Circles
Standard form:
(x – h)2 + (y – k)2 = r2
center: (h, k)
radius: r
Writing the Equation of a Circle
Write the equation of the circle.
the circle with center (0, 6) and radius r = 1
(x – h)2 + (y – k)2 = r2
(x – 0)2 + (y – 6)2 = 12
x2 + (y – 6)2 = 1
Finding Zeros or x-intercepts by Factoring
Find the zeros of the function by factoring.
g(x) = 3x2 + 18x
3x2 + 18x = 0
3x(x+6) = 0
3x = 0 or x + 6 = 0
x = 0 or x = –6
Set the function to equal to 0.
Factor: The GCF is 3x.
Apply the Zero Product Property.
Solve each equation.
Key Concept 1
Solving Rational Equations
Solve the equation x –
x(x) –
18
(x) = 3(x)
x
x2 – 18 = 3x
x2 – 3x – 18 = 0
(x – 6)(x + 3) = 0
x – 6 = 0 or x + 3 = 0
x = 6 or x = –3
18
x = 3.
Multiply each term by the LCD, x.
Simplify. Note that x ≠ 0.
Write in standard form.
Factor.
Apply the Zero Product Property.
Solve for x.
Solve the equation 6 + 5 = – 7 .
4
x
4
6 (4x) + 5 (4x) = – 7 (4x)
4
4
x
24 + 5x = –7x
24 = –12x
x = –2
Multiply each term by
the LCD, 4x.
Simplify. Note that x ≠ 0.
Combine like terms.
Solve for x.
Solve each equation.
5x
3x + 4
=
x–2
x–2
Multiply each term by
5x
3x
+
4
x – 2 (x – 2) = x – 2 (x – 2) the LCD, x – 2.
5x
(x – 2) = 3x + 4 (x – 2) Divide out common
x–2
x–2
factors.
Simplify. Note that x ≠ 2.
5x = 3x + 4
Solve for x.
x=2
The solution x = 2 is extraneous because it makes
the denominators of the original equation equal to 0.
Therefore, the equation has no solution.
Summer Sports Application
City Park Golf Course charges $20 to rent golf
clubs plus $55 per hour for golf cart rental.
Sea Vista Golf Course charges $35 to rent
clubs plus $45 per hour to rent a cart. For
what number of hours is the cost of renting
clubs and a cart the same for each course?
Example Continued
Step 1 Write an equation for the cost of renting clubs
and a cart at each golf course.
Let x represent the number of hours and y represent
the total cost in dollars.
City Park Golf Course: y = 55x + 20
Sea Vista Golf Course: y = 45x + 35
Because the slopes are different, the system is
independent and has exactly one solution.
When x =
, the y-values are both 102.5. The cost of
renting clubs and renting a cart for
hours is $102.50 at
either company. So the cost is the same at each golf
course for
hours.
A veterinarian needs 60 pounds of dog food
that is 15% protein. He will combine a beef
mix that is 18% protein with a bacon mix that
is 9% protein. How many pounds of each does
he need to make the 15% protein mixture?
Let x present the amount of beef mix in the mixture.
Let y present the amount of bacon mix in the mixture.
Example Continued
Write one equation based on the amount of dog food:
Amount of
beef mix
plus
amount of
bacon mix
equals
60.
+
y
=
60
x
Write another equation based on the amount of protein:
Protein of
beef mix
0.18x
plus
protein of
bacon mix
equals
protein in
mixture.
+
0.09y
=
0.15(60)
Example Continued
Solve the system.
x + y = 60
0.18x +0.09y = 9
x + y = 60
First equation
y = 60 – x
Solve the first equation for y.
0.18x + 0.09(60 – x) = 9
Substitute (60 – x) for y.
0.18x + 5.4 – 0.09x = 9 Distribute.
0.09x = 3.6 Simplify.
x = 40
Completing the Square
Complete the square for the expression. Write
the resulting expression as a binomial squared.
x2 – 14x +
Find
x2 – 14x + 49
(x – 7)2
Check
.
Add.
Factor.
Find the square of the binomial.
(x – 7)2 = (x – 7)(x – 7)
= x2 – 14x + 49
Complete the square for the expression. Write
the resulting expression as a binomial squared.
x2 + 9x +
Find
Add.
Factor.
.
Check Find the square
of the binomial.
Completing the Square – Steps (no decimals use improper fractions)
• 1. a must = 1, if not divide everything by a
• 2. Get the variables on one side and the constant on the other.
• 3. ADD a blank on both sides
• 4. Off to the side, take (b/2)2 (or you can multiply b by ½)2
• 5. Put the answer to step 4 in both blanks
• 6. Factor the LHS using shortcut and simplify the RHS
• 7. Square root both sides – don’t forget the + and –
• 8. If you know the square root, set up two equations and solve
for x. If not just solve for x.
Solving a Quadratic Equation by Completing the Square
Solve the equation by completing the square.
18x + 3x2 = 45
x2 + 6x = 15
x2 + 6x +
= 15 +
Divide both sides by 3.
Set up to complete the
square.
Add
x2 + 6x + 9 = 15 + 9
to both sides.
Simplify.
