5.8 Applications:

BeginningAlgebra

6.7 Applications:

1. To apply the Strategy for Problem Solving to applications whose solutions depend on solving equations by factoring.

English to Math

.

Word Problems

Strategy for Solving

Applications Applications involving quadratic equationsinvolving quadratic equations:

5.8 Applications with solutions involving solutions of quadratic equations.

1. Read the sentences carefullycarefully and determinedetermine the typetype of problem encountered.

4. Reread the problem to determinedetermine which are the correctcorrect

solutionssolutions for the question asked.

3. FactorFactor the equation and solvesolve each factor. There will always be two solutionstwo solutions to the equation.

2. Assign variable(s) to the unknown quantitiesunknown quantities and obtain

a quadratic equationquadratic equation with one variable in standard formstandard form.

Translating English Phrases and Sentences:

1. ... the product of two consecutive integers

x (x + 1) or x(x + 1)

Put ( ) around the object after the preposition “of”.

the product product of (two consecutive integers).

Let x be the first then (x + 1) is the second

2. The length is 5 inches more than twice the width.

(Whatever followsfollows “than” always comes firstcomes first.)

L = 2 W + 5 or L = 2W + 5

The length is 5 inches more thanthan twice the width.

3. The hypotenuse is 6 feet more than the short side.

The word hypotenusehypotenuse is implies a right triangle and is the slant side

READ IT:READ IT:

The hypotenuse, h, hypotenuse, h, is 6 feet more than6 feet more than the short side, short side, xx.

h = x + 6

4. The theorem for right triangles: hypotenuse-squaredhypotenuse-squared equals the sumsum of the short-side-squaredshort-side-squared and the long-side-squaredlong-side-squared.

hypotenuse2 2 = = short-sideshort-side22 + + long-sidelong-side22

or

short-sideshort-side22 = = hypotenuse2 2 – long-sidelong-side22

Translating English Phrases and Sentences:

or

long-sidelong-side22 = = hypotenuse2 2 – short-sideshort-side22

3. The demand is related to the price, usually a “new” price.

The The demand, , xx, for the , for the priceprice, , pp, is , is 800 800 – 100 100 pp

This becomes:

x = 800 – 100 pp

4. Revenue (money received) is price times demand.

Revenue,R, = priceprice, , pp, times , times demand, , x.x.

This becomes:

RR = pp (800 – 100p)

Translating English Phrases and Sentences:

R = R = pp

xx

Strategy for solving Word Problems involving quadratic equations involving quadratic equations:

1. Read the problem at least 3 times:

a) Look for key wordskey words to help decide the type of problem and the unknowns to be found.

b) Find the questionquestion - what, find, how much, etc.

c) List what is known and what is unknown and how the two unknowns are related.

2. Assign variables to the unknown quantities:

a) Assign x to one unknown quantityone unknown quantity and find relationships for the others.

b) Find the quadratic relationshipquadratic relationship in terms of xterms of x.

Strategy for solving Word Problems involving quadratic equations involving quadratic equations:

Zero-Factor Theorem: Zero-Factor Theorem: For real numbersreal numbers A and B,

if A·B = 0 then A = 0 or B = 0, or bothboth.

Solving Word Problems involving quadratic equationsinvolving quadratic equations:

3. SolveSolve and checkcheck the equation(s):

a) Solve the equation(s) for one variable. If needed, use this value to find the other unknown quantities .

b) Check the solution set by using the values as replacementsreplacements in all equationsall equations.

c) Reread the problem replacing found values in the original word sentence where they fit to make sure

you have a TRUE STATEMENT. IfIf the STATEMENT is TRUE, solution(s) are truth value(s).

The product of two consecutive positive integers is 156. Find the integers.

SolutionSolution:: Let x (x > 0) be the first integerfirst integer, then x + 1 will be the nextnext consecutive integerconsecutive integer.

Key words: productproduct and consecutive positive integersconsecutive positive integers:

Equation in x: x·(x+1) = 156

Number Problem

The productproduct of xx and x+1x+1 is 156156.

Equation in x: x·(x+1) = 156

x 2+ x = 156Distributive Property:

Add opposites Standard Equation:

x 2+ x – 156 = 0

(x + 13)(x – 12) = 0Factor:

Zero Product Theorem: (x + 13) = 0 or (x – 12) = 0

Solve equations: x = - 13 or x = 12

Since xx >> 00

CheckCheck the productproduct: 12·13 = 15612·13 = 156

x = 12 and x + 1 = 13

Number Problem, continued

The sum of two numbers is 17 and their product is 60. Find the two numbers.

Let n1 = n1 = xx then n2 = 17 n2 = 17 –– n1 n1 or n2 = n2 = 17 17 – x x.

