Synergy for Success in Mathematics 9 is designed for Grade 9 students. The textbook contains all the required learning competencies and is supplemented with some additional topics for enrichment. Lessons are presented using effective Singapore Math strategies that are intended for easy understanding and grasp of ideas for its target readers. Various exercises are also provided to help the learners acquire the necessary skills needed. The book is organized with the following recurring features in every chapter: Learning Goals Introduction Historical Note This gives the specific objectives that are intended to be achieved in the end. The reader is given a bird's eye view of the contents. A brief historical account of a related topic is included giving the reader an awareness of some important contributions of some great mathematicians or even stories of great achievements related to mathematics. Method/Exam Notes These additional tools help students recall important information, formulas, and shortcuts, needed in working out solutions. Examples Enhancing Skills Linking Together Chapter Test Chapter Project Making Connection PREFACE Step-by-step and detailed demonstrations of how a specific concept or technique is applied in solving problems. These are practice exercises found after every lesson, that will consolidate and reinforce what the students have learned. This visual tool can help the students realize the connection of all the ideas presented in the chapter. This is a summative test given at the end in preparation for the expected actual classroom examination containing the topics included in the chapter. A challenging task is designed for the learner giving him an opportunity to use what he/she has learned in the chapter. This may be a manipulative type of activity that is specifically chosen to enhance understanding of the concepts learned in the chapter. The students are exposed to facts and information that connect mathematics and culture. This is for the purpose of letting the learners appreciate the subject because of tangible or true-tolife stories that show how mathematics is useful and relevant. Every effort has been made in order for all the discussions in this book to be clear, simple, and straightforward. This book also gives opportunities for the readers to see the beauty of mathematics as an essential tool in understanding the world we live in. With this in mind, appreciation of mathematics goes beyond seeing; realizing its critical application to decision making in life completes the purpose of knowing and understanding mathematics.
Table of C ntents CHAPTER 1 QUADRATIC EQUATIONS AND THEIR ROOTS Introduction...1 Historical Note... 1.1 Quadratic Equations...3 1. Solving Quadratic Equations by Extracting Square Roots...9 Imaginary Roots... 1 1.3 Solving Quadratic Equations by Factoring... 16 1.4 Solving Quadratic Equations by Completing the Square... 4 1.5 The Quadratic Formula... 34 1.6 Studying the Roots of a Quadratic Equation... 43 The Nature of the Roots and the Discriminant of Quadratic Equations... 43 Deriving the Quadratic Equation Given its Roots... 45 1.7 Solving Problems Involving Quadratic Equations... 56 Rational Equations Leading to Quadratic Equations... 56 Equations Involving Integral Exponents Leading to Quadratic Equations... 59 Application of Quadratic Equations... 6 Linking Together... 7 Chapter Test... 71 Chapter Project... 73 Making Connection... 74 CHAPTER QUADRATIC FUNCTIONS AND THEIR GRAPHS Introduction... 75 Historical Note... 76.1 Quadratic Functions and Their Graphs... 77 The Standard Form of Quadratic Function... 77 The Graph of a Quadratic Function... 78. The Zeros of Quadratic Functions...19 The Zeros of a Quadratic Function...19 The Graphical Method of Finding the Zeros of a Quadratic Function...11
The Algebraic Method of Finding the Zeros of a Quadratic Function...114 Discriminant and the Nature of Zeros...14.3 The Vertex Form and Shifting Parabolas...139 The Vertex Form of a Quadratic Function...139 The Domain and Range of a Quadratic Function...14 Shifting Parabolas...144.4 Deriving Quadratic Functions...157.5 Problem Solving Involving Quadratic Functions...165.6 Quadratic Inequalities...17 Linking Together...183 Chapter Test...184 Chapter Project...187 Making Connection...188 CHAPTER 3 VARIATIONS Introduction...189 Historical Note...19 3.1 Ratio and Proportion...191 Ratio and Percentage...191 The Fundamental Property of Proportion...194 Rate...199 3. Direct and Inverse Variations...4 Variation...4 Direct Variation...4 Inverse Variation...15 3.3 Joint and Combined Variations...8 Joint Variation...8 Combined Variation...3 3.4 Solving Problems Involving Variations...39 Linking Together...51 Chapter Test...5 Chapter Project...55 Making Connection...58
