5.8 Applications: Beginning Algebra

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 involving quadratic equations Applications involving quadratic equations:

5.8 Applications with solutions involving solutions of quadratic equations. 1. Read the sentences carefully carefully and determine determine the type type of problem encountered. 4. Reread the problem to determine determine which are the correct solutions solutions for the question asked. 3. Factor Factor the equation and solve solve each factor. There will always be two solutions solutions to the equation. 2. Assign variable(s) to the unknown quantities quantities and obtain a quadratic equation equation with one variable in standard form form.

Translating English Phrases and Sentences: 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 (x + 1) is the second 2. The length is 5 inches more than twice the width. (Whatever follows follows “than” always comes first first.) L = 2  W + 5 or L = 2W + 5 The length is 5 inches more than than twice the width.

3. The hypotenuse is 6 feet more than the short side. hypotenuse The word hypotenuse is implies a right triangle and is the slant side READ IT: hypotenuse, h, 6 feet more thanshort side, x The hypotenuse, h, is 6 feet more than the short side, x. h = x The theorem for right triangles: hypotenuse-squared hypotenuse-squared equals the sum sum of the short-side-squared short-side-squared and the long-side-squared long-side-squared. 2 = short-side 2 + long-side 2 hypotenuse 2 = short-side 2 + long-side 2 or short-side 2 = 2 long-side 2 short-side 2 = hypotenuse 2 – long-side 2 Translating English Phrases and Sentences: or long-side 2 = 2 short-side 2 long-side 2 = hypotenuse 2 – short-side 2

3. The demand is related to the price, usually a “new” price. The, x, for the price, p, is p The demand, x, for the price, p, is 800 – 100  p This becomes: p x = 800 – 100  p 4. Revenue (money received) is price times demand. price, p, times, x. Revenue,R, = price, p, times demand, x. This becomes: Rp R = p  (800 – 100p) Translating English Phrases and Sentences: R = p x

involving quadratic equations Strategy for solving Word Problems involving quadratic equations: 1. Read the problem at least 3 times: key words a) Look for key words to help decide the type of problem and the unknowns to be found. question b) Find the question - 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 quantity quantity and find relationships for the others. b) Find the quadratic relationship relationship in terms of xx.xx. involving quadratic equations Strategy for solving Word Problems involving quadratic equations: Zero-Factor Theorem: For real numbers numbers A and B,B, if A·B A·B = 0 then A = 0 or B = 0, or both both.

involving quadratic equations Solving Word Problems involving quadratic equations: Solvecheck 3. Solve and check the equation(s): a) Solve the equation(s) for one variable. If needed, use this value to find the other unknown quantities. replacementsall equations b) Check the solution set by using the values as replacements in all equations. c) Reread the problem replacing found values in the original word sentence where they fit to make sure you have a TRUE STATEMENT. If If the STATEMENT is TRUE, solution(s) are truth value(s).

The product of two consecutive positive integers is 156. Find the integers. Solution:first integer next consecutive integer Solution: Let x (x > 0) be the first integer, then x + 1 will be the next consecutive integer. Key words: product product and consecutive positive integers integers: Equation in x: x·(x+1) = 156 Number Problem The product product of x x and x+1 x+1 is

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 x > 0 Since x > 0 Checkproduct12·13 = 156 Check the product: 12·13 = 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 = x x then n2 = 17 – n1 n1 or n2 = 17 – xx.xx. Key words: sum sum, product product and two numbers numbers: Equation in x: x·(17 – x) = 60 Number Problem The sum sum 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 Check Check the product product: 12·5 = 60 Check Check the sum sum: = 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 problems involving quadratic equations 1. Read the problem carefully carefully and determine the type type of problem encountered. 3. Use a known formula to find the equation equation for the problem. 2. Make a sketch sketch and label label the given dimensions dimensions. 4. Find standard form form quadratic equation equation in one variable variable. 6. Reread the problem to determine which are the correct solutions solutions for the question asked asked. 5. Find all solutions solutions for the equation.

Rectangle : L W Area = L·WL·W b c a h Area = b·hb·h 2 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 to 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)0) Then the length = 2x+5 width  length = area Equation in x: x (2x+5) = 52

Distributive Property: Area Equation: 2x 2 +5x = 52 x (2x+5) = 52 GN =   1 26  4 52  2 52  2 13  8 dif = 5 Rearrange equation: (x – 4)(2x+13) = 0 Scratch: 2x 2 +13x – 8x – 52 = 0 x(2x+13) – 4(2x+13) = 0 Width Width can’t be be negative negative, choose x = 4 Factor: using quadratic equations Area problem using quadratic equations Solutions are x = 4 and x = width = x; length = 2x+5; area = 52 2x 2 +5x – 52 = 0 width = 4 inches length = 13 inches length = 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 to to the base so assign the variable x to the base and find the relationship for the height. Make and label a sketch: Let the base base = xx xx, (x > 0)0) Then the height = x  x x+5 baseheightarea  base  height = area 1 2 x Equation in x: 1 2 x (x+5) = 102

