Circles

Standard form:

(x – h)2 + (y – k)2 = r2

center: (h, k)

radius: r

Write the equation of the circle.

Writing the Equation of a Circle

(x – 0)2 + (y – 6)2 = 12

x2 + (y – 6)2 = 1

the circle with center (0, 6) and radius r = 1

(x – h)2 + (y – k)2 = r2

Find the zeros of the function by factoring.

Finding Zeros or x-intercepts 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

Solve the equation x – = 3.

Solving Rational Equations

18x

x(x) – (x) = 3(x)18x Multiply each term by the LCD, x.

x2 – 18 = 3x Simplify. Note that x ≠ 0.

x2 – 3x – 18 = 0 Write in standard form.

(x – 6)(x + 3) = 0 Factor.

x – 6 = 0 or x + 3 = 0 Apply the Zero Product Property.

x = 6 or x = –3 Solve for x.

Multiply each term by the LCD, 4x.

24 + 5x = –7x Simplify. Note that x ≠ 0.

24 = –12x Combine like terms.

x = –2 Solve for x.

Solve the equation + = – . 54

6x

74

(4x) + (4x) = – (4x)6x

54

74

Solve each equation.

The solution x = 2 is extraneous because it makes the denominators of the original equation equal to 0. Therefore, the equation has no solution.

Divide out common factors.

Multiply each term by the LCD, x – 2.

Simplify. Note that x ≠ 2.

5x x – 2

3x + 4 x – 2

=

5x x – 2

3x + 4 x – 2

(x – 2) = (x – 2)

5x x – 2

3x + 4 x – 2

(x – 2) = (x – 2)

5x = 3x + 4

x = 2 Solve for x.

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?

Summer Sports Application

Example Continued

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.

Step 1 Write an equation for the cost of renting clubsand a cart at each golf course.

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

x y

60.

60+ =

Write another equation based on the amount of protein:

Protein of beef mix

plus protein of bacon mix

equals

0.18x 0.09y

protein in mixture.

0.15(60)+ =

Solve the system.x + y = 60

0.18x +0.09y = 9

x + y = 60

y = 60 – x

First equation

0.18x + 0.09(60 – x) = 9

0.18x + 5.4 – 0.09x = 9

0.09x = 3.6

x = 40

Solve the first equation for y.

Substitute (60 – x) for y.

Distribute.

Simplify.

Example Continued

x2 – 14x +

Complete the square for the expression. Write the resulting expression as a binomial squared.

Completing the Square

Add.

Factor.

Find .

x2 – 14x + 49

(x – 7)2

Check Find the square of the binomial.

(x – 7)2 = (x – 7)(x – 7)

= x2 – 14x + 49

Add.

Factor.

x2 + 9x +

Find .Check Find the square of the binomial.

Complete the square for the expression. Write the resulting expression as a binomial squared.

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.

Solve the equation by completing the square.

Solving a Quadratic Equation by Completing the Square

18x + 3x2 = 45

x2 + 6x = 15 Divide both sides by 3.

Simplify.

x2 + 6x + = 15 +

Add to both sides.

x2 + 6x + 9 = 15 + 9

Set up to complete the square.

Example Continued

Take the square root of both sides.

Factor.

Simplify.

(x + 3)2 = 24

Exact:

Approx: 1.9 and -7.9

Find the zeros of the function by completing the square.

Add to both sides.

g(x) = x2 + 4x + 12

x2 + 4x + 12 = 0

x2 + 4x + = –12 +

x2 + 4x + 4 = –12 + 4

Rewrite.

Set equal to 0.

Take square roots.

Simplify.

Factor.(x + 2)2 = –8

Express the number in terms of i.

Factor out –1.

Product Property.

Simplify.

Express in terms of i.

Product Property.

Add or subtract. Write the result in the form a + bi.

(4 + 2i) + (–6 – 7i)

Add real parts and imaginary parts.

(4 – 6) + (2i – 7i)

–2 – 5i

Multiply. Write the result in the form a + bi.

–2i(2 – 4i)

Distribute.

Write in a + bi form.

Use i2 = –1.

–4i + 8i2

–4i + 8(–1)

–8 – 4i

Multiply. Write the result in the form a + bi.

(3 + 6i)(4 – i)

Multiply.

Write in a + bi form.

Use i2 = –1.

12 + 24i – 3i – 6i2

12 + 21i – 6(–1)

18 + 21i

Simplify.

Multiply by the conjugate.

Distribute.

Dividing Complex Numbers

Simplify.

Use i2 = –1.

25(x – 2)2 = 9

(x – 2)2 =9

25

x – 2 = 9

25

3

5x – 2

=

EX:

or3

5x – 2 =

3

5–x – 2

=

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

x 0.19 or x –5.19

–b b2 – 4ac2a

2(1)

–5 52 – 4(1)(–1)

2–5 29

Solving Quadratics Using the Square Root Property

Subtract 11 from both sides.

4x2 + 11 = 59

Divide both sides by 4 to isolate the square term.

Take the square root of both sides.

Simplify.

x2 = 12

4x2 = 48

Approx: ±3.46

Exact:

Example

Factor: x3 – 2x2 – 9x + 18.

Group terms.(x3 – 2x2) + (–9x + 18)

Factor common monomials from each group.

x2(x – 2) – 9(x – 2)

Factor out the common binomial (x – 2).

(x – 2)(x2 – 9)

Factor the difference of squares.

(x – 2)(x – 3)(x + 3)

Solve the polynomial equation by factoring.

x4 + 25 = 26x2

Set the equation equal to 0.x4 – 26 x2 + 25 = 0

Factor the trinomial in quadratic form.

(x2 – 25)(x2 – 1) = 0

Factor the difference of two squares.

(x – 5)(x + 5)(x – 1)(x + 1)

Solve for x.

x – 5 = 0, x + 5 = 0, x – 1 = 0, or x + 1 =0

The roots are 5, –5, 1, and –1.

x = 5, x = –5, x = 1 or x = –1

Solve Radical Equations

A. Solve .Isolate the radical.

x – 5 = 0 or x + 1 = 0

x = 5 x = –1

You must check your answers

Solve Radical Equations

Solve .Isolate a radical.

(x – 8)(x – 24) = 0x – 8 = 0 or x – 24 = 0

Solve the equation.

Solving Absolute-Value Equations

Rewrite the absolute value as a disjunction.

This can be read as “the distance from k to –3 is 10.”

Add 3 to both sides of each equation.

|–3 + k| = 10

–3 + k = 10 or –3 + k = –10

k = 13 or k = –7

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) The set of real numbers between but not including 3 and 5.

-2 -1 0 1 2 3 4 5 6 7 8

3 < x < 5

Use interval notation to represent the set of numbers.

7 < x ≤ 12

(7, 12]

Interval Notation

7 is not included, but 12 is.

Solve the compound inequality.

|2x +7| ≤ 3 Multiply both sides by 3.

Subtract 7 from both sides of each inequality.

Divide both sides of each inequality by 2.

Rewrite the absolute value as a conjunction.

2x + 7 ≤ 3 and 2x + 7 ≥ –3

2x ≤ –4 and 2x ≥ –10

x ≤ –2 and x ≥ –5

Solve linear inequalities in one variable.

5

22x

–5x –22 Subtract 32.

Divide by –5;reverse inequality sign.

x –22

–5

32 5x 10

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) areboth negative when x = –2.

f (x) = (x – 5)(x – 3)

Example 1

Answer: [3, 5]

Solve a Polynomial Inequality