[email protected] MTH55_Lec-26_sec_5-7_PolyNom_Eqns-n-Apps.ppt 1 Bruce Mayer, PE Chabot College Mathematics Bruce Mayer, PE Licensed Electrical

[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

[email protected]

Chabot Mathematics

§5.7 PolyNomial§5.7 PolyNomialEqns & AppsEqns & Apps



Review §Review §

Any QUESTIONS About• §5.6 → Factoring Strategies

Any QUESTIONS About HomeWork• §5.6 → HW-16

5.6 MTH 55



§5.7 Solving PolyNomial Eqns§5.7 Solving PolyNomial Eqns

The Principle of Zero Products

Factoring to Solve Equations

Algebraic-Graphical Connection



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



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



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.



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.



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}



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



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”



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



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



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



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.



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



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.



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.



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



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



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.



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



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



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



−−xx2 2 − − x x + 6 = 0 + 6 = 0 Solve with Graph Solve with Graph

-4

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-5 -4 -3 -2 -1 0 1 2 3 4 5

M65_§7-1_Graphs_0607.xls5_Graphs_0607.xls

x

y

62 xxy

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



Example Example Find x-Intercepts Find x-Intercepts

-12

-5 5


x

y

932 2 xxy

Find the x-intercepts for graph (at Left) of Equation

At interceptsy = 0; so Use ZERO Products:

9320 2 xxy



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

-12

-5 5


x

y



Example Example x-Intercepts x-Intercepts

-12

-5 5


x

y

0,2

3 0,3



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

-5 -4 -3 -2 -1 1 2 3 4 5

4

3

6

2

5

1

-1

8

7

y = 8



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

-10

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0

1

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-6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6

file =XY_Plot_0211.xls

xy

822 xxxg

Root-1 Root-2



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.



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



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




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




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




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



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




Now LET x ≡ the Border Width Translate using Diagram The OverAll

• Width is 6ft + 2x

• Length is 10ft + 2x







Carry Out

1or9 xx Zero Products

Since a Length can Never be Negative, Discard −9 as solution




State: The reflecting pool border width should be 1 ft

12 ft

8 ft

Check: 2·(12ft·1ft) + 2·(6ft·1ft) = 24ft2 + 12ft2 = 36ft2



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




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




• 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




• 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




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



WhiteBoard WorkWhiteBoard Work

Problems From §5.7 Exercise Set• 74, 80, 82

ModelRockets



All Done for TodayAll Done for Today

BlastOff!



Bruce Mayer, PELicensed Electrical & Mechanical Engineer

[email protected]

Chabot Mathematics

AppendiAppendixx

–

srsrsr 22



Graph Graph yy = | = |xx||

Make T-tablex y = |x |

-6 6-5 5-4 4-3 3-2 2-1 10 01 12 23 34 45 56 6

x

y

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x

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M55_§JBerland_Graphs_0806.xls -10

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xy



Quadratic FormulaQuadratic Formula

Documents

[email protected] MTH55_Lec-26_sec_5-7_PolyNom_Eqns-n-Apps.ppt 1 Bruce Mayer, PE Chabot College Mathematics Bruce Mayer, PE Licensed Electrical