Example Continued
(x + 3)2 = 24
Factor.
Take the square root of
both sides.
Exact:
Approx: 1.9 and -7.9
Simplify.
Find the zeros of the function by completing the square.
g(x) = x2 + 4x + 12
x2 + 4x + 12 = 0
Set equal to 0.
x2 + 4x +
Rewrite.
= –12 +
x2 + 4x + 4 = –12 + 4
(x + 2)2 = –8
Add
to both sides.
Factor.
Take square roots.
Simplify.
Express the number in terms of i.
Factor out –1.
Product Property.
Product Property.
Simplify.
Express in terms of i.
Add or subtract. Write the result in the form a + bi.
(4 + 2i) + (–6 – 7i)
(4 – 6) + (2i – 7i)
–2 – 5i
Add real parts and imaginary
parts.
Multiply. Write the result in the form a + bi.
–2i(2 – 4i)
–4i + 8i2
Distribute.
–4i + 8(–1)
Use i2 = –1.
–8 – 4i
Write in a + bi form.
Multiply. Write the result in the form a + bi.
(3 + 6i)(4 – i)
12 + 24i – 3i – 6i2
Multiply.
12 + 21i – 6(–1)
Use i2 = –1.
18 + 21i
Write in a + bi form.
Dividing Complex Numbers
Simplify.
Multiply by the conjugate.
Distribute.
Use i2 = –1.
Simplify.
EX:
25(x – 2)2 = 9
(x – 2)2 =
x–2= 
9
25

9
25
3
x–2= 
5
3
3
x–2=
or x – 2 = –
5
5
EXACT: 13/5 and 7/5
Use the quadratic formula to find the real roots of quadratic equations.
x2 + 5x – 1 = 0
a = 1, b = 5, c = –1
–b  b2 – 4ac
2a
–5  52 – 4(1)(–1)
2(1)
x  0.19 or x  –5.19
–5  29
2
Solving Quadratics Using the Square Root Property
4x2 + 11 = 59
4x2 = 48
x2 = 12
Subtract 11 from both sides.
Divide both sides by 4 to isolate the
square term.
Take the square root of both sides.
Exact:
Approx: ±3.46
Simplify.
Example
Factor: x3 – 2x2 – 9x + 18.
(x3 – 2x2) + (–9x + 18)
x2(x – 2) – 9(x – 2)
Group terms.
Factor common monomials
from each group.
(x – 2)(x2 – 9)
Factor out the common
binomial (x – 2).
(x – 2)(x – 3)(x + 3)
Factor the difference of
squares.
Solve the polynomial equation by factoring.
x4 + 25 = 26x2
x4 – 26 x2 + 25 = 0
(x2 – 25)(x2 – 1) = 0
Set the equation equal to 0.
Factor the trinomial in
quadratic form.
(x – 5)(x + 5)(x – 1)(x + 1) Factor the difference of two
squares.
x – 5 = 0, x + 5 = 0, x – 1 = 0, or x + 1 =0
x = 5, x = –5, x = 1 or x = –1
The roots are 5, –5, 1, and –1.
Solve for x.
Solve Radical Equations
A. Solve
.
Isolate the radical.
You must check your answers
x – 5 = 0 or x + 1 = 0
x =5
x = –1
Solve Radical Equations
Solve
.
(x – 8)(x – 24) = 0
x – 8 = 0 or x – 24 = 0
Isolate a radical.
Solving Absolute-Value Equations
Solve the equation.
|–3 + k| = 10
This can be read as “the
distance from k to –3 is 10.”
–3 + k = 10 or –3 + k = –10
Rewrite the absolute
value as a
disjunction.
k = 13 or k = –7
Add 3 to both sides of
each equation.
An interval is the set of all numbers between
two endpoints, such as 3 and 5. In interval
notation the symbols [ and ] are used to
include an endpoint in an interval, and the
symbols ( and ) are used to exclude an
endpoint from an interval.
(3, 5)
-2
-1
The set of real numbers between but not
including 3 and 5.
0
1
2
3
4
5
3<x<5
6
7
8
Interval Notation
Use interval notation to represent the set of
numbers.
7 < x ≤ 12
(7, 12]
7 is not included, but 12 is.
Solve the compound inequality.
|2x +7| ≤ 3
Multiply both sides by 3.
2x + 7 ≤ 3 and 2x + 7 ≥ –3
Rewrite the absolute
value as a conjunction.
2x ≤ –4 and
x ≤ –2 and
2x ≥ –10
Subtract 7 from both
sides of each inequality.
x ≥ –5
Divide both sides of
each inequality by 2.
Solve linear inequalities in one variable.
32  5x  10
–5x  –22
–22
x
–5
22
x
5
Subtract 32.
Divide by –5;
reverse inequality sign.
Solve a Polynomial Inequality
Solve
x2 – 8x + 15 ≤ 0
(x – 3)(x – 5) ≤ 0
Solve a Polynomial Inequality
f(x) = (x – 5)(x – 3)
Think: (x – 5)
and (x – 3) are
both negative
when x = –2.
f(x) = (x – 5)(x – 3)
Solve a Polynomial Inequality
Example 1
Answer:
[3, 5]