[email protected] mth55_lec-26_sec_5-7_polynom_eqns-n-apps.ppt 1 bruce mayer, pe chabot...
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[email protected] • MTH55_Lec-26_sec_5-7_PolyNom_Eqns-n-Apps.ppt1
Bruce Mayer, PE Chabot College Mathematics
Bruce Mayer, PELicensed Electrical & Mechanical Engineer
Chabot Mathematics
§5.7 PolyNomial§5.7 PolyNomialEqns & AppsEqns & Apps
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Bruce Mayer, PE Chabot College Mathematics
Review §Review §
Any QUESTIONS About• §5.6 → Factoring Strategies
Any QUESTIONS About HomeWork• §5.6 → HW-16
5.6 MTH 55
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Bruce Mayer, PE Chabot College Mathematics
§5.7 Solving PolyNomial Eqns§5.7 Solving PolyNomial Eqns
The Principle of Zero Products
Factoring to Solve Equations
Algebraic-Graphical Connection
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Bruce Mayer, PE Chabot College Mathematics
Quadratic EquationsQuadratic Equations
Second degree equations such as 9t2 – 4 = 0 and x2 + 6x + 9 = 0 are called quadratic equations
A quadratic equation is an equation equivalent to one of the form
ax2 + bx + c = 0
• where a, b, and c are constants, with a ≠ 0
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Bruce Mayer, PE Chabot College Mathematics
Principle of Zero ProductsPrinciple of Zero Products
An equation AB = 0 is true if and only if A = 0 or B = 0, or both = 0.
That is, a product is 0 if and only if at LEAST ONE factor in the multiplication-chain is 0• i.e.; Need a Zero-FACTOR to create
a Zero-PRODUCT
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Bruce Mayer, PE Chabot College Mathematics
Example Example Solve Solve ((xx + 4)( + 4)(xx −− 3) = 3) = 00 In order for a product to be 0, at least
one factor must be 0. Therefore, either
x + 4 = 0 or x − 3 = 0 Solving each equation:
x + 4 = 0 or x − 3 = 0
x = −4 or x = 3 Both −4 and 3 should be checked in the
original equation.
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Bruce Mayer, PE Chabot College Mathematics
Check for Check for ((xx + 4)( + 4)(xx −− 3) = 0 3) = 0
For x = −4: For x = 3:
(x + 4)(x − 3) = 0 (x + 4)(x − 3) = 0
(−4 + 4)(−4 − 3) (3 + 4)(3 − 3)
0(−7) 7(0)
0 = 0 0 = 0
True True
The solutions are −4 and 3.
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Bruce Mayer, PE Chabot College Mathematics
Solve Solve 4(34(3xx + 1)( + 1)(xx −− 4) = 0 4) = 0
Since the factor 4 is constant, the only way for 4(3x + 1)(x − 4) to be 0 is for one of the other factors to be 0. That is,
3x + 1 = 0 or x − 4 = 0
3x = −1 or x = 4
x = −1/3 So the solutions to the Equation are
x = −1/3 and x = 4
{−1/3 , 4}
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Bruce Mayer, PE Chabot College Mathematics
Check Check 4(34(3xx + 1)( + 1)(xx –– 4) = 0 4) = 0 For −1/3: 4(3x + 1)(x − 4) = 0
4(3•[−1/3] + 1)([−1/3] − 4) = 0 4(−1 + 1)(−1/3 − 12/3) = 04(0)(−13/3 ) = 0 0 = 0
For 4: 4(3x + 1)(x − 4) = 0 4((3(4) + 1)(4 − 4) = 0
4(13)(0) = 00 = 0
The solutions are −1/3 and 4
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Bruce Mayer, PE Chabot College Mathematics
Solve Solve 3 3yy((yy −− 7) = 0 7) = 0
SOLUTION
3 y (y − 7) = 0
y = 0 or y − 7 = 0
y = 0 or y = 7
The solutions are 0 and 7• The Check is Left to the Student
– Should be easily “EyeBalled”
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Bruce Mayer, PE Chabot College Mathematics
Factoring to Solve EquationsFactoring to Solve Equations
By factoring and using the principle of zero products, we can now solve a variety of quadratic equations.