Multiply Reciprocal Multiply Reciprocal: x(x+5) = 204 base base = x; height height = x+5; area area = 102 GN =  1 51   2 34  6 68  3 17  12 dif = 5 Rearrange equation equation: x 2 + 5x – 204 = 0 Factor Factor: (x + 17)(x -12) = 0 Scratch: x 2 +17x – 12x – 204 = 0 x(x+17) – 12(x+17) = 0 Solutions Solutions are x = -17 and x = 12 Width can’t be negative Width can’t be negative, choose x = 12 using quadratic equations Area problem using quadratic equations base = 12 inches height = 17 inches height height = x + 5 = 17 Area Equation Area Equation: 1 2 x (x+5) = 102 Check: Check:  12  17 =

Pythagorean Theorem: a b c 90  In any triangle, the square square of the longest side (called hypotenuse hypotenuse) is equal to the sum sum of the squares squares of the other two sides sides ( called legs legs). a2 a2 a2 a2 + b2 b2 b2 b2 = c2c2c2c2 The sketch at the right illustrates the relationship geometrically. The proof is one of the applications found in geometry. There are many special values for a, b, and c which are always whole numbers but many applications result in finding square roots of numbers. For applications in this section all values are whole numbers. a2 a2 a2 a2 + b2 b2 b2 b2 = c2c2c2c2

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? a 2 + b 2 = c 2 Pythagorean Theorem: a 2 + b 2 = c 2 The long leg leg is compared to to the short leg leg so we will assign the variable x to the short leg leg and find the relationship relationship for the long leg leg. Make and label a sketch: Let the short leg (a) = x 90  2x x NOTE: 0<x<13 Hypotenuse (c) = 13 Then the long leg (b) = 2x+2 Pythagorean Theorem: x2 x2 + (2x+2) 2 = 13 2

Square the Binomial Binomial: Equation Equation: x 2+ (2x+2) 2 (2x+2) 2 = 13 2 GN =  5 33  25 dif = 8 Rearrange equation: equation: 5x 2 + 5x 2 + 8x – 165 = 0 (5x + 33)(x – 5) = 0 Scratch: 5x 2 +33x 5x 2 +33x – 25x – = 0 x(5x+33) x(5x+33) – 5(5x+33) 5(5x+33) = 0 Solutions Solutions are can’t be negative Leg1 can’t be negative, choose x = 5 Factor: and x = 5 Leg1 = 5 inches Leg2 = 12 inches Leg2 = 2x+2 = = 13 2 Check: = 13 2 x 2+ 4x 2 4x 2 + 8x + 4 = x = a 2 + b 2 = c 2 Pythagorean Theorem: a 2 + b 2 = c 2

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 inches inches. What are the dimensions of the triangle? a 2 + b 2 = c 2 Pythagorean Theorem: a 2 + b 2 = c 2 The hypotenuse hypotenuse is compared to to the short leg leg so we will assign the variable x to the short leg leg and find the relationship relationship for the long leg 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 x = (2x –– –– 3) 2

Square Binomial Binomial: x = 4x 2 4x 2 – 12x + 9 Equation Equation: x = (2x – 3) 2 GN =  3 9595 dif = 4 Rearrange equation: equation: 3x 2 3x 2 – 12x – 135 = 0 3(x – 9)(x + 5) = 0 Scratch: x 2 x 2 – 9x 9x + 5x – = 0 x(x x(x – 9) 9) + 5(x 5(x – 9) 9) = 0 can’t be negative Leg1 can’t be negative, choose x = 9 Factor: Solutions Solutions are x = 9 and x = - 5 Short Leg = 9 inches Hypotenuse = 15 inches – Hypotenuse = 2x – 3 = = 15 2 Check: = 15 2 a 2 + b 2 = c 2 Pythagorean Theorem: a 2 + b 2 = c 2 Distributive Property Property: 3(x 2 3(x 2 – 4x – 45) = 0

Business Problems Cost, though a necessity, is considered a negative quantity quantity. Cost can be fixed, such as rent on the building, or variable, depending on the items, x, produced. Sometimes the price price, 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 expressions expressions are called the Demand Demand. Revenuex items price, p Revenue, R, is the money earned on x items, each sold at a certain price, p. Business problems problems will be encountered in economics economics, statistics statistics, applied calculus calculus or just everyday life life. It is necessary to learn to read and interpret these problems. Vocabulary in business includes cost cost and revenue revenue.

number price price revenue A company manufactures diskettes for home computers. It knows from past experience that the number of diskettes it can sell each day, x, is related to the price p per diskette by the equation x = 800  100·p. At what price, p, should the company sell the diskettes if it wants the daily revenue to be $1,200? Business Problem Formula for Revenue Revenue: R = p ·x or p ·x ·x = R Demand for diskettes diskettes: x = 800  100·p Revenue Equation: Equation: p ·(800  100·p )= $1,200

Replacement Property Property: 800p  100p 2 = 1200 Distributive property property: Formula for Revenue Revenue: p · x = R p ·(800  100·p ) = $1, p 2  800p = 0Add opposites opposites: p 2  8p + 12 = 0Multiply reciprocal reciprocal of : (p  6)(p  2) = 0Factor: p = 6 or or p = 2 Solve the equation 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 THE END