Bruce Mayer, PE Licensed Electrical & Mechanical Engineer [email protected]

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[email protected] • MTH55_Lec49_sec_82_Derive_Quadratic_Eqn.ppt1
Bruce Mayer, PE Chabot College Mathematics
Bruce Mayer, PELicensed Electrical & Mechanical Engineer
Chabot Mathematics
§8.2 Quadratic§8.2 QuadraticEquationEquation
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Bruce Mayer, PE Chabot College Mathematics
Review §Review §
Any QUESTIONS About• §8.1 → Complete the Square
Any QUESTIONS About HomeWork• §8.1 → HW37
8.1 MTH 55
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Bruce Mayer, PE Chabot College Mathematics
The Quadratic FormulaThe Quadratic Formula
The solutions of ax2 + bx + c = 0 are given by
a
acbbx
2
42
This is one of theMOST FAMOUSFormulas in allof Mathematics
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Bruce Mayer, PE Chabot College Mathematics
§8.2 Quadratic Formula§8.2 Quadratic Formula
The Quadratic Formula
Problem Solving with the Quadratic Formula
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Bruce Mayer, PE Chabot College Mathematics
Derive Quadratic Formula  1Derive Quadratic Formula  1 Consider the
General Quadratic Equation
• Where a, b, c are CONSTANTS
Solve This Eqn for x by Completing the Square
First; isolate the Terms involving x
Next, Divide by “a” to give the second degree term the coefficient of 1
Now add to both Sides of the eqn a “quadratic supplement” of (b/2a)2
02 cbxax
cbxax 2
a
cx
a
bx 2
a
cababx
a
bx
222
22
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Bruce Mayer, PE Chabot College Mathematics
Derive Quadratic Formula  2Derive Quadratic Formula  2 Now the LeftHandSide
(LHS) is a PERFECT Square
a
c
a
b
a
bx
a
cababx
a
bx
22
222
22
22
Take the Square Root of Both Sides
a
c
a
b
a
bx
2
22
Combine Terms inside the Radical over a Common Denom
2
2
2
2
2
2
4
4
2
4
4
42
42
a
acb
a
bx
aa
ac
a
b
a
bx
a
c
a
b
a
bx
[email protected] • MTH55_Lec49_sec_82_Derive_Quadratic_Eqn.ppt7
Bruce Mayer, PE Chabot College Mathematics
Derive Quadratic Formula  4Derive Quadratic Formula  4 Note that Denom is,
itself, a PERFECT SQ
Next, Isolate x
But this the Renowned QUADRATIC FORMULA
Note That it was DERIVED by COMPLETING theSQUARE
a
acb
a
bx
a
acb
a
bx
2
4
2
4
4
22
2
2
a
acb
a
bx
2
4
2
2
a
acbbx
2
42
Now Combine over Common Denom
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Bruce Mayer, PE Chabot College Mathematics
Example a) 2Example a) 2xx22 + 9 + 9xx −− 5 = 0 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
Now Know a, b, and c
a b c
[email protected] • MTH55_Lec49_sec_82_Derive_Quadratic_Eqn.ppt9
Bruce Mayer, PE Chabot College Mathematics
Solution a) 2Solution a) 2xx22 + 9 + 9xx −− 5 = 0 5 = 0
Using a = 2, b = 9, c = −52 4
2
b b acx
a
2( ) 49 9 2( )( )
2 )2(
5x
9 81 ( 40)
4x
Be sure to write the fraction bar ALL the way across.
9 121
4x
Recall the Quadratic Formula→ Sub for a, b, and c
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Bruce Mayer, PE Chabot College Mathematics
Solution a) 2Solution a) 2xx22 + 9 + 9xx −− 5 = 0 5 = 0
From Last Slide: 9 121
4x
9 11
4x
9 11 9 11 or
4 4x x
2 20 or
4 4x x
1 or 5
2x x
So:
The Solns:
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Bruce Mayer, PE Chabot College Mathematics
Example b) Example b) xx22 = = −−1212xx + 4 + 4
Soln b) write x2 = −12x + 4 in standard form, identify a, b, & c, and solve using the quadratic formula:
1x2 + 12x – 4 = 0
a
acbbx
2
42
a b c
2( ) 412 12 1( )( 4)
)12(x
12 144 16 12 160
2 2x
12 16 10
2 2x
4 106
2x 6 2 10
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Bruce Mayer, PE Chabot College Mathematics
Example c) 5Example c) 5xx22 −− xx + 3 = 0 + 3 = 0
Soln c) Recognize a = 5, b = −1, c = 3 → Sub into Quadratic Formula
a
acbbx
2
42
2( ) ( ) 4( )( )
2(
1 1 5
5)
3x
1 1 60
10x
1 59
10x
Since the radicand, –59, is negative, there are NO realnumber solutions.
