**Chapter 11: Quadratic Equations, Inequalities and
Functions**

• Discuss different methods of solving quadratic equations

• Methods for graphing quadratic equations

• Applications of quadratic equations

**Section 11.1: Solving Quadratic Equations using the
Square Root Property
**• Quadratic equations come in the form of ax

• To be able to solve a quadratic equation, it must be in standard form

**Zero Factor Property
**• If two numbers have a product of zero, then at least one of the numbers
must be zero.

• We use this property to solve equations after they have been factored

• Once the equation is factored we set each factor equal to zero and then solve the

individual equations.

**Example:** Use the zero factor property to solve the
given quadratic equation.

2x^{2} − 3x = −1

**Square Root Property
**• If k is a positive number and if x

The solution can also be written as

• If the quadratic equation is of the form x

**Examples: **Solve each equation. Write radicals in
simplified form.

a) z ^{2} = 49

b) x^{2} = 12

c) 3x^{2} − 8 = 88

d) (p − 4)^{2} = 3

e) (5m +1)^{2} = 7

f) x^{2} = −17

g) (x + 5)^{2} = −100

**Section 11.2: Solving Quadratic Equations by Completing
the Square
**• A method for solving quadratic equations

• The goal is to factor the left side of the quadratic equation so that it is a perfect square

• The right side of the equation is a constant

• The square root property is the used to finish the solving of the equation

• We start with an equation in standard form ax

• The end result is of the form (x + k )

**Steps to Complete the Square
**1. Make sure a is 1. If a is not 1, then perform the proper division

2. Write in the form ax

3. Complete the square using the formula

4. Add the value found in step 3 to both sides of the equation

5. Factor the left side as a perfect square, simplify the right side

6. Use the square root property to solve

**Examples: **Solve the given equations using
completing the square.

a) x^{2} − 2x −10 = 0

b) x^{2} + 4x = 1

c) 4k^{2} − 24k +11 = 0

d) 5t ^{2} −15t +12 = 0

**Section 11.3: Solving Quadratic Equations by the Quadratic Formula
**• Another method for solving quadratic equations

• Before the quadratic formula can be used the quadratic equation must be in standard form

**Quadratic Formula
**

**Examples:** Use the quadratic formula to solve the
given quadratic equations.

a) 4x^{2} −11x − 3 = 0

b) 2x^{2} +19 = 14x

c) (x + 5)(x +1) = 10x

**Discriminant
**• A part of the quadratic formula; b

• The discriminant can be used to determine the number and type of solutions a quadratic

equation has

• The following table lays out the types of solutions

Discriminant |
Number and Type ofSolutions |

Positive, and the square of an integer | 2 rational solutions |

Positive, but not the square of an integer | 2 irrational solutions |

Zero | 1 rational solution |

Negative | 2 non-real complex solutions |

**Examples:** Find each discriminant. Use it to predict
the number and type of solutions for each

equation.

a) 10x^{2} − x − 2 = 0

b) 3x^{2} − x = 7

c) 16x^{2} + 25 = 40x

**Example: **Find k so that the equation will have
exactly one rational solution.

x^{2} − kx + 64 = 0

**Section 11.4: Equations in Quadratic Form
**• We can solve quadratic equations by 4 different methods

-Factoring

-Square Root Property

-Completing the Square

-Quadratic Formula

• Some equations can be simplified down to quadratic form

**Example: **Solve the given equations.

**Example:** Solve the given application problem using
the six step method discussed previously.

1. In 1 ¾ hours Sammy rows his boat 5 miles up the river and comes back. The
speed of the

current is 3 mph. How fast does Sammy row?

**Section 11.5: Formulas and Further Applications
**• This section illustrates the many uses of quadratic equations

• We also look at formulas

**Examples:** Solve the given formulas for the
specified variable.

a) A =πr ^{2} for r

c) 2t^{2} − 5t + k = 0 for t

• Recall the Pythagorean Relation, c^{2} = a^{2} + b^{2}

**Example:** A ladder is leaning against a house. The
distance from the bottom of the ladder to the

house is 5 ft. The distance from the top of the ladder to the ground is 1 ft
less than the length of

the ladder. How long is the ladder?

**Example:** A ball is projected upward from the
ground. Its distance in feet from the ground at t

seconds is s(t) = −16t^{2} + 64t .

a) At what time will the ball hit the ground?

b) At what time will the ball be 32 ft from the ground?

**Section 11.6: Graphs of Quadratic Functions
**• We now look at how to graph quadratic equations

• In this section we look at horizontal shifts and vertical shifts associated with parabolas

• We can also use a table of values to plot points and determine the graph of quadratic

equations

• A quadratic function is of the form f (x) = ax

• A quadratic function can also be in the form of f (x) = a(x − h)

• Functions of the form f (x) = x

k units down if k is negative

• Functions of the form f (x) = (x − h)

positive and h units to the left if h is negative

• Recall the ideas of domain and range; interval notation

• Recall the general shape of quadratic functions; f (x) = x

**General Principles
**1. Graph the quadratic function defined by f (x) = a(x − h)

(h, k)

2. The graph opens up if a is positive and down if a is negative

3. The graph is wider than that of f (x) = x

f (x) = x

shift, the vertex, how the graph opens and its width in terms of f (x) = x

a) f (x) = x

b) f (x) = −x

c) f (x) = x

d) f (x) = (x + 2)

e) f (x) = (x − 3)

f) f (x) = −(x −1)

g) f (x) = (x − 2)

h) f (x) = −3(x +1)

• To determine the vertex of the parabola, we use the following formula

a < 0the parabola opens down.

2. Find the vertex using the vertex formula.

3. Find the x and y intercepts. Recall that an x intercept is found when y is replaced with

zero and the equation is solved for x. A y intercept is found when x is replaced with zero

and the equation is solved for y; this usually involves factoring.

4. Graph the vertex and intercepts. Plot additional points as needed.

**Example: **Graph the given quadratic equation. Use
the general principles as described above.

Also, determine the domain and range.

f (x) = x^{2} − 6x + 5

• We can use the vertex formula to solve applications of quadratic equations

• Application problems which involve maximum or minimum areas require the use of
the

vertex formula

• Recall that the vertex of the parabola can be thought of as its maximum point
or its

minimum point which enables us to use the vertex formula when determining a

maximum or minimum value of a quadratic equation

**Example:** Use the vertex formula to solve the
following application problems.

1. A farmer has 100 ft of fencing. He wants to put a fence around the
rectangular field next

to a building. Find the maximum area he can enclose and the dimensions of the
field

when the area is maximized.

2. A toy rocket is launched from the ground so that its distance in feet above
the ground

after t seconds is s(t) = −16t ^{2} + 208t . Find the maximum height it
reaches and the

number of seconds it takes to reach that height.