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the Section 5.8 homework?
Pass your worksheets for this
assignment to the middle
aisle for pickup now.
Remember the problem like this one
from the homework that was due today?
Wouldn’t it be nice if there was an
easier way to do it than by factoring?
Leave factoring up on board: (4x - 9)(3x + 8)
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Sample
Problems
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(Dr. Bruce Johnston)
Section 8.2
The Quadratic Formula
The Quadratic Formula
The quadratic formula is another technique we
can use for solving quadratic equations.
Remember, quadratic equations are polynomial
equations of degree 2, such as
x2 + 3x -7 = 0
or
5x2 – 14 = 0.
The quadratic formula is derived from a process
called “completing the square” for a general
quadratic equation.
– See Section 8.1 if you’re interested in seeing how
this formula is derived.
– This will also be covered in Math 120 in more
detail, along with the technique called
“completing the square”.
The Quadratic Formula:
The solutions to the equation
ax2 + bx + c = 0
are given by the formula
 b  b  4ac
x
2a
2
Note: This formula IS on the pink formula sheet, but you’ll
probably have it memorized by the time you’ve done the
first few homework problems.
The Big Question:
How can we tell
when we should use
factoring and when we should
use the quadratic formula?
Example 1
Solve x2 + 4x + 3 = 0 by
• Factoring
• The quadratic formula.
Which way works best?
Solve x2 + 4x + 3 = 0 by Factoring:
This one is pretty easy to factor.
The factoring is (x + 3)(x + 1) = 0,
so the solutions are given by
x + 3 = 0, or x = -3,
and x + 1 = 0, which gives x = -1.
Now, solve x2 + 4x + 3 = 0 by the quadratic formula:
a = 1, b = 4, c = 3, so the formula gives:
 4  4 2  4 1 3  4  16  12
x

2 1
2
4 4 42 42 2




 1
2
2
2
2
42 6
or

 3
2
2
Which way works best for this
problem?
In this case, the factoring method is much
quicker, although BOTH methods give the same
answer.
Example 2
Solve x2 + 5x + 12 = 0 by
• Factoring
• The quadratic formula.
Which way works best?
Solve x2 + 5x + 12 = 0 by Factoring:
This one looks pretty easy to factor, but when you
start trying to find two factors of 12 that add up to
5, nothing works.
(1+12=13, 2+6=8, 3+4=7).
What does this mean?
It means that the polynomial is PRIME, and there
are no rational solutions. (Remember, a rational
number is either an integer or a fraction.)
Solve x2 + 5x + 12 = 0 (continued):
• Let’s see what the quadratic formula gives in this case:
a = 1, b = 5, c = 12
so the formula gives:
 5  52  4 112  5  25  48  5   23
x


2 1
2
2
Notice that the number under the radical sign is negative,
which means there are no real answers. If the number under
the square root sign comes out to be positive but it’s not a
perfect square, this means the answer is a real number, but
is irrational because it can’t be simplified to remove the
radical. In either of these cases, we’d say the polynomial is
prime, and therefore has no rational roots.
So which way works best for solving
x2 + 5x + 12 = 0?
Either way works fine, but if you think a
polynomial is prime, a good way to check is by
calculating the discriminant (b2 – 4ac). If the
discriminant is either negative or not a perfect
square, then you know for sure that your
polynomial is prime and there are no rational
solutions.
Now re-do this problem from the 5.8
homework using the quadratic formula:
Answers: -8/3, 9/4
Which way works best in this case?
Either way works, but the quadratic formula
approach is probably going to be faster than
factoring for most people.
Moral of the story: For a quadratic equation
with a leading coefficient other than 1, it’s
probably going to be quicker to solve it using the
quadratic formula than it would be to factor the
polynomial.
Question: What if some coefficients in your
quadratic equation are fractions?
ANSWER: Clear them first by multiplying all terms by the LCD:
Solve
1
8
x2
+x–
5
2
= 0 by the quadratic formula.
x2 + 8x – 20 = 0 (multiply both sides by 8)
a = 1, b = 8, c = -20
 8  (8)  4(1)(20)  8  64  80  8  144
x



2
2
2(1)
2
 8  12  20
4

or ,  10 or 2
2
2
2
•
The expression under the radical sign in the
quadratic formula (b2 – 4ac) is called the
discriminant.
•
The discriminant will take on a value that is
positive, 0, or negative.
•
The value of the discriminant indicates two
distinct real solutions (if it’s positive), one real
solution (if it’s zero), or two complex, but not real
solutions (if it’s negative – a topic to be discussed
in Math 120).
The Discriminant and the Kinds of Solutions
to ax2 + bx +c = 0
Discriminant
b2 – 4ac
Kinds of solutions
to ax2 + bx + c = 0
b2 – 4ac > 0
Two unequal real solutions
(If b2 – 4ac is a perfect square,
the two solutions will be
rational numbers. If not, they’re
both irrational.)
b2 – 4ac = 0
Graph of
y = ax2 + bx + c
Two x-intercepts
One real solution
(a repeated solution)
(If b2 – 4ac is a perfect square,
the solution will be a rational
number. If not, it’s irrational.)
One x-intercept
b2 – 4ac < 0
No real solution;
two complex imaginary
solutions
No x-intercepts
Example
Use the discriminant to determine the number
and type of solutions for the following equation.
5 – 4x + 12x2 = 0
a = 12, b = -4, and c = 5
b2 – 4ac = (-4)2 – 4(12)(5)
= 16 – 240
= -224
Since the discriminant is negative, there are no
real solutions.
Question: What would this graph look like?
Example
Use the discriminant to determine the number
and type of solutions for the following equation.
25x2 - 4 = 0
a = 25, b = 0 (why?) , and c = -4
b2 – 4ac = (0)2 – 4(25)(-4) = 0 – -400 = 400
Since the discriminant is positive, there are two real
solutions.
(You could go on to show that the solutions are 2/5 and -2/5, either
by factoring or using the quadratic formula.)
Example
Use the discriminant to determine the number and
type of solutions for the following equation.
5 – 4x + 12x2 = 0
a = 12, b = -4, and c = 5
b2 – 4ac = (-4)2 – 4(12)(5) = 16 – 24 = -224
Since the discriminant is negative, there are no
real solutions.
Example
Use the discriminant to determine the number
and type of solutions for the following equation.
x2 – 8x + 16 = 0
a = 1, b = -8, and c = 16
b2 – 4ac = (-8)2 – 4(1)(16) = 64 – 64 = 0
Since the discriminant is zero, there is one real
solution. (You could go on to show that the solution is 4,
either by factoring or using the quadratic formula.)
Question: What would this graph look like?
How do you figure out the answers if the
discriminant is positive but not a perfect square?
2 possible approaches:
1. Exact answer: The exact answer will contain a radical, i.e. it
will be an irrational number. (More on this in Chapter 7...)
2. Approximate answer: Use your calculator to get an
approximate decimal answer.
REMINDER!!!
IMPORTANT NOTE:
Use the quadratic formula technique to solve all
problems in this homework assignment.
There are a couple of word problems at the
end of the assignment in which the online
learning aids will show factoring as the
solution method. You should use the quadratic
formula instead (and you will find it to be
easier and quicker than factoring.)
Reminder:
This homework assignment
on Section 8.2 is due
at the start of
next class period.
You may now
OPEN
your LAPTOPS
and begin working on the
homework assignment
(if there’s any time left...)
But remember, you can always work in the
JHSW 203 open lab after class (or before
tour next class session) if you want some help
on this homework.