Transcript 7.1 Quadratic Equations • Quadratic Equation: ax2  bx  c  0 • Zero-Factor Property: If a and b are real numbers and.

7.1 Quadratic Equations
• Quadratic Equation:
ax2  bx  c  0
• Zero-Factor Property:
If a and b are real numbers and if ab=0
then either a = 0 or b = 0
7.1 Quadratic Equations
•
Solving a Quadratic Equation by factoring
1. Write in standard form – all terms on one side
of equal sign and zero on the other
2. Factor (completely)
3. Set all factors equal to zero and solve the
resulting equations
4. (if time available) check your answers in the
original equation
7.1 Quadratic Equations
• Example:
2x  5  7x
2
standardform: 2 x 2  7 x  5  0
factored: (2 x  5)(x  1)  0
2 x  5  0 or x  1  0
solutions: x  2.5, x  1
7.1 Quadratic Equations
• If 2 resistors are in series the resistance is 8 ohms
and in parallel the resistance is 1.5 ohm. What are
the resistances?
x y  8 y  8 x
xy
x(8  x)
 1 .5 
 1.5
x y
8
12  8 x  x 2  x 2  8 x  12  0
( x  6)(x  2)  0
x  2, y  6 or x  6, y  2
7.2 Completing the Square
• Square Root Property of Equations:
If k is a positive number and if a2 = k, then
a k
or
a- k
and the solution set is: {
k ,- k}
7.2 Completing the Square
• Example:
5 x  2
2
3
5 x  2  3 or 5 x  2   3
5 x  2  3 or 5 x  2  3
2 3
2 3
x
or x 
5
5
7.2 Completing the Square
• Example of completing the square:
x  6 x  7  0 cannotbe factored
2
x  6 x  9  2  0 (completethesquare)
2
( x  3)  2  0  ( x  3)  2
2
2
x  3   2  x  3  2
7.2 Completing the Square
•
Completing the Square (ax2 + bx + c = 0):
1. Divide by a on both sides
(lead coefficient = 1)
2. Put variables on one side, constants on the
other.
3. Complete the square (take ½ of x coefficient
and square it – add this number to both sides)
4. Solve by applying the square root property
7.2 Completing the Square
2
2
x

y
 ( x  y )(x  y )
• Review:
x2  y2
(prime)
x 3  y 3  ( x  y )(x 2  xy  y 2 )
x  y  ( x  y )(x  xy  y )
3
3
2
2
x  y  ( x  y )(x  y )  ( x  y)(x  y)(x  y )
4
4
2
2
2
2
2
• x4 + y4 – can be factored by completing the square
2
7.2 Completing the Square
• Example:
Complete
the square:
   y 
 x   2 x y  y   2 x y
 x  y    2 xy
x y  x
4
4
2 2
2 2
2

2 2
2
2

 x  y  2xy x  y  2xy
2
2
2
2 2
2
Factor the difference
of two squares: 2
2 2
2
2

7.3 The Quadratic Formula
• Solving ax2 + bx + c = 0:
Dividing by a: x2  b x  c  0
a
a
Subtract c/a:
x  ba x   ac
2
Completing the
2
square by
x 
adding b2/4a2:
b
a
x
b2
4a 2
 
c
a
b2
4a 2
7.3 The Quadratic Formula
• Solving ax2 + bx + c = 0 (continued):
2
Write as a
2
b
 4ac
b 2
4 ac
b
square: x  2 a    4 a 2  4 a 2  4a 2
Use square root
property:
b  b2  4ac
x

2a
2a
 b  b2  4ac
Quadratic formula: x 
2a
7.3 The Quadratic Formula
• Quadratic Formula:
 b  b  4ac
x
2a
2
b  4ac
2
is called the discriminant.
If the discriminant is positive, the solutions are
real
If the discriminant is negative, the solutions are
imaginary
7.3 The Quadratic Formula
• Example:
x2  5x  6  0
x
a  1, b  -5, c  6
 (5) 
5
( 5)  4(1)(6)
2(1)
2
25  24 5 1

 
2
2 2
x  3, x  2
7.3 The Quadratic Formula
•
Complex Numbers and the Quadratic Formula
Solve x2 – 2x + 2 = 0
 (2)  (2)  4(1)(2)
x
2(1)
2
2   4 2  4i 2  2i



