Chabot Mathematics
§8.4 Eqns w/
Quadratic Form
Bruce Mayer, PE
Licensed Electrical & Mechanical Engineer
BMayer@ChabotCollege.edu
Chabot College Mathematics
1
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-53_Fa08_sec_8-4_Eqns_Quadratic_in_Form.ppt
Review § 8.3
MTH 55
 Any QUESTIONS About
• §8.3 → Quadratic Fcn Applications
 Any QUESTIONS About HomeWork
• §8.3 → HW-39
Chabot College Mathematics
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-53_Fa08_sec_8-4_Eqns_Quadratic_in_Form.ppt
Recognizing Eqns in Quadratic
Form
 Certain equations that are not really
quadratic can be thought of in such
a way that they can be solved as
quadratic. For example, because
the square of x2 is x4, the equation
x4 − 5x2 + 4 = 0 is said to be
“quadratic in x2”
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3
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-53_Fa08_sec_8-4_Eqns_Quadratic_in_Form.ppt
Quadratic Form
x4 – 5x2 + 4 = 0
 Continuing from
the previous Slide
 The last equation
(x2)2 – 5(x2) + 4 = 0
can be solved by
u2 – 5u + 4 = 0.
factoring or by the
quadratic formula. Then, remembering
that u = x2, we can solve for x
 Equations that can be solved like this
are reducible to, or are in, quadratic
form.
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-53_Fa08_sec_8-4_Eqns_Quadratic_in_Form.ppt
Example  Solve x4 – 5x2 + 4 = 0
 SOLUTION
 Let u = x2. Then we solve by
substituting u for x2 and u2 for x4:
u2 – 5u + 4 = 0
(u – 1)(u – 4) = 0
Factoring
u – 1 = 0 or u – 4 = 0
Principle of
zero products
u = 1 or u = 4
Chabot College Mathematics
5
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-53_Fa08_sec_8-4_Eqns_Quadratic_in_Form.ppt
Example  Solve x4 – 5x2 + 4 = 0
 Recall
u = x2
= 1 or = 4
x  1 or x  2.
x2
x2
Replace u with x2
 To check, note that for both x = 1 and
x = −1, we have x2 = 1 and x4 = 1.
Similarly, for both x = 2 and x = −2, we
have x2 = 4 and x4 = 16.
 Thus instead of making four checks, we
need make only two.
Chabot College Mathematics
6
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-53_Fa08_sec_8-4_Eqns_Quadratic_in_Form.ppt
Example  Solve x4 – 5x2 + 4 = 0
 CHECK:
x = 1:
x4 – 5x2 + 4 = 0
x = 2:
x4 – 5x2 + 4 = 0
(1) – 5(1) + 4 = 0
(16) – 5(4) + 4 = 0
TRUE
TRUE
 STATE:
The solutions are 1, −1, 2, and −2.
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-53_Fa08_sec_8-4_Eqns_Quadratic_in_Form.ppt
Quadratic Form Quandry
 Caution!
 A common error on problems like
the previous example is to solve
for u but forget to solve for x.
 Remember to solve for the
original variable!
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-53_Fa08_sec_8-4_Eqns_Quadratic_in_Form.ppt
Radical and Rational Equations
 Sometimes rational equations,
radical equations, or equations
containing exponents that are
fractions are reducible to quadratic.
 It is especially important that
answers to these equations be
CHECKED in the original equation.
Chabot College Mathematics
9
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-53_Fa08_sec_8-4_Eqns_Quadratic_in_Form.ppt
Example  Solve x  8 x  9  0.
 SOLUTION: Let u 
 Next Substitute: u for
x
2
x and u for x
u2 – 8u – 9 = 0
(u – 9)(u +1) = 0
u – 9 = 0 or u + 1 = 0
u = 9 or u = –1
Chabot College Mathematics
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-53_Fa08_sec_8-4_Eqns_Quadratic_in_Form.ppt
Example  Solve x  8 x  9  0.
 Then by the Principle of Zero Products
x  9 or x  1
x  81 or x 1
by
 CHECK
x = 81:
x = 1:
x 8 x 9  0
x 8 x 9  0
1 8 1  9  0
81  8 81  9  0
81  8(9)  9  0
81  72  9  0
Chabot College Mathematics
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 x   1
1 8  9  0
TRUE
FALSE
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-53_Fa08_sec_8-4_Eqns_Quadratic_in_Form.ppt
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2
To Solve an Equation That is
Reducible to Quadratic Form
1. The equation is quadratic in form
if the variable factor in one term is
the square of the variable factor in
the other variable term.
2. Note Carefully any substitutions
made
Chabot College Mathematics
12
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-53_Fa08_sec_8-4_Eqns_Quadratic_in_Form.ppt
To Solve an Equation That is
Reducible to Quadratic Form
3. Whenever you making a
substitution, remember to solve
for the variable that is used in the
original equation.
4. Check possible answers in the
ORIGINAL equation.
Chabot College Mathematics
13
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-53_Fa08_sec_8-4_Eqns_Quadratic_in_Form.ppt
WhiteBoard Work
 Problems From §8.4 Exercise Set
• 6, 12, 18, 24, 28, 48
 Lyngen kirke (Lyngen Church)
A steeple was added to the
west side of the church during
the reconstruction period in
1840-46. It is a majestic
steeple with a NeoGothic
character, and is
approximately
QUADRATIC IN FORM.
Chabot College Mathematics
14
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-53_Fa08_sec_8-4_Eqns_Quadratic_in_Form.ppt
All Done for Today
Prob48
“Cleans
Up”
Chabot College Mathematics
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-53_Fa08_sec_8-4_Eqns_Quadratic_in_Form.ppt
Chabot Mathematics
Appendix
r  s  r  s r  s 
2
2
Bruce Mayer, PE
Licensed Electrical & Mechanical Engineer
BMayer@ChabotCollege.edu
–
Chabot College Mathematics
16
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-53_Fa08_sec_8-4_Eqns_Quadratic_in_Form.ppt
Graph y = |x|
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 Make T-table
x
-6
-5
-4
-3
-2
-1
0
1
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5
6
Chabot College Mathematics
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y = |x |
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5
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0
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y
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x
0
-6
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file =XY_Plot_0211.xls
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-53_Fa08_sec_8-4_Eqns_Quadratic_in_Form.ppt
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y
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-10
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-1
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x
0
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M55_§JBerland_Graphs_0806.xls
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Chabot College Mathematics
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M55_§JBerland_Graphs_0806.xls
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-53_Fa08_sec_8-4_Eqns_Quadratic_in_Form.ppt
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