Algebra I
Module 3
Lesson 9: Solving Quadratic Equations
Name
Period
Date
Day #1
In order to solve a quadratic equation you must…
1.) Make sure the equation is written in standard
form and set equal to zero.
2.) Factor
3.) Set each term equal to zero
4.) Solve
1.)
x2 – 11x + 24 = 0
2.)
x2 + x – 20 = 0
3.)
x2 – 25 = 0
4.)
x2 – 5x – 10 = 4
5.)
x2 + 4x = 12
6.)
x2 + 5x = 24
7.)
x2 + 5x – 35 = 3x
8.)
x2 – 3x + 11 = 26 – 5x
CLASSWORK: Factor and solve the following equations.
1.) x2 + 8x + 12 = 0
2.) x2 – 5x – 14 = 0
3.)
x2 + 8x = 20
4.)
x2 – x = 20
5.)
x2 = 36
6.)
x2 + 21x = 100
7.)
x2 – 48 = 2x
8.)
x2 = 100
Day #2
Important things to remember!
 Your equation must be written in
and set equal to
before you factor.
 When you factor, you must ALWAYS look for a
first. Trust me. It
makes factoring SO MUCH EASIER if you do.
 Another name for the solutions to a quadratic equation are the
or
the
or the
of the equation.
Solve the following equations.
1.) 2x2 + 18x + 40 = 0
2.)
3x2 + 9x – 54 = 0
3.)
5x2 – 15x – 20 = 0
4.)
3x2 – 75 = 0
5.)
3x2 + 16x + 5 = 0
6.)
2x2 + 7x + 5 = 0
7.)
4x2 – 12x = 7
8.)
x(x + 7) = 30
9.)
4x2 – 100 = 0
10.) (x – 3)2 = 4x – 12
CLASSWORK: Solve the following equations.
1.) 4x2 + 12x – 40 = 0
2.)
2x2 – 7x – 15 = 0
3.)
4.)
(x + 4)2 = 5x + 26
x(x + 8) = 20
Day #3
Opening Question
Is x = 11 the solution to 3x – 7 = 26? Explain how you determined
your answer.
If a number is the solution to an equation, that means that, when you substitute the value
into the equation, it makes the equation
. So you can either
the equation and see if your answers match, OR you can
in the
values and see if it makes the equation true... OR there is one more way…
Ex #1:
(1)
(2)
(3)
(4)
The zeros of the function f(x) = x2 – 10x – 24 are
-6 and 4
6 and -4
12 and -2
-12 and 2
You can use your calculator to find zeros of a given function (as long as the zeros are
integers). Plug the function into Y1 in your calculator. Then go into the table of values and
look to see what values make the equation equal to zero. Give it a try!
1.)
Keith determines the zeros of the function f(x) to be -6 and 5. What could be Keith’s
function?
(a)
(b)
(c)
(d)
2.)
f(x) = (x + 5)(x + 6)
f(x) = (x + 5)(x – 6)
f(x) = (x – 5)(x – 6)
f(x) = (x – 5)(x + 6)
What are the zeros of the function f(x) = x2 – 13x – 30?
(a)
(b)
(c)
(d)
-10 and 3
10 and -3
-15 and 2
15 and -2
3.)
The zeros of the function f(x) = 3x2 – 3x – 6 are
(a)
(b)
(c)
(d)
4.)
The zeros of the function f(x) = 2x2 – 4x – 6 are
(a)
(b)
(c)
(d)
5.)
-1 and -2
1 and -2
1 and 2
-1 and 2
3 and -1
3 and 1
-3 and 1
-3 and -1
John and Sarah are each saving money for a car. The total amount of money John
will save is given by the function f(x) = 60 + 5x. The total amount of money Sarah
will save is given by the function g(x) = x2 + 46. After how many weeks, x, will they
have the same amount of money saved? Explain how you arrived at your answer.
6.)
Write an equation that defines m(x) as a trinomial where
m(x) = (3x – 1)(3 – x) + 4x2 + 19.
Solve for x when m(x) = 0.
Day #4
Last day of Regents questions dealing with solving quadratic equations. These came up
real recently, so get used to them!
1.)
2.)
What are the solutions to the equation 3x2 + 10x = 8?
(1)
2
(2)
−
2
(3)
4
and -2
(4)
−
3
and -4
3
4
3
and 4
and 2
What is the solution of the equation 2(x + 2)2 – 4 = 28?
(1)
(2)
(3)
(4)
3.)
3
6, only
2, only
2 and -6
6 and -2
Find the zeros of f(x) = (x – 3)2 – 49, algebraically
4.)
Solve the equation for y:
5.)
Amy solved the equation 2x2 + 5x – 42 = 0. She stated that the solutions to the
7
equation were and -6. Do you agree with Amy’s solutions? Explain why or why not.
2
(y – 3)2 = 4y – 12