Chabot Mathematics
§8.2 Quadratic
Equation
Bruce Mayer, PE
Licensed Electrical & Mechanical Engineer
BMayer@ChabotCollege.edu
Chabot College Mathematics
1
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-49_sec_8-2_Derive_Quadratic_Eqn.ppt
Review § 8.1
MTH 55
 Any QUESTIONS About
• §8.1 → Complete the Square
 Any QUESTIONS About HomeWork
• §8.1 → HW-37
Chabot College Mathematics
2
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-49_sec_8-2_Derive_Quadratic_Eqn.ppt
The Quadratic Formula
 The solutions of
ax2 + bx + c = 0
are given by
This is one of the
MOST FAMOUS
Formulas in all
of Mathematics
 b  b  4ac
x
2a
2
Chabot College Mathematics
3
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-49_sec_8-2_Derive_Quadratic_Eqn.ppt
§8.2 Quadratic Formula
 The Quadratic
Formula
 Problem Solving
with the
Quadratic
Formula
Chabot College Mathematics
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-49_sec_8-2_Derive_Quadratic_Eqn.ppt
Derive Quadratic Formula - 1
 Consider the
General
Quadratic Equation
 Next, Divide by “a” to
give the second degree
term the coefficient of 1
b
c
x  x
a
a
ax  bx  c  0
2
• Where a, b, c are
CONSTANTS
 Solve This Eqn for x by
Completing the Square
 First; isolate the Terms
involving x
ax  bx  c
2
Chabot College Mathematics
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2
 Now add to both Sides
of the eqn a “quadratic
supplement” of (b/2a)2
2
2
b
b a b a c
x  x
 
 
a
 2   2  a
2
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-49_sec_8-2_Derive_Quadratic_Eqn.ppt
Derive Quadratic Formula - 2
 Now the Left-Hand-Side  Combine Terms inside
(LHS) is a PERFECT
the Radical over a
Square
Common Denom
2
2
b
b a b a c
2
2
x  x
 
 
b
b
c
a
 2   2  a
x


2
2a
2
b   b  c

x    
2a   2a  a

 Take the Square Root
of Both Sides
2
b
 b  c
x
   
2a
 2a  a
Chabot College Mathematics
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4a 2
a
b
b 2 c 4a
x


2
2a
4a
a 4a
b
b 2  4ac
x

2a
4a 2
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-49_sec_8-2_Derive_Quadratic_Eqn.ppt
Derive Quadratic Formula - 4
 Note that Denom is,
itself, a PERFECT SQ
b
b 2  4ac
x

2a
4a 2
b
b 2  4ac
x

2a
2a
 Next, Isolate x
b
b 2  4ac
x 
2a
2a
Chabot College Mathematics
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 Now Combine over
Common Denom
 b  b  4ac
x
2a
2
 But this the Renowned
QUADRATIC FORMULA
 Note That it was
DERIVED by
COMPLETING the
SQUARE
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-49_sec_8-2_Derive_Quadratic_Eqn.ppt
Example a) 2x2 + 9x − 5 = 0
 Solve using the Quadratic Formula:
2x2 + 9x − 5 = 0
 Soln a) Identify a, b, and c and
substitute into the quadratic formula:
2x2 + 9x − 5 = 0
a
b
c
 Now Know a, b, and c
Chabot College Mathematics
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-49_sec_8-2_Derive_Quadratic_Eqn.ppt
Solution a) 2x2 + 9x − 5 = 0
 Using a = 2, b = 9, c = −5
b  b2  4ac
x
2a
Recall the Quadratic Formula
→ Sub for a, b, and c
9  (9)2  4(2)(5)
x
2(2)
9  81  (40)
x
4
9  121
x
4
Chabot College Mathematics
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Be sure to write the
fraction bar ALL the
way across.
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-49_sec_8-2_Derive_Quadratic_Eqn.ppt
Solution a) 2x2 + 9x − 5 = 0
 From Last Slide: x  9  121
 So:
9  11
x
4
9  11
x
4
2
x
4
1
 The Solns: x 
2
Chabot College Mathematics
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4
or
or
9  11
x
4
20
x
4
or
x  5
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-49_sec_8-2_Derive_Quadratic_Eqn.ppt
Example b) x2 = −12x + 4
 b  b 2  4ac
x
2a
 Soln b) write
12  (12)2  4(1)(4)
x
x2 = −12x + 4
2(1)
in standard form,
identify a, b, & c,
12  144  16 12  160
x