Key words: sumsum, productproduct and two numberstwo numbers:

Equation in x: x·(17 – x) = 60

Number Problem

The sumsum of n1 and n2 is 17.Solution:

The product of n1 and n2 is 60

Distributive Property:

x 2 – 17x + 60 = 0

(x – 12)(x – 5) = 0

(x – 12) = 0 or (x – 5) = 0

x = 12 or x = 5

CheckCheck the productproduct: 12·5 = 6012·5 = 60CheckCheck the sumsum: 12 + 5 = 1712 + 5 = 17

Equation in x: x·(17 – x) = 60

Number Problem, continued

17x – x 2= 60

Add opposites Standard Equation:

Factor:

Zero Product Theorem:

Solve equations:

Area problemsArea problems involving quadratic equationsinvolving quadratic equations

1. Read the problem carefullycarefully and determine the typetype of problem encountered.

3. Use a known formula to find the equationequation for the problem.

2. Make a sketchsketch and labellabel the given dimensionsgiven dimensions.

4. Find standard formstandard form quadratic equationquadratic equation in one variableone variable.

6. Reread the problem to determine which are the correctcorrect solutionssolutions for the question askedquestion asked.

5. Find all solutionsall solutions for the equation.

Rectangle:

L

WArea = L·W

b

cah

Area = b·h2

Triangle:

The length of a rectangle is 5 inches more than twice the width. The area of the rectangle is 52 square inches. What are the dimensions of the rectangle?

Area problem using quadratic equations

. 2x + 5

. x

The length is compared tocompared to the width so assign the variable x to the width and find the relationship for the length.

Make and label a sketch:Let the width = x , (x > 0)

Then the length = 2x+5

width length = area

Equation in x: x (2x+5) = 52

Distributive Property:

Area Equation:

2x2+5x = 52

x (2x+5) = 52 GN =104GN =104

1041041 261 2644

52522 2 131388

dif = 5dif = 5

Rearrange equation:

(x – 4)(2x+13) = 0

Scratch:Scratch:

2x2+13x – 8x – 52 = 0

x(2x+13) – 4(2x+13) = 0

WidthWidth can’t becan’t be negativenegative, choose x = 4

Factor:

Area problem using quadratic equationsusing quadratic equations

Solutions are x = 4 and -132

x =

width = x; length = 2x+5; area = 52

2x2+5x – 52 = 0

width = 4 incheslength = 13 incheslength = 2x+5 = 13

The height of a triangle is 5 inches more than its base. The area is 102 square inches. Find the dimensions of the triangle.

Area problem using quadratic equations

The height is compared tocompared to the base so assign the variable x to the base and find the relationship for the height.

Make and label a sketchMake and label a sketch::Let the basebase = xx , (x > 0)

Then the height = x + 5

90 x

x+5

basebase heightheight = areaarea

12

Equation in xx: 12

x (x+5) = 102

Multiply ReciprocalMultiply Reciprocal: x(x+5) = 204

basebase = x; heightheight = x+5; areaarea = 102 GN =204

2041 514

1022 346

683 1712

dif = 5Rearrange equationRearrange equation: x2+ 5x – 204 = 0

FactorFactor: (x + 17)(x -12) = 0 Scratch:Scratch:

x2+17x – 12x – 204 = 0

x(x+17) – 12(x+17) = 0SolutionsSolutions are x = -17 and x = 12

Width can’t be negativeWidth can’t be negative, choose x = 12

Area problem using quadratic equationsusing quadratic equations

base = 12 inchesheight = 17 inches

heightheight = x + 5 = 17

Area EquationArea Equation: 12

x (x+5) = 102

Check:Check: 12 17 = 102

12

Pythagorean Theorem:

a

bc

90

In any triangle, the squaresquare of the longest sidelongest side (called hypotenusehypotenuse) is equal to the sumsum of the squaressquares of the other two sidestwo sides ( called legslegs).

aa22 + + bb22 = = cc22

The sketch at the right illustrates therelationship geometrically. The proof is one of the applications found in geometry. There are many specialvalues for a, b, and c which arealways whole numbers but many applications result in finding square roots of numbers. For applicationsin this section all values are whole numbers.

aa22 + + bb22 = = cc22

The length of a one leg of a right triangle is 2 more than twice the other. The hypotenuse of the triangle is 13 inches. What are the dimensions of the triangle?

Pythagorean Theorem: aa22 + b + b22 = c = c22

The long leglong leg is compared tocompared to the short legshort leg so we will assign the variable x to the short legshort leg and find the relationshiprelationship for the long leglong leg.

