Download Solving Equations That Are Quadratic in Form

11.3
Solving Equations That Are Quadratic in
Form
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.
Solve:
Fractions
LCD
Clear Fractions
1
10
1
 2
 
y  5 y  25
5
Restrictions
1
10
1

 
y  5 y  5y  5
5
LCD: 5(y – 5)(y + 5)
R: y ≠ 5, y ≠ -5
10


 1 
 1
5y  5 y  5  
  5y  5 y  5    
  5y  5 y  5  
 5
 y  5
 y  5 y  5  
5y  5  510   1y  5y  5

5y  25  50   1 y 2  25
5y  25   y 2  25
y 2  5y  50  0
y
 10 y  5   0
y  10
y 5

 10
Solve:
Square Root
Isolate
Check solutions
3x  4 x  4  0
3 x  4  4 x
3 x  42
  4 x 
2
9 x 2  24 x  16  16 x
9 x 2  40x  16  0
9 x  4x
 4  0
4
x 
9
x  4
4
 
9
Check:
x 
4
9
4
4
3   4
4 0
9
9
4
 2  12
 4  
0
3
3 3
4 8 12
 
0
3 3 3
Check:
True
x 4
34   4 4  4  0
12  42  4  0
12  8  4  0
False
Solve:
Fractions
LCD
Clear Fractions
4
8
9 
m 1
m
Restrictions
4
8
9 
m 1
m
LCD: m (m – 1)
R: m ≠ 0, m ≠ 1
 4 
  8

m m  1  9m m  1  
m m  1
 m  1
 m 
4m  9mm  1  8m  1
4m  9m2  9m  8m  8
9m2  5m  8m  8
9m2  3m  8  0
9m2  3m  8  0
a = 9, b = 3, c = -8
m 
 3 
32  49 8
29
 3  9  288
m 
18
 3  297
m 
18
1
9*33
1
 3  3 33
m 
18 6
 1  33
m 
6
  1  33  1  33 
,


6
6


Solve:
1  5x
1
 6x
5
6
1  2  0
x x
2
0
Fractions
LCD
Clear Fractions
LCD: x2
Restrictions
R: x ≠ 0
x
5
2  6 
x 1  x    x  2   x 2 0
 x1
x 
2
2
x 2  5x  6  0
x
 2x  3   0
x  2
x  3
 3,  2
Solve the equation. 8x  7  x  4  0
a) 9i
b) 3i
c) 3i
d) 3
11.3
Copyright © 2011 Pearson Education, Inc.
Slide 11- 8
Solve the equation. 8x  7  x  4  0
a) 9i
b) 3i
c) 3i
d) 3
11.3
Copyright © 2011 Pearson Education, Inc.
Slide 11- 9
Solve the equation.
8
8

3
x2 x2
a) 2, 6
b)
c)
3
 , 6
2
2
 ,6
3
d) 2 , 6
3
11.3
Copyright © 2011 Pearson Education, Inc.
Slide 11- 10
Solve the equation.
8
8

3
x2 x2
a) 2, 6
b)
c)
3
 , 6
2
2
 ,6
3
d) 2 , 6
3
11.3
Copyright © 2011 Pearson Education, Inc.
Slide 11- 11
Solve:
x  32
 2x  3   15  0
Repeated binomial
Substitution
Goal: x =
Let u = (x – 3)
1st Substitution
u 2  2u  15  0
Make a simpler problem
u  5u  3 
2nd
Substitution
Find the final solution.
u  5
0
u 3
x–3= –5
x–3=3
x =–2
x=6
 2, 6 
Solve:
x

2


Repeated binomial
Substitution
Goal: x =
 2x  2 x  2x  3
Let u = (x2 – 2x)
1st Substitution
u 2  2u  3
u 2  2u  3  0
1  2 ,1  2 ,1  i

u  3u  1  0
2nd Substitution
u  3 u  1
x 2  2 x  3
x2  2 x  1
x2  2 x  3  0
x2  2 x  1  0
2
2
2 ,1  i 2
  2    2 2  413
x
21
2 8
x
2
2  2i 2
x
2
  2   2 2  41 1
x
21
2 8
x
2
22 2
x
2
x  1 i 2
x  1 2
Solve:
x 4  13 x 2  36  0
(x2)2
4 solutions
First exponent is double
the exponent of the
middle term
Let u = x2
u 2  13u  36  0
1st Substitution
u  4u  9 
0
u  4 u 9
2nd Substitution
x2  4
x2  9
x2   4
x2   9
x  2
x  3
 3,  2, 2, 3 
Solve the equation. x 4  7 x 2  12  0
a) 2,  2 3
b)
2,  3
c)
2,  3
d) 2, 3
11.3
Copyright © 2011 Pearson Education, Inc.
Slide 11- 15
Solve the equation. x 4  7 x 2  12  0
a) 2,  2 3
b)
2,  3
c)
2,  3
d) 2, 3
11.3
Copyright © 2011 Pearson Education, Inc.
Slide 11- 16