Example: Solve x2 + 9x + 14 = 0 SOLUTION: This equation requires us
to FIRST factor the polynomial since there are no like terms to combine and there is a squared term. THEN we use the principle of zero products
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Bruce Mayer, PE Chabot College Mathematics
Solve Solve xx22 + 9 + 9xx + 14 = 0 + 14 = 0 Factor the Left Hand Side (LHS) by
Educated Guessing (FOIL Factoring):x2 + 9x + 14 = 0(x + 7)(x + 2) = 0x + 7 = 0 or x + 2 = 0
x = −7 or x = −2.
The Tentative Solutions are −7 and −2• Let’s Check
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Bruce Mayer, PE Chabot College Mathematics
Check Check xx = = −7−7 & & xx = −2 = −2
For −7: For −2:
x2 + 9x + 14 = 0 x2 + 9x + 14 = 0
(−7)2 + 9(−7) + 14 (–2)2 + 9(–2) + 14
49 − 63 + 14 4 − 18 + 14
−14 + 14 −14 + 14
0 = 0 0 = 0
True True Thus −7 and −2 are VERIFIED as
Solutions to x2 + 9x + 14 = 0
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Bruce Mayer, PE Chabot College Mathematics
Example Example Solve Solve xx22 + 9 + 9xx = 0 = 0
SOLUTION: Although there is no constant term, because of the x2-term, the equation is still quadratic. Try factoring:
x2 + 9x = 0 → see GCF = x
x(x + 9) = 0
x = 0 or x + 9 = 0
x = 0 or x = −9 The solutions are 0 and −9.
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Bruce Mayer, PE Chabot College Mathematics
Caveat Mathematicus Caveat Mathematicus
CAUTIONCAUTION: We must have 0 on one side of the equation before the principle of zero products can be used.
Get all nonzero terms on one side of the equation and 0 on the other
Example: Solve: x2 − 12x = −36
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Bruce Mayer, PE Chabot College Mathematics
Solve Solve xx22 −− 12 12xx = = −−3636
SOLUTION: We first add 36 to BOTH Sides to get 0 on one side:
x2 − 12x = −36
x2 − 12x + 36 = −36 + 36 = 0
(x − 6)(x − 6) = 0
x − 6 = 0 or x − 6 = 0
x = 6 or x = 6 There is only one solution, 6.
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Bruce Mayer, PE Chabot College Mathematics
Standard FormStandard Form
To solve a quadratic equation using the principle of zero products, we first write it in standard form: with 0 on one side of the equation and the leading coefficient POSITIVE.
We then factor and determine when each factor is 0.
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Bruce Mayer, PE Chabot College Mathematics
Example Example Functional Eval Functional Eval
Given f(x) = x2 + 10x + 26. Find a such that f(a) = 1.
SOLUTIONSet f (a) = 1f (a) = a2 + 10a + 26 = 1
a2 + 10a + 25 = 0 (a + 5)(a + 5) = 0 a + 5 = 0 so a = −5
The check is left for you & I to do Later
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Bruce Mayer, PE Chabot College Mathematics
Example Example Solve Solve 99xx22 = 49 = 49 SOLUTION: 9x2 = 49
9x2 − 49 = 0 ► Diff of Sqs: (3x)2 & 72
(3x − 7)(3x + 7) = 0 3x − 7 = 0 or 3x + 7 = 0 3x = 7 or 3x = −7
x = 7/3 or x = −7/3
The solutions are 7/3 & −7/3
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Bruce Mayer, PE Chabot College Mathematics
Solve: 14Solve: 14xx22 + 9 + 9xx + 2 = 10 + 2 = 10xx + 6 + 6 SOLUTION: Be careful with an equation like this!
Since we need 0 on one side, we subtract 10x and 6 from Both Sides to get the RHS = 0
14x2 + 9x + 2 = 10x + 614x2 + 9x − 10x + 2 − 6 = 0 14x2 − x − 4 = 0
(7x − 4)(2x + 1) = 0 7x − 4 = 0 or 2x + 1 = 0 7x = 4 or 2x = −1 x = 4/7 or x = −1/2 The solutions are 4/7 and −1/2.