CHAPTER 4 RATIONAL EXPONENTS AND RADICAL EXPRESSIONS Introduction...59 Historical Note...6 4.1 Roots of Numbers and Radical Expressions...61 Squares and Square Roots...61 Radical Expressions...65 Cubes and Cube Roots...68 The Principal nth Root...69 4. Expressions with Rational Exponents...76 Rational Exponents...76 Positive Rational Exponents...78 Negative Rational Exponents...8 4.3 Operations Involving Radical Expressions...89 Simplifying Radical Expressions Using Rational Exponents...89 Multiplying Radical Expressions...91 Simplifying Radical Expressions Using Factoring...9 Dividing Radical Expressions...95 Rationalizing Denominators with One Term...97 4.4 More about the Operations Involving Radical Expressions...33 Adding and Subtracting Radical Expressions...33 Multiplying Radical Expressions with Two or More Terms...36 Multiplying and Dividing Radical Expressions with Different Indices...37 Rationalizing the Denominator with Two Terms...31 4.5 Equations Involving Radical Expressions and Applications...315 Radical Equations Leading to Linear Equations...315 Radical Equations that Lead to Quadratic Equations...318 Solving Problems Involving Expressions with Rational Exponents and Radical Expressions...31 Linking Together...36 Chapter Test...37 Chapter Project...33 Making Connection...33
CHAPTER 5 SIMILARITY AND RIGHT TRIANGLES Introduction...333 Historical Note...334 5.1 Proportionality and Similarity...335 Review on Ratio and Proportion...335 Geometric Mean...339 The Side-Splitter Theorem...341 Similarity of Polygons...345 5. Triangle Similarity Theorems...353 5.3 About Right Triangles...37 Proportions in a Right Triangle...37 The Pythagorean Theorem...376 5.4 Special Right Triangles...386 Linking Together...395 Chapter Test...396 Chapter Project...43 Making Connection...44 CHAPTER 6 QUADRILATERALS Introduction...45 Historical Note...46 6.1 Review on Quadrilaterals...47 The Sum of the Measures of the Interior Angles of a Quadrilateral...49 Types of Quadrilateral...411 6. Parallelograms...419 Quadrilaterals That Are Parallelograms...44 6.3 On Some Special Parallelograms...43 Rectangles...43 Rhombi...433 Squares...434 6.4 Trapezoids and Kites...438 Trapezoids...438 Kites...44 Linking Together...451 Chapter Test...45 Chapter Project...457 Making Connection...458
CHAPTER 7 TRIGONOMETRIC FUNCTIONS Introduction...459 Historical Note...46 7.1 Angle Measure...461 Angles...461 Degree Measure...46 Coterminal Angles...464 Radian Measure...467 Relation of Degree Measure and Radian Measure...469 Reference Angles...471 Degree-Minute-Second Form of an Angle...474 7. The Six Trigonometric Functions...481 Trigonometric Functions of 45º...483 Trigonometric Functions of 3º and 6º...484 Trigonometric Identities...485 7.3 Trigonometric Functions of any Angle...495 7.4 Applications of Trigonometry...56 Solving a Right Triangle...56 Angles of Elevation and Depression...57 7.5 Laws of Sines and Cosines...515 Linking Together...531 Chapter Test...53 Chapter Project...535 Making Connection...536 Glossary...537 Index...544 Bibliography...548 Photo Credits...55
1 QUADRATIC EQUATIONS AND THEIR ROOTS Learning Goals At the end of the chapter, you should be able to: 1.1 1. 1.3 The famous and remarkable shoe industry in Marikina gives pride to the Philippines. Known as the shoe capital of the country, the city was once recognized by the Guinness World Records because of the world s largest pair of shoes that the shoemakers in the city made. Also, the shoe industry helps alleviate the unemployment problem. The production of shoes depends on the number of employees and profit. To help the manufacturers and shoemakers to maximize production and profit, a quadratic equation can be used as a model. In this chapter we study the fundamental concepts revolving around quadratic equations. We find ways and methods to solve problems involving quadratic equations. We also look at the connection of the quadratic equation and its roots. 1.4 1.5 1.6 1.7 Define quadratic equation and simplify quadratic equations in one variable into their standard form Establish the square root property and solve quadratic equations by extracting square roots Recall factoring techniques and apply those techniques to solve quadratic equations Establish the method of completing the square and solve quadratic equations by completing the square Derive the quadratic formula using completing the square and solve quadratic equations by applying the quadratic formula Describe the nature of the roots using the discriminant of a quadratic equation, establish the relationship of the quadratic equation to the sum and product of its roots, and form quadratic equations using the roots Solve problems involving quadratic equations using different methods
Historical Note As early as BC, Babylonians used cuneiform, a system of writing on clay tablets, to write algebraic and geometric problems. There are clay tablets showing evidence that the Babylonians were familiar with today s way of solving quadratic equations. Their solution is generally known today as "completing the square" method of solving quadratic equations. Aside from the Babylonians, other ancient civilizations formulated and studied quadratic equations. The ancient Egyptians listed the areas of possible sides and shapes. Pythagoras (569 5 BC), a Greek philosopher and mathematician, formulated the Pythagorean theorem which states that the square of the length of the hypotenuse of a right triangle is equal to the sum of the squares of the lengths of the other two sides.