and solve using
2
2
the quadratic
12
16 10
x

formula:
2
2
1x2 + 12x – 4 = 0
a
b
Chabot College Mathematics
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c
4 10
x  6 
 6  2 10
2
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-49_sec_8-2_Derive_Quadratic_Eqn.ppt
Example c) 5x2 − x + 3 = 0
 b  b 2  4ac
x
2a
 Soln c) Recognize a = 5, b = −1, c = 3
→ Sub into Quadratic Formula
(1)  (1)2  4(5)(3)
x
2(5)
 The COMPLEX No.
1  1  60
x
10
1  59
x
10
Chabot College Mathematics
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Soln
1
59
x  i
10
10
Since the radicand, –59,
is negative, there are
NO real-number solutions.
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-49_sec_8-2_Derive_Quadratic_Eqn.ppt
Quadratic Equation Graph
 The graph of a quadratic
eqn describes a “parabola”
which has one of a:
y  x 2  8x  12
x intercepts
• Bowl shape
• Dome shape
y  3x 2  6 x  1
vertex
 The graph, depending
on the “Vertex” Location,
may have different numbers of
of x-intercepts: 2 (shown), 1, or NONE
Chabot College Mathematics
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-49_sec_8-2_Derive_Quadratic_Eqn.ppt
The Discriminant
 It is sometimes enough to know what
type of number (Real or Complex) a
solution will be, without actually solving
the equation.
 From the quadratic formula, b2 – 4ac, is
known as the discriminant.
 The discriminant determines what type
of number the solutions of a quadratic
equation are.
• The cases are summarized on the next sld
Chabot College Mathematics
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-49_sec_8-2_Derive_Quadratic_Eqn.ppt
Soln Type by Discriminant
Discriminant
b2 – 4ac
Nature of Solutions
xIntercepts
0
Only one solution; it is a
real number
Only one
Positive
Two different real-number
solutions
Negative
Two different NONreal
complex-number solutions
(complex conjugates)
Chabot College Mathematics
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Two
different
None
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-49_sec_8-2_Derive_Quadratic_Eqn.ppt
Example  Discriminant
 Determine the nature of the solutions of:
5x2 − 10x + 5 = 0
 SOLUTION
 Recognize a = 5, b = −10, c = 5
 Calculate the Discriminant
b2 − 4ac = (−10)2 − 4(5)(5) = 100 − 100 = 0
 There is exactly one, real solution.
• This indicates that 5x2 − 10x + 5 = 0 can
be solved by factoring  5(x − 1)2 = 0
Chabot College Mathematics
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-49_sec_8-2_Derive_Quadratic_Eqn.ppt
Example  Discriminant
 Determine the nature of the solutions of:
5x2 − 10x + 5 = 0
 SOLUTION
Examine Graph
• Notice that the Graph
crosses the x-axis
(where y = 0) at
exactly ONE point as
predicted by the
discriminant
Chabot College Mathematics
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-49_sec_8-2_Derive_Quadratic_Eqn.ppt
Example  Discriminant
 Determine the nature of the solutions of:
4x2 − x + 1 = 0
 SOLUTION
 Recognize a = 4, b = −1, c = 1
 Calculate the Discriminant
b2 – 4ac = (−1)2 − 4(4)(1) =1 − 16 = −15
 Since the discriminant is negative,
there are two NONreal
complex-number solutions
Chabot College Mathematics
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-49_sec_8-2_Derive_Quadratic_Eqn.ppt
Example  Discriminant
 Determine the nature of the solutions of:
4x2 − 1x + 1 = 0
 SOLUTION
Examine Graph
• Notice that the Graph
does NOT cross the
x-axis (where y = 0)
indicating that there are
NO real values for x that
satisfy this Quadratic Eqn
Chabot College Mathematics
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-49_sec_8-2_Derive_Quadratic_Eqn.ppt
Example  Discriminant
 Determine the nature of the solutions of:
2x2 + 5x = −1
 SOLUTION: First write the eqn in Std
form of ax2 + bx + c = 0 →
2x2 + 5x + 1 = 0
 Recognize a = 2, b = 5, c = 1
 Calculate the Discriminant
b2 – 4ac = (5)2 – 4(2)(1) = 25 – 8 = 17
 There are two, real solutions
