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Solving Quadratic
Equations
by Factoring
Zero Product Property
For any real numbers a and b,
if ab=0, then either
a=0, b=0, or both.
In this first example, the equation is
already factored and is set equal to zero.
To solve, simply set the individual factors
equal to zero.
x 32x 1 0
x 3 0 or 2x 1 0
x 3 or 2x 1
1
x 3 or x
2
The solutions are -3 and 1/2.
In this example, you must first factor
the equation. Notice the familiar
pattern. After factoring, set the
individual factors equal to zero.
2
9x 4 0
Factor using “difference of two squares.”
3x 23x 2 0
3x 2 0 or 3x 2 0
3x 2 or 3x 2
2
2
x or x
3
3
In the next example, you must set the
equation equal to zero before
factoring. Then set the individual
factors equal to zero and solve.
2
x 27 6x
2
x 6x 27 0
x 9x 3 0
x 9 0 or x 3 0
x 9 or x 3
Re-write this example in the proper
form. Notice that the leading coefficient
is not one. Use an appropriate factoring
technique. Then solve as you have
done before.
2
2x 3 5x
2
2x 5x 3 0
2x 1x 3 0
2x 1 0 or x 3 0
x
1
2
or x 3
This one uses a different technique
than the previous ones. Really, this is
something you should consider at the
beginning of every factoring problem.
See if you can solve it.
2
2x 8x 0
2x x 4 0
2x 0 or x 4 0
x 0 or x 4
Did you take out GCF?
Now, try several problems. Write these
on your own paper, showing all steps
carefully.
1. 3y 52y 7 0
2
2. x x 12
2
3. d 5d 0
4. 4c 2 25
5. 18u 1 3u
2
After completion,
click here.
Here are the answers.
For help, click on the numbers.
1.
2.
3.
4.
5.
y 5 3 or y 7 2
x 4 or x 3
d 0 or d 5
c 5 2 or c 5 2
u 1 6 or u 1 3
If all are correct,
you’re finished!
3y 52y 7 0
3y 5 0 or 2y 7 0
3y 5 or 2y 7
y 5 3 or y 7 2
Back to questions
2
x x 12
2
x x 12 0
x 4 x 3 0
x 4 0 or x 3 0
x 4 or x 3
Back to questions
d 5d 0
2
d d 5 0
d 0 or d 5 0
d 0 or d 5
Back to questions
4c2 25
4c 25 0
2
2c 52c 5 0
2c 5 0 or 2c 5 0
2c 5 or 2c 5
c 5 2 or c 5 2
Back to questions
2
18u 3u 1
2
18u 3u 1 0
6u 13u 1 0
6u 1 0 or 3u 1 0
6u 1or 3u 1
u 1 6 or u 1 3
Back to questions