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Solving Equations That Are Quadratic in Form11.311.3
1. Solve equations by rewriting them in quadratic form.2. Solve equations that are quadratic in form by using
substitution.3. Solve application problems using equations that are
quadratic in form.
51
25
105
12
yy
Solve:Solve:
51
5510
51
yyy
51
55555
10555
51
555 yyyy
yyy
yy
55110555 yyy
25150255 2 yy
05052 yy
FractionsLCDClear Fractions
LCD: 5(y – 5)(y + 5)
R: y ≠ 5, y ≠ -5
25255 2 yy
0510 yy
510 yy
10
Restrictions
Linear or
quadratic?
Solve:Solve: Square RootIsolateCheck solutions
Check:
Linear or
quadratic?
0443 xx
xx 443
22 443 xx
xxx 1616249 2
016409 2 xx
0449 xx
494 xx
Check:
94x
0494
494
3
True
4x
04812
042412
044443
False
94
03
1232
434
03
1238
34
18
1911
4
mm
mmmmm
m
mm8
91
4
Solve:Solve: FractionsLCDClear Fractions
LCD: m (m – 1)
R: m ≠ 0, m ≠ 1
Restrictions
Linear or
quadratic?
mm8
91
4
18194 mmmm
88994 2 mmmm
8859 2 mmm
0839 2 mm
183333 m
0839 2 mm
a = 9, b = 3, c = -8
92
89433 2 m
1828893 m
182973 m
9*33
6331 m
6331
,6
3311 1
6
065
1 22
222 xx
xx
xx
Solve:Solve: FractionsLCDClear Fractions
LCD: x2
R: x ≠ 0
Restrictions
Linear or
quadratic?
0651 21 xx
065
12
xx
0652 xx
032 xx
32 xx 2,3
x
1
Slide 11- 8Copyright © 2011 Pearson Education, Inc.
Solve the equation.
a) 9i
b) 3i
c) 3i
d) 3
8 7 4 0x x
11.3
Slide 11- 9Copyright © 2011 Pearson Education, Inc.
Solve the equation.
a) 9i
b) 3i
c) 3i
d) 3
8 7 4 0x x
11.3
Slide 11- 10Copyright © 2011 Pearson Education, Inc.
Solve the equation.
a) 2, 6
b)
c)
d)
8 83
2 2x x
3, 6
2
2,6
3
2, 6
3
11.3
Slide 11- 11Copyright © 2011 Pearson Education, Inc.
Solve the equation.
a) 2, 6
b)
c)
d)
8 83
2 2x x
3, 6
2
2,6
3
2, 6
3
11.3
Solve:Solve: 015323 2 xx
01522 uu
Let u = (x – 3)
1st Substitution
035 uu
35 uu
x – 3 = – 5 x – 3 = 32nd Substitution
x = – 2 x = 6
6,2
Repeated binomialSubstitutionGoal: x =
Make a simpler problem
Find the final solution.
Solve:Solve: 3222 222 xxxx
322 uuLet u = (x2 – 2x)
1st Substitution
013 uu13 uu2nd Substitution
Repeated binomial
SubstitutionGoal: x =
0322 uu
322 xx
0322 xx
1231422 2 x
122 xx
282 x
2222 i
x
21 ix
0122 xx
1211422 2 x
282 x
2222 x
21 x
21212121 ii ,,,
Solve:Solve:
Let u = x2
1st Substitution
2nd Substitution
03613 24 xx
036132 uu
94
094
uu
uu
94 22 xx
94 22 xx
32 xx
3,2,2,3
First exponent is double the exponent of the middle term
4 solutions
(x2)2
Slide 11- 15Copyright © 2011 Pearson Education, Inc.
Solve the equation.
a)
b)
c)
d)
4 27 12 0x x
2, 2 3
2, 3
2, 3
2, 3
11.3
Slide 11- 16Copyright © 2011 Pearson Education, Inc.
Solve the equation.
a)
b)
c)
d)
4 27 12 0x x
2, 2 3
2, 3
2, 3
2, 3
11.3