Chabot College Mathematics
20
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-49_sec_8-2_Derive_Quadratic_Eqn.ppt
Example  Discriminant
 Determine the nature of the solutions of:
0.3x2 − 0.4x + 0.8 = 0
 SOLUTION
 Recognize a = 0.3, b = −0.4, c = 0.8
 Calculate the Discriminant
b2 − 4ac = (−0.4)2 − 4(0.3)(0.8) =0.16–0.96 = −0.8
 Since the discriminant is negative, there are
two NONreal complex-number solutions
Chabot College Mathematics
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-49_sec_8-2_Derive_Quadratic_Eqn.ppt
Writing Equations from Solns
 The principle of zero products
informs that this factored equation
(x − 1)(x + 4) = 0 has solutions
1 and −4.
 If we know the solutions of an
equation, we can write an equation,
using the principle of Zero Products
in REVERSE.
Chabot College Mathematics
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-49_sec_8-2_Derive_Quadratic_Eqn.ppt
Example  Write Eqn from solns
 Find an eqn for which 5 & −4/3 are solns
 SOLUTION
x = 5 or x = –4/3
x – 5 = 0 or x + 4/3 = 0
Using the principle of
zero products
(x – 5)(x + 4/3) = 0
x2 – 5x + 4/3x – 20/3 = 0
3x2 – 11x – 20 = 0
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Get 0’s on one side
Multiplying
Combining like terms
and clearing fractions
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-49_sec_8-2_Derive_Quadratic_Eqn.ppt
Example  Write Eqn from solns
 Find an eqn for which 3i & −3i are solns
 SOLUTION
x = 3i or x = –3i
x – 3i = 0 or x + 3i = 0
Get 0’s on one side
(x – 3i)(x + 3i) = 0
Using the principle
of zero products
x2 – 3ix + 3ix – 9i2 = 0
x2 + 9 = 0
Chabot College Mathematics
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Multiplying
Combining like terms
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-49_sec_8-2_Derive_Quadratic_Eqn.ppt
WhiteBoard Work
 Problems From §8.2 Exercise Set
• 18, 30, 44, 58
Solving Quadratic Equations
1. Check to see if it is in the form ax2 = p or (x + c)2 = d.
•
If it is, use the square root property
2. If it is not in the form of (1), write it in standard form:
•
ax2 + bx + c = 0 with a and b nonzero.
3. Then try factoring.
4. If it is not possible to factor or if factoring seems difficult, use the
quadratic formula.
•
The solns of a quadratic eqn cannot always be found by factoring.
They can always be found using the quadratic formula.
Chabot College Mathematics
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-49_sec_8-2_Derive_Quadratic_Eqn.ppt
All Done for Today
The
Quadratic
Formula
Chabot College Mathematics
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Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-49_sec_8-2_Derive_Quadratic_Eqn.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
27
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-49_sec_8-2_Derive_Quadratic_Eqn.ppt
Graph y = |x|
6
 Make T-table
x
-6
-5
-4
-3
-2
-1
0
1
2
3
4
5
6
Chabot College Mathematics
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5
y = |x |
6
5
4
3
2
1
0
1
2
3
4
5
6
y
4
3
2
1
x
0
-6
-5
-4
-3
-2
-1
0
1
2
3
-1
-2
-3
-4
-5
file =XY_Plot_0211.xls
-6
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-49_sec_8-2_Derive_Quadratic_Eqn.ppt
4
5
6
5
5
y
4
4
3
3
2
2
1
1
0
-10
-8
-6
-4
-2
-2
-1
0
2
4
6
-1
0
-3
x
0
1
2
3
4
5
-2
-1
-3
-2
M55_§JBerland_Graphs_0806.xls
-3
Chabot College Mathematics
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-4
M55_§JBerland_Graphs_0806.xls
-5
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-49_sec_8-2_Derive_Quadratic_Eqn.ppt
8
10
Quadratic Equation Graph
 The graph of a quadratic eqn
describes a “parabola”
which has one of a:
y  x2  4x  5
• Bowl shape
• Dome shape
 The graph, depending
on the “Vertex” Location
may have different numbers of
x-intercepts: 2 (shown), 1, or NONE
Chabot College Mathematics
30
Bruce Mayer, PE
BMayer@ChabotCollege.edu • MTH55_Lec-49_sec_8-2_Derive_Quadratic_Eqn.ppt