Quadratic Techniques to Solve
Polynomial Equations
CCSS: F.IF.4 ; A.APR.3
CCSS: F.IF.4
• For a function that models a relationship between
two quantities, interpret key features of graphs and
tables in terms of the quantities, and sketch graphs
showing key features given a verbal description of
the relationship. Key features include: intercepts;
intervals where the function is increasing,
decreasing, positive, or negative; relative
maximums and minimums.
CCSS: A.APR.3
• Identify zeros of polynomials when suitable
factorizations are available, and use the
zeros to construct a rough graph of the
function defined by the polynomial.
Standards for
Mathematical Practice
• 1. Make sense of problems and persevere in solving them.
• 2. Reason abstractly and quantitatively.
• 3. Construct viable arguments and critique the reasoning
of others.
• 4. Model with mathematics.
• 5. Use appropriate tools strategically.
• 6. Attend to precision.
• 7. Look for and make use of structure.
• 8. Look for and express regularity in repeated reasoning.
Essential Question:
• How do we use quadratic techniques
solving equations?
Objectives
• Solve third and fourth degree equations that
contain quadratic factors, and
• Solve other non-quadratic equations that
can be written in quadratic form.
Intro
• Some equations are not quadratic but can be
written in a form that resembles a quadratic
equation. For example, the equation x4 –
20x2 + 64 = 0 can be written as (x2)2 – 20x2
+ 64 = 0. Equations that can be written this
way are said to be equations in quadratic
form.
Key Concept:
• An expression that is quadratic form can be
written as:
• au² +bu + c for any numbers a, b, and c,
a≠0, where u is some expression in x.
• The expression au² +bu + c is called
quadratic form of the original expression.
Once an equation is written in quadratic form,
it can be solved by the methods you have
already learned to use for solving quadratic
equations.
Ex. 1: Solve x4 – 13x2 + 36 = 0
x 4  13x 2 36  0
( x 2 ) 2  13( x 2 )  36  0
( x 2  9)( x 2  4)  0
( x  3)( x  3)( x  2)( x  2)  0
x3 0
x  3
x 3  0
x3
x20
x  2
x2 0
x2
The solutions or roots are -3, 3, -2, and 2.
The graph of x4 – 13x2 + 36 = 0
looks like:
The graph of y = x4 – 13x2
+ 36 crosses the x-axis 4
times. There will be 4 real
solutions.
• Recall that (am)n = amn for any positive
number a and any rational numbers n and
m. This property of exponents that you
learned in chapter 5 is often used when
solving equations.
Ex. 2: Solve
1
2
1
2
1
4
x  6x  8  0
1
4
x  6x  8  0
1
4 2
1
4
( x )  6( x )  8  0
1
4
1
4
( x  2)( x  4)  0
1
check
4
check
1
2
1
4
x  6x  8  0
1
2
1
4
16  6(16 )  8  0
4  6( 2)  8  0 ?
4  12  8  0 ?
12  12  0
00
1
2
1
4
x 2 0
( x  4)  0
1
4
x  6x  8  0
1
4
2 )8  0
256 x 6(256
16  6(41)  8  0 ?
1
2
1
4
4 04?
24 (24
x ) 
00
x  16
(2)
4
1
4
x 4
1
4 4
( x )  44
x  256
2
3
Ex. 3: Solve t  16
2
3
t  16
2
3 3
(t )  (16) 3
t 2  163 or 4 6
1
6 2
t  (4 )
t  43 or  64
check
2
3
t  16
2
3
check
2
3
t  16
2
3
64  16 ?
(64)  16 ?
(3 64 ) 2  16 ?
(3  64 ) 2  16 ?
4 2  16 ?
16  16
(4) 2  16 ?
16  16
Ex. 4: Solve
x 7 x 8  0
x 7 x 8  0
( x ) 2  7( x )  8  0
 b  b 2  4ac
x
2a
 (7 (7) 2  4(1)( 8)
x
2(1)
7  81

2
7  9 16


8
2
2
79 2


 1
2
2
x 8
( x)  8
x  64
2
2
Ex. 4: Solve x  7 x  8  0
check
x 7 x 8  0
64  7 64  8  0
64  7(8)  8  0
64  56  8  0
00
There is no real number x such that is = - x  1
1.Since principal root of a number can not be
negative, -1 is not a solution. The only solution
would be 64.
• Some cubic equations can be solved using
the quadratic formula. First a binomial
factor must be found.
Ex. 5: Solve
x  27  0
3
x  27  0
x 2  3x  9  0
( x  3)( x  3 x  9)
x 3  0
 b  b 2  4ac

2a
3
2
x3
 3  32  4(1)(9)

2(1)
 3  9  36

2
 3   27

2
 3  3i 3

2
Quadratic form
ax2 + bx + c = 0 2x2 – 3x – 5 = 0
This also would be a quadratic form of an
equation 2 y 4  3 y 2  5  0
or
2x  3 x  5  0
or
2
3
1
3
2 x  3x  5  0
How would you solve
2 y  3y  5  0
or
4
2
2x  3 x  5  0
or
2
3
1
3
2 x  3x  5  0
Use Substitution
How would you solve
2y  3y  5  0
4
2
2u  3u  5  0
2
Use Substitution
Let u = y2
u2 = y4
How would you solve
2u  3u  5  0
2
2u  5u  1  0
5
u  ; u  1
2
Use Substitution
Let u = y2
u2 = y4
How would you solve
5
u  ; u  1
5
2
2
2
Then y  ; y  1 Use Substitution
2
Let u = y2
2 = y4
u
5
y   ; y   1
2
10
y
; y  i
2
How would you solve
Then
5
u  ; u  1
2
5
x ;
2
x  1
25
x  ; x 1
4
Use Substitution
Let u  x
u2  x
Do both answers
work?
How would you solve
5
u  ; u  1
2
Then 2 x  3 x  5  0 Use Substitution
Let u  x
2
 25 
 25 
u
x
2   3    5 
 4
 4
Do both answers
25  5 
work?
 3   5 
2
2
25 15 10
  0
2 2 2
21  3 1  5  0
How would you solve
Then
1
3
1
3
5
x 
2
3
  5
x   
  2
1
3
125
x
8
x  1
3
3
Use Substitution
Let u  x
 13 
3
 x    1
 
x  1
5
u  ; u  1
2
1
3
u x
2
2
3
How would you solve
5
u  ; u  1
2
Then
Use Substitution
2
1
Let
u x
 125  3  125  3
2
  3
 5  0
 8 
 8 
u x
Do both answers
work?
2
1
2(1) 3  3(1) 3  5  0
1
3
2
2
3