The COMPLEX No. Soln
10
59
10
1ix
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Bruce Mayer, PE Chabot College Mathematics
Quadratic Equation GraphQuadratic Equation Graph
The graph of a quadratic eqn describes a “parabola” which has one of a:• Bowl shape
• Dome shape
The graph, dependingon the “Vertex” Location,may have different numbers of of xintercepts: 2 (shown), 1, or NONE
1282 xxy
163 2 xxy
x intercepts
vertex
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Bruce Mayer, PE Chabot College Mathematics
The DiscriminantThe 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
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Bruce Mayer, PE Chabot College Mathematics
Soln Type by DiscriminantSoln Type by Discriminant
Discriminant b2 – 4ac
Nature of Solutionsx
Intercepts
0Only one solution; it is a real number
Only one
Positive
Two different realnumber solutions
Two different
NegativeTwo different NONreal complexnumber solutions (complex conjugates)
None
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Bruce Mayer, PE Chabot College Mathematics
Example Example Discriminant 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
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Bruce Mayer, PE Chabot College Mathematics
Example Example Discriminant Discriminant
Determine the nature of the solutions of:
5x2 − 10x + 5 = 0 SOLUTION
Examine Graph
• Notice that the Graphcrosses the xaxis (where y = 0) atexactly ONE point aspredicted by the discriminant
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Bruce Mayer, PE Chabot College Mathematics
Example Example Discriminant 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 complexnumber solutions
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Bruce Mayer, PE Chabot College Mathematics
Example Example Discriminant Discriminant
Determine the nature of the solutions of:
4x2 − 1x + 1 = 0 SOLUTION
Examine Graph
• Notice that the Graphdoes NOT cross the xaxis (where y = 0) indicating that there are NO real values for x that satisfy this Quadratic Eqn
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Bruce Mayer, PE Chabot College Mathematics
Example Example Discriminant 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
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Bruce Mayer, PE Chabot College Mathematics
Example Example Discriminant 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 complexnumber solutions
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Bruce Mayer, PE Chabot College Mathematics
Writing Equations from Solns 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.
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Bruce Mayer, PE Chabot College Mathematics
Example Example Write Eqn from solns 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
(x – 5)(x + 4/3) = 0
x2 – 5x + 4/3x – 20/3 = 0
3x2 – 11x – 20 = 0
Get 0’s on one side
Using the principle of zero products
Multiplying
Combining like terms and clearing fractions
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Bruce Mayer, PE Chabot College Mathematics
Example Example Write Eqn from solns 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
(x – 3i)(x + 3i) = 0
x2 – 3ix + 3ix – 9i2 = 0
x2 + 9 = 0
Get 0’s on one side
Using the principle of zero products
Multiplying
Combining like terms
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Bruce Mayer, PE Chabot College Mathematics
WhiteBoard WorkWhiteBoard 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.
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Bruce Mayer, PE Chabot College Mathematics
All Done for TodayAll Done for Today
TheQuadraticFormula
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Bruce Mayer, PE Chabot College Mathematics
Bruce Mayer, PELicensed Electrical & Mechanical Engineer
Chabot Mathematics
AppendiAppendixx
–
srsrsr 22
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Bruce Mayer, PE Chabot College Mathematics
Graph Graph yy =  = xx
Make Ttablex y = x 
6 65 54 43 32 21 10 01 12 23 34 45 56 6
x
y
6
5
4
3
2
1
0
1
2
3
4
5
6
6 5 4 3 2 1 0 1 2 3 4 5 6
file =XY_Plot_0211.xls
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Bruce Mayer, PE Chabot College Mathematics
3
2
1
0
1
2
3
4
5
3 2 1 0 1 2 3 4 5
M55_§JBerland_Graphs_0806.xls 5
4
3
2
1
0
1
2
3
4
5
10 8 6 4 2 0 2 4 6 8 10
M55_§JBerland_Graphs_0806.xls
x
y
[email protected] • MTH55_Lec49_sec_82_Derive_Quadratic_Eqn.ppt30
Bruce Mayer, PE Chabot College Mathematics
Quadratic Equation GraphQuadratic Equation Graph
The graph of a quadratic eqn describes a “parabola” which has one of a:• Bowl shape
• Dome shape
542 xxy
The graph, dependingon the “Vertex” Locationmay have different numbers of xintercepts: 2 (shown), 1, or NONE