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Bruce Mayer, PE Chabot College Mathematics
Solve Eqns by Zero ProductsSolve Eqns by Zero Products
1. Write an equivalent equation with 0 on one side, using the addition principle.
2. Factor the nonzero side of the equation
3. Set each factor that is not a constant equal to 0
4. Solve the resulting equations
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Bruce Mayer, PE Chabot College Mathematics
Parabola interceptsParabola intercepts Find the x-intercepts for
the graph of the equation shown.
y = x2 + 2x 8
The x-intercepts occur where the plot crosses y = 0
Parabola
Thus at the x-intercepts
y = 0 = x2 + 2x − 8
So Can use the principle of zero products
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Bruce Mayer, PE Chabot College Mathematics
Parabola intercepts cont.Parabola intercepts cont. SOLUTION: To find the
intercepts, let y = 0 and solve for x.
y = x2 + 2x 8
0 = x2 + 2x − 8
0 = (x + 4)(x − 2)
x + 4 = 0 or x − 2 = 0
x = −4 or x = 2
The x-intercepts are (−4, 0) and (2, 0).
Parabola
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Bruce Mayer, PE Chabot College Mathematics
−−xx2 2 − − x x + 6 = 0 + 6 = 0 Solve with Graph Solve with Graph
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Solve60 2 xx
Recall from Graphing that the x-axis is the Location where y = 0
Thus on Graph find x for y = 0
Solns: x = −3 and x = 2
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Bruce Mayer, PE Chabot College Mathematics
Example Example Find x-Intercepts Find x-Intercepts
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M65_§7-1_Graphs_0607.xls5_Graphs_0607.xls
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y
932 2 xxy
Find the x-intercepts for graph (at Left) of Equation
At interceptsy = 0; so Use ZERO Products:
9320 2 xxy
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Bruce Mayer, PE Chabot College Mathematics
Example Example Find x-Intercepts Find x-Intercepts
At Intercepts y = 0, so
y = 0 = 2x2 + 3x − 9 FOIL-factor the Quadratic expression
0 = 2x2 + 3x − 9 = (2x − 3)(x + 3) = 0 By ZERO PRODUCTS
(2x − 3) = 0 or (x + 3) = 0 Solving for x (the intercept values):
x = 3/2 or x = −3
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M65_§7-1_Graphs_0607.xls5_Graphs_0607.xls
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Bruce Mayer, PE Chabot College Mathematics
Example Example x-Intercepts x-Intercepts
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M65_§7-1_Graphs_0607.xls5_Graphs_0607.xls
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Bruce Mayer, PE Chabot College Mathematics
PolyNomial Fcns and GraphsPolyNomial Fcns and Graphs Consider, for
example the eqn, x2 − 2x = 8. One way to begin to solve this equation is to graph the function f (x) = x2 − 2x. Then look for any x-value that is paired with 8, as shown at Right
x
y 2( ) 2f x x x
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y = 8
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Bruce Mayer, PE Chabot College Mathematics
PolyNomial Fcns and GraphsPolyNomial Fcns and Graphs Equivalently, we
could graph the function given by g(x) = x2 − 2x − 8 and look for the values of x for which g(x) = 0. These values are what we call the roots, or zeros, of a polynomial function
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Root-1 Root-2
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Bruce Mayer, PE Chabot College Mathematics
Problem SolvingProblem Solving
Some problems can be translated to quadratic equations, which we can now solve.
The problem-solving process is the same as for other kinds of problems.
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Bruce Mayer, PE Chabot College Mathematics
The Pythagorean TheoremThe Pythagorean Theorem
Recall Pythagorus’ Great Discovery: In any right triangle, if a and b are the
lengths of the legs and c is the length of the hypotenuse, then
a2 + b2 = c2
a
b
c
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Bruce Mayer, PE Chabot College Mathematics
Example Example Screen Diagonal Screen Diagonal
A computer screen has the dimensions (in inches) shown below.
Find the length of the diagonal of the screen.
x
3x
6x
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Bruce Mayer, PE Chabot College Mathematics
Example Example Screen Diagonal Screen Diagonal
Familiarize. A right triangle is formed using the diagonal and sides of the screen. x + 6 is the hypotenuse with x and x + 3 as legs.
