Lesson Plan
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First Step
Applying the First Step
Second Step
Now You Try!
Part 2: Solving for x in a Quadratic Equation
Finding the Zeroes/Roots
Solve for x once again
Practice Problem
Quiz – Question #1
Question #2
Question #3
Question #4
Question #5
Question #6
Question #7
Question #8
 Lets
use the example x + 12 = 4x. In this
problem we are solving for the variable x.
 The
first thing we want to do is get all the
like terms together by putting all the
plain numbers on one side of the equal
sign, and put all the numbers with an x
attached to them on the other side of the
equal sign.

We will get all the like terms together through
adding and subtracting.

In our example of x + 12 = 5x we will subtract
the x from the left side of the equal sign to cross
out the x.

Then we subtract the x from the right side of the
equal sign as well.

Whatever we do to the left side of the equal
sign, we do the same operation to the right side.

We now have 12 = 4x after subtracting x from both
sides.

The next step is to get x alone in this equation.

We do this by dividing the right and left sides of the
equation by whatever number is multiplied by x.

In our example we would divide both sides by 4 to get
x = 3 because 4 divided by 4 equals x and 12 divided
by 4 equals 3.

x = 3 is our answer!
If 6x - 12 = 4x + 6 then solve for x.
 Click
on the correct answer!
a) x = 3
c) x = 9
b) x = -3
d) x = -6/10
 Go
to
Quiz:
 Go to
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Part 2 of the Lesson
Question #2
Question #3
Question #4
Question #5
Question #6
Question #7
Question #8

If you need help, go back to the Table of Contents to
find what you need to review.
Back
 Back
Quiz:
 Back
 Back
 Back
 Back
 Back
 Back
 Back
 Back

to You Try!
to Practice Problem
to
to
to
to
to
to
to
to
Question #1
Question #2
Question #3
Question #4
Question #5
Question #6
Question #7
Question #8

This is called finding the “zeroes” or “roots” of
the equation. In exact terms, we are finding
where the graph any given equation crosses the
x-axis.

Here we will work with the example:
x² + 5x + 6 = 0

First, we need to find 2 numbers that add to the
middle term in the equation and also multiply to
the third term in the equation.

In our example of x² + 5x - 6 = 0 these two
numbers are 6 and -1 because 6 + -1 = 5 (the
middle term) and 6 x -1 = -6 (the third term).

Now we insert the two numbers we found into
(x± ) (x± ) = 0

We just “unfoiled” the equation x² + 5x - 6 = 0
to be (x + 6) (x – 1) = 0

Now we simply solve for x treating the contents
of both parenthesis as separate equations
equaling zero.

Therefore, x + 6 = 0 and x – 1 = 0.

When we solve for x we get zeroes of x = -6 and
x = 1. That’s our answer! Now lets practice!
If x² + 5x + 6 = 0 then what are the zeroes of
this equation?
a) x = 2, x = 3
c) x = -2, x = -3
b) x = 6, x = -1
d) x = -6, x = 1
If 20x + 35 = 2 + 31x then solve for x.
a) x = 3
c) x = 37/51
b) x = 51/37
d) x = 1/3
If 50x + 200 = 300 + 25x then solve for x.
a) x = 20
c) x = 4
b) x = 1/4
d) x = 5
If x² + 3x - 28 = 0 then what are the zeroes of
this equation?
a) x = 4, x = 7
c) x = -4, x = 7
b) x = 4, x = -7
d) x = -4, x = -7
If x² - 8x + 12 = 0 then what are the zeroes of
this equation?
a) x = 6, x = 2
c) x = -6, x = 2
b) x = 6, x = -2
d) x = -6, x = -2
If 30x + 75 = 125 + 5x then solve for x.
a) x = 8
c) x = 200/35
b) x = 2
d) x = 50/35
If x² + 8x + 40 = 0 then what are the zeroes of
this equation?
a) x = -8, x = -5
c) x = 8, x = -1
b) x = 8, x = 5
d) x = -8, x = 1
If 8x + 7 – 4x = 27 – 12x + 5 then solve for x.
a) x = 25/39
c) x = -8/39
b) x = 2
d) x = 25/16
If x² + 8x - 48 = 0 then what are the zeroes of
this equation?
a) x = 12, x = 4
c) x = 12, x = -4
b) x = -12, x = -4
d) x = -12, x = 4
CONGRATULATIONS!!!
YOU FINISHED THE
LESSON!!!