Synergy for Success in Mathematics 1.1 Chapter 1 Quadratic Equations Certain problems in mathematics and science frequently lead to situations involving quadratic equations. Consider the following problem. May and Paul want to put a fence around their rectangular garden. The width of the garden is feet shorter than its length. Also, the area of the garden is 8 square feet. How long should the fence be? Let x be the length of the garden. Thus, the width measures ( x ) feet. Since the area of the rectangular garden is 8 square feet, x ( x ) = 8. When the distributive property and the addition property of equality are applied, the simplest form of the equation is shown as follows: Exam Note The area of a rectangle is equal to the product of its length l and width w; that is, A = lw. x ( x ) = 8 x x = 8 x x 8 = Notice that the highest degree of the resulting equation is, which makes the equation not linear. The resulting equation is an example of a second-degree equation, also known as quadratic equation. However, this chapter focuses on quadratic equation in one variable. Exam Note A linear equation is an algebraic equation whose highest degree among the terms is 1. 3
A Quadratic Equation in One Variable A quadratic equation in one variable is a seconddegree equation written in the standard form ax + bx + c = where a, b, and c are real numbers and a ¹. Exam Note The term quadratic is derived from the Latin word quadratus, which means squared. Other examples of quadratic equations in one variable are as follows: 3x + x 5= This signifies that the equation is quadratic. The quadratic equation has only one variable, x. 5m m+ 1= This signifies that the equation is quadratic. The quadratic equation has only one variable, m. Example 1 Identify whether the equation is a quadratic equation in one variable or not. (a) x + 5x 7+ x = (b) x + y = 1 (c) y( 3y 5y + 3)= (d) a( a 1)= a 1 (e) ( b+ 3) ( b 3)= b 9 ( ) SOLUTION Simplify each equation to determine whether the equation is a quadratic equation in one variable or not. (a) x + x + 5x 7= 3x + 5x 7= The resulting equation is a quadratic equation in one variable. Method Note Always express an algebraic equation in standard form to determine if the equation satisfies the characteristics of a quadratic equation in one variable. 4
Synergy for Success in Mathematics Chapter 1 (b) The equation x + y = 1 is a quadratic equation since the highest degree among the terms is. However, it is not expressed in one variable. (c) y( 3y 5y + 3)= 3 6y 1y + 6y = Exam Note Not all quadratic equations are expressed in one variable. Some have two. The highest degree among the terms is 3. Hence, it is not a quadratic equation. (d) a( a 1)= a 1 a a a= a 1 a+ 1= The resulting equation is a quadratic equation in one variable. (e) ( b+ 3) ( b 3)= b 9 b 9= b 18 b + 9= b 9= ( ) The highest degree among the terms is. The equation is expressed in one variable. Hence, it is a quadratic equation in one variable. When a quadratic equation in one variable is written in standard form, you can easily identify the numerical coefficients of each term. For example, the equation x + 5x 6= is already in standard form. The numerical coefficients of each term of the equation are indicated below. x + 5x 6= The constant term is -6. The numerical coefficient of 5x. is 5. The numerical coefficient of x is 1. 5
Based on the standard form of quadratic equation in one variable, you can represent the numerical coefficient of the first term (the term with the highest degree) as a, the numerical coefficient of the middle term (the term in linear degree) as b, and the constant term as c. Hence, the values of a, b, and c in the equation x + 5x 6= are 1, 5, and -6, respectively a= 1; b= 5; and c = 6. Example Simplify each quadratic equation into its standard form. Identify the values of a, b, and c. (a) 3x x+ 7= (b) 4 3x x = (c) 5+ x = 3x SOLUTION (a) The quadratic equation 3x x+ 7= is already in standard form. Thus, a= 3, b= 1, and c = 7. (b) The terms in 4 3x x = need to be rearranged. The commutative property of addition should be applied. 4 3x x = x 3x + 4=. Thus, a=, b= 3, and c = 4. (c) Based on the addition property of equality and the commutative property of addition, the standard form of the equation is indicated as follows: 5+ x = 3x 5+ x + 3x= x + 3x+ 5= Hence, a=, b= 3, and c = 5. 6