Translate. Applying the Pythagorean Theorem, Use the Diagram to translate as follows:
x2 + (x + 3)2 = (x + 6)2
a2 + b2 = c2
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Bruce Mayer, PE Chabot College Mathematics
Example Example Screen Diagonal Screen Diagonal
Carry out. Solve the equation by:
(x + 3)(x – 9) = 0
x + 3 = 0 or x – 9 = 0
x = –3 or x = 9
2x2 + 6x + 9 = x2 + 12x + 36
x 2 – 6x – 27 = 0
x2 + (x2 + 6x + 9) = x2 + 12x + 36
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Bruce Mayer, PE Chabot College Mathematics
Example Example Screen Diagonal Screen Diagonal
Check. The integer −3 cannot be the length of a side because it is negative. For x = 9, we have x + 3 = 12 and x + 6 = 15. Since 81 + 144 = 225, the lengths determine a right triangle.
Thus 9, 12,and 15 check. State. The length of the diagonal of the
screen is 15 inches
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Bruce Mayer, PE Chabot College Mathematics
Example Example Area Allotment Area Allotment
A LandScape Architect designs a flower bed of uniform width around a small reflecting pool. The pool is 6ft by 10ft. The plans call for 36 ft2 of plant coverage. How WIDE should the Border be?
FamiliarizewithDiagram
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Bruce Mayer, PE Chabot College Mathematics
Example Example Area Allotment Area Allotment
Now LET x ≡ the Border Width Translate using Diagram The OverAll
• Width is 6ft + 2x
• Length is 10ft + 2x
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Bruce Mayer, PE Chabot College Mathematics
Example Example Area Allotment Area Allotment
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Bruce Mayer, PE Chabot College Mathematics
Example Example Area Allotment Area Allotment
Carry Out
1or9 xx Zero Products
Since a Length can Never be Negative, Discard −9 as solution
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Bruce Mayer, PE Chabot College Mathematics
Example Example Area Allotment Area Allotment
State: The reflecting pool border width should be 1 ft
12 ft
8 ft
Check: 2·(12ft·1ft) + 2·(6ft·1ft) = 24ft2 + 12ft2 = 36ft2
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Bruce Mayer, PE Chabot College Mathematics
Example Example Rocket Ballistics Rocket Ballistics
A Model Rocket is fired UpWards from the Ground. The Height, h, in feet of the rocket can be found from this equation:
Find the time that it takes for the Rocket to reach a height of 48 feet
21680 ttth – Where t is time in seconds
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Bruce Mayer, PE Chabot College Mathematics
Example Example Rocket Ballistics Rocket Ballistics
Familiarize: We must find t such that h(t) = 48.• Thus Substitute 48 for h(t) in
the ballistics equation
Carry Out:• Subtract 48 from both sides
2168048 tt
4816800 2 tt
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Bruce Mayer, PE Chabot College Mathematics
Example Example Rocket Ballistics Rocket Ballistics
• Divide Both Sides by −16
• Write in Standard form
350 2 tt
4850 2 tt
• Use QUADRATIC Formula with a = 1, b = −5, and c = 3
Prime Not Factorable
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Bruce Mayer, PE Chabot College Mathematics
Example Example Rocket Ballistics Rocket Ballistics
• Approximate the Sq-Roots
Since “What goes UP must come DOWN” The Rocket will reach 48ft TWICE; Once while blasting-UP and again while Free-Falling DOWN
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Bruce Mayer, PE Chabot College Mathematics
Example Example Rocket Ballistics Rocket Ballistics
State: The rocket will climb to 48ft in about 0.7 seconds, continue its climb, and then it will descend (fall) to a height of 48ft after a total flite time of 4.3 seconds as it continues its FreeFall to the Ground
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Bruce Mayer, PE Chabot College Mathematics
WhiteBoard WorkWhiteBoard Work
Problems From §5.7 Exercise Set• 74, 80, 82
ModelRockets
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Bruce Mayer, PE Chabot College Mathematics
All Done for TodayAll Done for Today
BlastOff!
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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 T-tablex y = |x |
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Bruce Mayer, PE Chabot College Mathematics
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Bruce Mayer, PE Chabot College Mathematics
Quadratic FormulaQuadratic Formula