2
2
2
 1 i
7.3 The Quadratic Formula
Method
Advantages
Factoring
Fastest method
Disadvantages
Not always
factorable
2
Square root
Not always this
form : ( x  a)  b
property
form
Completing the Can always be Requires a lot
square
used
of steps
Quadratic
Can always be Slower than
Formula
used
factoring
7.4 The Graph of the Quadratic
Function
•
•
A quadratic function is a function that can be
written in the form:
f(x) = ax2 + bx + c
The graph of a quadratic function is a parabola.
The vertex is the lowest point (or highest point if
the parabola is inverted
7.4 The Graph of the Quadratic
Function
•
Vertical Shifts:
f ( x)  x  k
2
The parabola is shifted upward by k units or
downward if k < 0. The vertex is (0, k)
•
Horizontal shifts:
f ( x)  x  h
2
The parabola is shifted h units to the right if h >
0 or to the left if h < 0. The vertex is at (h, 0)
7.4 The Graph of the Quadratic
Function
•
Horizontal and Vertical shifts:
f ( x)  x  h  k
2
The parabola is shifted upward by k units or
downward if k < 0. The parabola is shifted h
units to the right if h > 0 or to the left if h < 0
The vertex is (h, k)
7.4 The Graph of the Quadratic
Function
•
Graphing:
f ( x)  ax  h  k
2
1. The vertex is (h, k).
2. If a > 0, the parabola opens upward.
If a < 0, the parabola opens downward (flipped).
3. The graph is wider (flattened) if 0  a  1
The graph is narrower (stretched) if a  1
7.4 The Graph of the Quadratic
Function
f ( x)  x  h  k
2
Vertex = (h, k)
7.4 The Graph of the Quadratic
Function
•
Vertex Formula:
The graph of f(x) = ax2 + bx + c has vertex
 b

,
 2a
  b 
f
 
 2a  
7.4 The Graph of the Quadratic
Function
•
Graphing a Quadratic Function:
1. Find the y-intercept (evaluate f(0))
2. Find the x-intercepts (by solving f(x) = 0)
3. Find the vertex (by using the formula or by
completing the square)
4. Complete the graph (plot additional points as
needed)
18.1 Ratio and Proportion
• Ratio – quotient of two quantities with the
a
same units
b
Note: Sometimes the units can be converted
to be the same.
18.1 Ratio and Proportion
• Proportion – statement that two ratios are
equal:
a
c
b
d

Solve using cross multiplication:
ad  bc
18.1 Ratio and Proportion
• Solve for x:
Solution:
81
x 3

9
7
81 7  9  ( x  3)
567  9 x  27
540  9 x
x  60
18.1 Ratio and Proportion
• Example: E(volts)=I(amperes)  R(ohms)
How much current for a circuit with 36mV and
resistance of 10 ohms?
36m V
mV
3 V
I 
 3.6
 3.6  10
10


I  3.6  103 A  3.6m A
18.2 Variation
•
Types of variation:
1. y varies directly as x: y  kx
k
2. y varies inversely as x:
y
x
3. y varies directly as the
y  kx2
square of x:
4. y varies directly as the y  k x
square root of x:
18.2 Variation
•
Solving a variation problem:
1. Write the variation equation.
2. Substitute the initial values and solve for k.
3. Rewrite the variation equation with the value
of k from step 2.
4. Solve the problem using this equation.
18.2 Variation
•
Example: If t varies inversely as s and
t = 3 when s = 5, find s when t = 5
1. Give the equation: t  k
s
k
2. Solve for k: 3   k  15
5
3.
15
Plug in k = 15: t 
s
4. When t = 5: 5  15  5s  15  s  3
s
B.1 Introduction to the Metric System
• Metric system base units:
length  meter(m)
mass(weight )  kilogram(kg)
time  second(s)
temperature  degrees Celsius (C)
amount of substance  mole(mol)
electricalcurrent  ampere(A)
luminousintensity candela (cd)
volume liter (L)
B.1 Introduction to the Metric System
Multiple in
decimal form
Power of 10
Prefix
Symbol
1000000
1000
100
106
103
102
mega
kilo
hecto
M
k
h
10
1
101
100
deka
base unit
da
0.1
0.01
10-1
10-2
deci
centi
d
c
0.001
10-3
milli
m
0.000001
10-6
micro

B.1 Introduction to the Metric System
• 1 gram = weight of 1 ml of water
• Unit of weight = 1Kg = 1000 grams
• 1 liter of water weighs 1 Kg
B.2 Reductions and Conversions
• Conversions:
1 cm  0.394 inches
1 m  39.37 inches
1 km  0.621m iles
5 km  3.1 m iles
Note: In Canada, speed is in kph
instead of mph
B.2 Reductions and Conversions
• Conversions
1 liter  1000cm  1000cc
1 cc  1 m l
1 liter  1.057 quarts
1 teaspoon 5 m l
3
B.2 Reductions and Conversions
• Try these:
– 3.5 liters = ________ ml
–
1
8
liter = ________ cc