Synergy for Success in Mathematics Chapter 1 ENHANCING SKILLS A Write QUADRATIC if the equation is a quadratic equation in one variable. Otherwise, write EQUATION. Justify your answer. (1) 4x + x 7= : () (3) x y = 1 : 4 9 x x 4 + 6 8= : ( ) = (4) x+ 5 x 7 (5) 4a + 4ab+ b = : (6) ( x+ y) ( x y)= : (7) ( x+ 1) ( x 1)= 4 : (8) 5x( x 8)= 1 : (9) 5( a 1) ( a+ 3)= 7 : (1) 5x( x + 5x 6)= : : B Simplify each quadratic equation into its standard form. Then, identify the values of the numerical coefficients a, b, and c. Express the leading coefficient a as a nonnegative real number. (11) 5x x = 9 Standard form: a = b = c = (1) x( x+ 1)= 8 Standard form: a = b = c = (13) 8 x + 13x = Standard form: a = b = c = 7
(14) ( x+ 1) ( x 3)= Standard form: a = b = c = (15) ( x 1) ( x+ 3)+ 6= Standard form: a = b = c = (16) ( y+ 7) ( y 7)= Standard form: a = b = c = ( ) = (17) y 7 4 Standard form: a = b = c = ( ) = (18) y + 4 16 Standard form: a = b = c = (19) mm ( 7)= Standard form: a = b = c = () ( m+ ) ( m )= ( m+ 1) m 1 Standard form: a = b = c = ( ) 8
Synergy for Success in Mathematics Chapter 1 1. Solving Quadratic Equations by Extracting Square Roots The values that satisfy the quadratic equation are called solutions. Solutions are called roots of the equation. Consider the equation x = 4. The number 4 has two real square roots, and -. These values are the solutions or roots of the equation. You may verify if the values satisfy the equation by substituting x. = 4 4= 4 ( ) = 4 4= 4?? Square roots provide a fast solution in determining the roots of a quadratic equation of the form X = d. Square Root Property Let X be an algebraic expression and d be any real number. If X = d, then X = d or X = d. The expression ± d represents two roots: d and - d. Exam Note Note that X can be a monomial, binomial, or any polynomial. Solving Quadratic Equations by Extracting Square Roots Step 1 Write the given quadratic equation in the form X = d. Step Apply the square root property. Step 3 Substitute the acquired values for the X variable in the given equation to check if they satisfy the equation. 9
Example 1 Solve x = 16. SOLUTION Follow the steps in solving quadratic equations by extracting the square roots. Step 1 The given equation is already in the form X = d. Step Apply the square root property. x = 16 x =± 16 x =± 4 Step 3 Substitute the acquired values for x in the given equation. x = 16 x = 16? 4 = 16 ( 4) =? 16 16 = 16 16 = 16 Method Note Make sure to always check the values you acquired. Some values may not be the solution for the given equation. Both -4 and 4 satisfy the original equation. Hence, the roots of x = 16 are -4 and 4. Example Solve 5x = 3. SOLUTION Step 1 Translate the equation in the form X = d. 5x = 3 5x 3 = Divide both sides by 5. 5 5 x = 6 Step Apply the square root properly x = 6 x =± 6. 1
Step 3 Check the acquired values if they satisfy the given equation. Substitute - 6 for x in the equation. 5x = 3? ( ) = 5 6 3? 56 ( )= 3 = 3 3 Synergy for Success in Mathematics Chapter 1 Substitute 6 for x in the equation. 5x = 3? ( ) = 5 6 3? 56 ( )= 3 3 3 = Hence, the roots of the equation are - 6 and 6. A solution set is the collection of all the roots of a given equation. Every quadratic equation in one variable has at most two solutions or roots. Therefore, the solution set of a quadratic equation consists of at most two elements. Example 3 ( ) = Solve a 1 3. SOLUTION Step 1 The equation is already in the form X = d. where X = a 1. Step Apply the square root property. Then, add 1 to both sides. ( a 1) = 3 a 1=± 3 a 1+ 1=± 3+ 1 a = 1± 3 11
