Transcript No Slide Title

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  32x 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  23x  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  9x  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 1x  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  52y  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  52y  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
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d  5d  0
2
d d  5  0
d  0 or d  5  0
d  0 or d  5
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4c2  25
4c  25 0
2
2c  52c  5  0
2c  5  0 or 2c  5  0
2c  5 or 2c  5
c   5 2 or c  5 2
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2
18u  3u  1
2
18u  3u  1  0
6u  13u  1  0
6u  1  0 or 3u  1  0
6u  1or 3u  1
u  1 6 or u  1 3
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