Solving
Quadratic Equations:
Case: x2
+ bx + c = 0, c < 0
Previous
Homework:
Chapter 8,
Section 4: problems 1-20 even on solving quadratic equations by factoring when
c > 0.
I. Review:
Go
over homework problems and use these problems to give a final review of
c > 0 case
before we move on to new case.
Write on
board: ÒSummary: Problems of the form:
x2
+ bx + c = 0, when c > 0.
This will lead
to both terms in the factoring having the same sign: both positive when b is
positive, both negative when b is negative. Ò
Teacher does
the following two examples from homework on the board.
a. x2
+ 5x + 6 = 0
Here c = 6
which is positive so both factors will be same sign.
Since b = 5 is
positive, both factors will be positive.
List the
possible positive factors of 6.
Positive
Factors of 6: 6 and 1 , 3 and 2, 2 and 3, 1 and 6.
Since factors
have the same sign, order does not matter in the factoring.
Which of these
factors will let the middle term add up to 5?
(x + 3) (x +
2) = 0 or (x + 1) (x + 6) =0.
Use FOIL to
check. Must be
(x + 3) (x +
2) = 0
x + 3 = 0 or x
+ 2 = 0
x = - 3 or x =
-2 are solutions
Ask students
if they have any questions.
b. x2 - 7x + 12 = 0.
Here c = 12
which is positive so both factors will be same sign.
Since b = -7,
both factors will be negative.
List the
possible negative factors of 12.
Since factors
have the same sign, order does not matter in the factoring.
-12 and -1, -6
and -2, -4 and -3.
x2 - 7x + 12 = ( x - ) ( x - ) =0
Do FOIL in
your head to decide which factors to choose:
( x - 4) (x -
3) = 0
x - 4 = 0 or
x
- 3 = 0
x = 4 or
x
= 3.
Ask students
if they have any questions.
Have students
had in their homework to teacher.
II. New
Material: Write on
board: ÒSolving x2 + bx + c = 0 when c < 0 Ò
Ex. x2 + x - 12 = 0
Note that c =
-12 < 0 so we are in a new situation here.
Write on
board: ÒRule: When c < 0, then one factor must be positive and one factor
must be negative and the order
matters. Ò
Factors for
-12 (order matters so list them all):
-12 and 1, 12 and -1, - 6 and 2, 6 and
-2, - 4 and 3, 4 and 3.
Check all
cases:
(x -12) (x +
1) (x
+ 12) (x -1)
(x - 6) (x +
2) (x
+ 6) (x - 2)
(x - 4) (x +
3) (x
+ 3) (x - 4)
In all cases,
see that FOIL gives - 12. But the
middle term does not always work out.
(x - 12) (x +
1) = x2 - 11x - 12 = 0
(x + 12) (x -
1) = x2 +11x - 12 = 0
(x - 6) (x + 2) = x2 - 4x - 12 = 0
(x + 6) (x
- 2) = x2 +4x - 12 = 0
(x - 4) (x + 3) = x2 - x - 12 = 0
(x + 4) (x
- 3) = x2 +x - 12 = 0
Finally see
which case works out.
Short cut: +4
- 3 = +1
Summary: To
solve: x2 + bx + c =0 when c < 0,
look for
factors of c of opposite sign such that:
(first factor) + (second factor) = b.
Problem for
students to do individually: Using this example on the board as a model, solve
x2
+ x - 110 =0.
While students
are doing this, the teacher works at his desk, going over their homework they
have just handed in. He checks whether their final answer is correct or not;
puts an X next to the problems that do not have the correct answer, puts their
total score on at the top of the paper, enters the scores in his grade
book. When done, will give the
papers back to students. If time
permits, teacher walks around room seeing what students are doing, making sure
they are not talking or looking at one anotherÕs work.
When students
are done problem, teacher works through the problem on the board showing
carefully how its done. Then asks students to raise their hand if they did it
correctly.
Assigns homework
on this new topic: Chapter 3, Section 5, problems 8 - 24 even. Students can use the remaining time in class to
start working on the homework. Teacher is at his desk and students can come up
and ask questions. Teacher continues to grade homework if not yet done.