Make and label a sketch:

Let the short leg (a) = x

90 2x + 2

13x

NOTE: 0<x<13

Hypotenuse (c) = 13

Then the long leg (b) = 2x+2

Pythagorean Theorem:

x2 + (2x+2)2 = 132

Square the BinomialSquare the Binomial:

EquationEquation: xx 2 2 + (2x+2) + (2x+2)22 = 13 = 1322 GN = 825

1655 3325

dif = 8

Rearrange equation:Rearrange equation: 5x5x22+ 8x – 165 = 0+ 8x – 165 = 0

(5x + 33)(x – 5) = 0 (5x + 33)(x – 5) = 0 Scratch:Scratch:

5x5x22+33x+33x – – 25x – 16525x – 165 = 0 = 0

x(x(5x+335x+33) – 5() – 5(5x+335x+33) = 0) = 0SolutionsSolutions are

Leg1 can’t be negativecan’t be negative, choose x = 5

Factor:Factor:

and x = 5

Leg1 = 5 inchesLeg2 = 12 inches

Leg2 = 2x+2 = 12

Check: 5522 + 12 + 1222 = 13 = 1322

xx 2 2 + 4x + 4x22 + 8x + 4 = 169 + 8x + 4 = 169

-335

x =

Pythagorean Theorem: aa22 + b + b22 = c = c22

The length of a hypotenuse of a right triangle is 3 less than twice the short side. The long side of the triangle is 12 inchesinches. What are the dimensions of the triangle?

Pythagorean Theorem: aa22 + b + b22 = c = c22

The hypotenusehypotenuse is compared tocompared to the short legshort leg so we will assign the variable x to the short legshort leg and find the relationshiprelationship for the long leglong leg.

Make and label a sketch:

Let the short leg (a) = x

90

12

2x-3 x

NOTE: 0<x<12

Then the Hypotenuse (c) = 2x –– 3

The long leg (b) = 12

Pythagorean Theorem:

x2 + 122 = (2x –– 3)2

Square BinomialSquare Binomial: xx 2 2 + 144 = 4x + 144 = 4x2 2 – 12x + 9– 12x + 9

EquationEquation: xx 2 2 + 12 + 1222 = (2x – 3) = (2x – 3)22

GN = 45

153 95

dif = 4Rearrange equation:Rearrange equation: 3x3x22 – 12x – 135 = 0 – 12x – 135 = 0

3(x – 9)(x + 5) = 0 3(x – 9)(x + 5) = 0

Scratch:Scratch:

xx2 2 – 9x– 9x + + 5x – 455x – 45 = 0 = 0

xx((x – 9x – 9) + 5() + 5(x – 9x – 9) = 0) = 0

Leg1 can’t be negativecan’t be negative, choose x = 9

Factor:Factor:

SolutionsSolutions are x = 9 and x = - 5

Short Leg = 9 inchesHypotenuse = 15 inches

Hypotenuse = 2x –– 3 = 15

Check: 9922 + 12 + 1222 = 15 = 1522

Pythagorean Theorem: aa22 + b + b22 = c = c22

Distributive PropertyDistributive Property: 3(x3(x22 – 4x – 45) = 0 – 4x – 45) = 0

Business Problems

Cost, though a necessity, is considered a negativenegativequantityquantity. Cost can be fixed, such as rent on the building,

or variable , depending on the items, x, produced.

Sometimes the priceprice, p, is given in terms of x p = p(x). Other times the number of items, x, is given in terms of p x = x(p). These expressionsexpressions are called the DemandDemand.

RevenueRevenue, R, is the money earned on xx items items, each sold at a certain price, price, pp.

Business problemsBusiness problems will be encountered in economicseconomics, statisticsstatistics, applied calculusapplied calculus or just everyday lifeeveryday life. It is necessary to learn to read and interpret these problems.

Vocabulary in business includes costcost and revenuerevenue.

A company manufactures diskettes for home computers. It knows from past experience that the numbernumber of diskettes it can sell each day, x, is related to the priceprice p per diskette by the equation x = 800  100·p. At what priceprice, p, should the company sell the diskettes if it wants the daily revenuerevenue to be $1,200?

Business Problem

Formula for RevenueFormula for Revenue: R = p ·x or p ·x = R

Demand for diskettesDemand for diskettes: x = 800  100·p

Revenue Equation:Revenue Equation: p ·(800  100·p )= $1,200

Replacement PropertyReplacement Property:

800p 100p 2 = 1200Distributive propertyDistributive property:

Formula for RevenueFormula for Revenue: p·x = R

p ·(800  100·p ) = $1,200

100p 2 800p + 1200 = 0Add oppAdd oppositesosites:

p 2 8p + 12 = 0Multiply recipMultiply reciprocal of 100rocal of 100:

(p 6)(p 2) = 0Factor:

p = 6 oror p = 2Solve the equationSolve the equation:

If the company assigns the price of $2 or the price $6 to each diskette the revenue is $1,200

Business Problem

5.8

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