Step 3 Check the acquired values if they satisfy the given equation. Substitute 1-3 for a. a 1 3 ( ) =? ( 1 3 1) = 3? ( 3) = 3 3= 3 Substitute 1+ 3 for a. ( a 1) = 3? ( + ) = 1 3 1 3? ( ) = 3 3 3 3 Hence, the solution set is 1 3, 1+ 3. = { } Imaginary Roots Consider the equation x = 4. When the square roots of the equation are extracted, the resulting equation is x =± 4. In this case, the roots of the equation are not real numbers. To resolve this problem, mathematicians formulate a new kind of number. We call such numbers as imaginary numbers. Imaginary Number An imaginary number is a number that can be expressed as the product of a real number and an imaginary unit i. The imaginary unit i is defined by i = 1, or simply i= 1. Examples of imaginary numbers are i, 5i, - i, and 15i and many more. 1
You can simplify -4 as follows: 4= ( 4) ( 1) =( 4) ( 1) =( )() i = i Synergy for Success in Mathematics Chapter 1 Exam Note Since imaginary numbers are not real numbers, they cannot be plotted on a real number line. Hence, the square root of -4 is i. The roots of the equation x = 4, are -i and i. Example 4 Find the roots of 5x + 8=. SOLUTION Step 1 Translate 5x + 8=. into the form X = d. 5x + 8= 5x = 8 5x 8 = 5 5 x = 16 Step Apply the square root property. x = 16 x =± 16 Step 3 Substitute the acquired values. 16 = ( 16) ( 1) = 4i Thus, the roots of 5x + 8= are 4i and 4i. The equation has two imaginary roots. 13
ENHANCING SKILLS A Find the roots of each quadratic equation. (1) x = 5 ( ) = (9) x + 1 1 () x 36 = 45 ( ) = (1) 5 x 1 45 (3) x = 7 ( ) = (11) x + 3 81 (4) x 1= ( ) + = (1) x + 5 1 (5) x 196 = x (6) 5x = 15 ( ) = (13) x 36 5 (7) x 98= ( ) = (14) x + 1 5 (8) 3x + 15= 6 x (15) 3x + 1= 6 14
B Solve each problem. Synergy for Success in Mathematics Chapter 1 (16) Five times the square of a positive integer is 65. What is the integer? (17) The square of the sum of a number and 1 is 5. What is the smallest possible value of the number? (18) Solve 4x + 4x 5= 6 by extracting square roots. (19) Gena is working on her math project. She has trimmed the sides of a square cardboard by 7 decimeters. The area of the cardboard then has decreased by 81 square decimeters. What was the original length of a side of the cardboard? () Drei wanted a bigger garden lot to accommodate more variety of plants. So, she extended each side of the square lot by 5 meters. Now, the lot has an area of 5 square meters. What was the original area of the garden lot? 15
1.3 Solving Quadratic Equations by Factoring The following example shows some factoring techniques that can be used in solving quadratic equations. Example 1 Factor each polynomial. (a) x + 4x (b) x + 6x+ 9 (c) x -4x- 1 (d) 6x -15x- 9 (e) 3x -18 SOLUTION (a) Factor the binomial by common monomial factoring. x 4x x x x + = ( )+ () ( ) = x x+ (b) The quadratic trinomial x + 6x+ 9 is a perfect square. Therefore, x 6x 9 x x 3 3 + + = + ( )()+ ( ) = x + 3 Method Note Common monomial factoring is based on the distributive property. ax ( + y+ z)= ax + ay + az Method Note Recall that ( ) = ± + x± y x xy y. (c) Factor the polynomial by using the cross method. x 4x 1 x 7 7x x + 3 3x x( x) 3( 7) 4x Hence, x 4x 1= ( x 7) ( x+ 3). (d) Apply common monomial factoring and the cross method in factoring 6x -15x- 9. 16
Synergy for Success in Mathematics Chapter 1 ( ) 6x 15x 9= 3x 5x 3 x + 1 x x 3 6x ( x 3) 5x = 3x+ 1 x 3 ( )( ) (e) Apply common monomial factoring and the special product formula for the difference of two squares. 3x 18 3 x 3 36 = ( ) ( ) ( ) = 3 x 36 ( )( ) = 3 x+ 6 x 6 Not all quadratic equations can be solved using the square root property. Instead, you can apply the factoring techniques and the zero product property to find the roots of the given quadratic equation. Zero Product Property If the product of two algebraic expressions is zero, then at least one of the factors is equal to zero. In other words, if AB =, then A= or B=. Below are the steps in solving quadratic equations by factoring. Solving Quadratic Equations by Factoring Step 1 Write the equation in standard form ax + bx + c =. Step Factor the quadratic trinomial of the equation. Step 3 Apply the zero product property. Step 4 Substitute the acquired values in the original equation to check if they satisfy the equation. 17
