = 4
3. Check: x=4 only works so x = -2 is extraneous
Using Power Principle Twice:
√x+2 - √2x-3 = 1
Steps 1: Get 1 radical on each side √x+2 = √2x-3 + 1
x+2 = (√2x-3)²+ 2•√2x-3 + 1²
x+2 = 2x -3 + 2√2x-3 + 1
x+2 = 2x + 2√2x-3 -2
(-x+4)²= (2√2x-3)²
(x²-8x+16)=4(2x-3)
(x²-8x+16) = 8x-12
x²-16x+28=0
(x-14)(x-2)=0
x=14 x=2
Check: and only 2 works, so 14 is extraneous
2. Square both sides:
Fractional Equations: if f(x) and g(x) are algebraic expressions, the quotient:
Numerator f(x)
Denominator g(x)
**denominator cannot = 0 will be undefined
Solving Fractional Eq.
Ex. 2x²+x-3
2x²+5x+3 = 0
Steps: 1. Factor the numerator
2x²+x-3=0
(2x+3)(x-1)=0
x = -3/2 x = 1
Step 2 Discard any solution that makes the denominator =0
x = -3/2 for 2x²+5x+3 --- doesn’t work to make it =0
then x =1 for 2x²+5x+3 --- does work so x =1
p. 116 #2-24 even, 28-62 even
-----------------------------------------------------------------------------------------------------------2.5 Inequalities
Inequalities = for any 2 real #’s, a and b, exactly one of the following statements is true
a<b a=b a>b
Compound Inequality b<c<d mean b<c and c<d
Interval Notation
Interval = the set of all 3’s lying between 2 fixed #’s
[c,d] = c≤ x≤ d
(c,d) = c<x<d
[c,d) = c≤x<d
(c,d] = c<x≤d
*c and d serve as ENDPOINTS of the interval
[c,d] is CLOSED INTERVAL b/c both endpoints are included [ ]
(c,d) is OPEN INTERVAL b/c neither endpoint is included ( )
[c,d) is HALF OPEN INTERVAL where the bracket indicates which point is included
(c,d] “
“
[b,∞) all real #’s x such that x≥b
(b,∞) all real #’s x such that x>b
(-∞,b] all real #’s x such that x≤b
(-∞,b) all real #’s x such that x<b
entire # line (-∞,∞) ** ∞ never has brackets only ( )
Basic Principles for Solving Inequalities:
1. +/- the same quantity on both sides of the inequality
2. x/÷ both sides of the inequality by the same positive quantity
3. x/÷ both sides of the inequality by the same negative quantity and reverse the
direction of the inequality
Compound Linear Inequalities
Ex. Solve -5<2x-1≤x+7
Steps: 1. Break into 2 equations: -5<2x-1 and 2x-1≤x+7
-2<x and x≤8 so -2<x≤8 (-2,8]
Ex. -2<4-3x<6
a) -2<4-3x
2>x
(-⅔,2)
and
and
4-3x<6
x>-⅔
2>x>-⅔
**The solutions of an inequality of the form f(x)<g(x) consists of intervals on the x-axis,
where the graph of f is below the graph of g
**The solutions of f(x)>g(x) consists of intervals on the x-axis where the graph of f is
above the graph of g
**The graph of g=f(x)-g(x) lies above the x axis when f(x)-g(x)>0 and below the x-axis
when f(x)-g(x)<0
Ex. Solve x4-x3-12x²>4x+10 rewrite as x4-x3-12x²-4x-10>0 then graph
Quadratic and Factorable Inequalities: -- can use quad. Formula b/c x² and not
factorable
Ex. x²+3x-2≤0 where graph lies below x axis a=1 b=3 c=-2 plug in quad formula and
solve
Answer: -3-√17 ≤ x ≤ -3 + √17
2
2
Factorable Inequality:
Ex. (x-3)(x+6)(x+1)4 ≤ 0
x=3 x=-6 x=1
Where graph lies below x-axis must graph on calc. and see that
graph is below x-axis on interval -6≤x≤3
Solving Inequalities Summary:
1. Write the inequality in 1 of the following forms:
F(x)>0 f(x)≥0 f(x)<0 f(x) ≤0
2. Determine the zeros of f, exactly if possible, approximate otherwise
3. Determine the interval, or intervals, on the x-axis where the graph of f is above or
below the x-axis.
HW. P. 124 #2-72 even
2.5A Absolute Value Inequalities:
#1. X intercept Method:
a) put the absolute value equation as the first y=
b) then put whatever is on the other side of the inequality as the 2nd y=
c) Use the zoom and trace to determine the x intervals of the absolute value
graph is below the graph of whatever is on the other side of the inequality
Ex. │x4 + x² -2x +3│< 5x
A. f(x) = x4 + x² -2x +3
B. g(x) = 5x
Then graph and see the x intervals below 5x line
#2 If │r│≤ k
is -k ≤ r ≤ k
“and” inequalities
Ex. Solve │2x - 5│≤ 8
-8 ≤2x -5 ≤ 8
#3 If │r│≥ k
is r ≤ -k or r ≥ k
Ex. │4x + 3│ > 5
4x + 3 >5
or 4x + 3 < -5
“or” inequalities
** Quadratics follow the same rules as above
2