Example Determine the roots of the equation x SOLUTION x =. Step 1 The given equation is already in standard form. Step Factor the quadratic trinomial of the equation. x x = x 5 ( x 4)= ( ) + = Step 3 Apply the zero product property. x 5= or x+ 4= x= 5 or x= 4 Step 4 Substitute the acquired values in the original equation to check if they satisfy the equation. Substitute 5 for x. x x =? () ( ) = Substitute -4 for x. 5 5 5 5 = x?? 5 5= ( ) = x =? 4 ( 4) = Hence, the roots of x 16+ 4 =?? = = x = are 5 and -4. Method Note Remember that the quadratic trinomial of the equation must be in standard form before the zero product property is applied. 18
Synergy for Success in Mathematics Chapter 1 Example 3 Find the solution sets of the following quadratic equations. (a) 3x + x= 8 (b) 5x 5x= SOLUTION (a) Rewrite the equation in standard form. 3x + x= 8 3x + x 8= 8 8 3x + x 8= Factor the quadratic trinomial. You may use the cross method. 3x + x= 8 3 x + x 8= x 6x 3x 4 4x 3x 8 x Hence, ( x+ ) ( 3x 4)=. Find the values of x by applying the zero product property. x + = or 3x 4= x = 3x = 4 4 x = 3 Thus, the solution set is, 4. 3 (b) Extract the common factor 5x. 5x 5x= 5x( x 1)= 5x = or x 1= x = x = 1 The solution set is { 1, }. Method Note To check the solution set if its values satisfy the given equation, always substitute the acquired values in the equation. 19
We can also use the factoring method of solving quadratic equations in solving word problems. Example 4 Paulo plans to build a rectangular swimming pool in his backyard. The area of the swimming pool must be 96 square feet, and its perimeter must be 4 feet. Find the length and the width of the pool. SOLUTION Let l and w be the length and the width of the rectangular swimming pool, respectively. w l The area of the pool must be 96 square feet. Thus, lw = 96. (equation 1) Moreover, the perimeter of the pool must be 4 feet. Thus, l+ w= 4. (equation ) Using equation 1, find l in terms of w. lw = 96 lw 96 = Multiply both sides by 1 w w w. 96 l = w Translate the resulting equation into a quadratic equation. l+ w= 4 96 w 4 Apply the substitution w + = property. 19 + w = 4 Multiply by 96 w w. 19 w + w w Multiply both sides by w. w = ( 4 ) 19 w w w w 4 w + ( )= ( ) 19+ w = 4w w 4w+ 19= Apply the distributive property. Simplify the equation. Add -4w to both sides. Method Note The area of a rectangle is the product of its length and width. A = lw Method Note The perimeter of a rectangle is the sum of all its sides. P = l + w
Synergy for Success in Mathematics Chapter 1 Find the roots of the derived quadratic equation using factoring. w 4w+ 19= ( w w+ 96)= ( ) = w w+ 96 Bring out the common factor from the lefthand side. Divide both sides by. w w+ 96 = ( w 8) ( w 1)= Factor the quadratic trinomial. Now, find the value of w by applying the zero product property. ( w 8) ( w 1)= w 8= w = 8 or w 1= w = 1 There are two possible values of the width. Substitute each possible value for w in equation (1). l lw = 96 l( 8)= 96 ( ) = 8 96 8 8 l = 1 lw = 96 l( 1)= 96 ( ) = l 1 96 1 1 l = 8 Check the acquired values if they satisfy the equation. A= lw 1 8 =( )( ) = 96 ft P= l+ w = 1 ( )+ ( 8) = 4 ft Based on the acquired values, we have obtained the correct dimensions. The length must be 1 feet, and the width must be 8 feet. 1
ENHANCING SKILLS A Solve each quadratic equation by factoring. (1) x( 5x 3)= (9) 3x 5x = () ( 7x 5) ( 7x+ 5)= (1) 15x + 4x 3= (3) 4x 3 x 3 ( ) 6 = (11) 6x + 3x+ 7= (4) x x= (1) 5x 6x+ 5= (5) x 5x+ 6= (13) ( x+ 5) ( x 5)= 39 (6) x + 4x 1= (7) x + 1x+ 16 = (14) 9 x 1 = 16 (8) x + 3x = (15) ( 3x 4) ( x 7)=