Equations 5 The aim of this unit is to equip the learners with the concept of equations. The principal foci of this unit are degree of an equation, inequalities, quadratic equations, simultaneous linear equations, graphical equation and their applications in solving usiness prolems followed y eamples.
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Bangladesh Open University Lesson 1: Equation and Identity After studying this lesson, you should e ale to: Eplain the nature and characteristics of equations; Eplain the nature and characteristics of identities; Solve the equations; Solve the inequalities. Introduction Many applications of mathematics involve solving equation. In this lesson we will discuss the equation, identities and uses of equations. Equation An equation is a statement which says that two quantities are equal to each other. An equation consists of two epressions with a sign etween them. In other words, if two sides of an equality are equal only for particular value of the unknown quantity or quantities involved, then the equality is called an equation. For eample, 4 8 is true only for. Hence, it is an equation. An equation which does not contain any variale is either a true statement, such as + 3 5, or a false statement, such as 3 + 5 1. If an equation contains a variale, the solution set of the equation is the set of those values for the variale which gives a true statement when sustituted into the equation. For eample, the solution set of y 4 is (, ), ecause ( ) 4, and 4, ut y 4 if y is any numer other than or. Identities The equations signify relation etween two algeraic epressions symolized y the sign of equality. If two sides of an equality are equal for all values of the unknown quantity or quantities involved, then the equality is called an identity. For eample, y ( + y) ( y) is an identity. We can prove that identities hold true for whatever are values of the variales sustituted in these. It we use and y 3 in the aove identity, we have () (3) ( + 3) ( 3) or, 4 9 (5) ( 1) or, 5 5 Again, y sustituting the values of 4 and y 6, we have ( 4) ( 6) ( 4 6) ( 4 + 6) or, 16 36 ( 10) () or, 0 0 Hence, identities hold true for whatever value is put for variales. An equation is a statement which says that two quantities are equal to each other. If two sides of an equality are equal for all values of the unknown quantity or quantities involved, then the equality is called an identity. Business Mathematics Page-91
School of Business Derived identities are the identities derived y transposing the values in the asic identities and are very useful in tackling some prolems in mathematics. Derived Identities Derived identities are the identities derived y transposing the values in the asic identities and are very useful in tackling some prolems in mathematics. For eample, (1) Identity ( + y) + y + y... (i) Derived Identities + y ( + y) y and y ( + y) ( + y ) () Identity ( y) y y... (ii) Derived Identities + y ( y) + y and y + y ( y) By adding (i) and (ii) ( + y) + ( y) ( + y )... (iii) By sustituting (ii) from (i), we get ( + y) ( y) 4y... (iv) By dividing oth (i) and (ii) y 4 and then sutracting (ii) from (i), we have [( + y) / 4] [( y) ] The following section of this lesson contains some model applications of equations. Eample 1: Solve, 1 + 1 + + 1 1 1 3 1 + 1 + + 1 1 1 3 By cross multiplying, we have 3 1 + 3 1 1 + + 1 or, or, 3 1 + 3 1 + 1 + 1 1 + 4 1 Squaring oth sides, we have, 4(1 + ) 16 (1 ) or, 4 + 4 16 16 Transposing the term 16 and 4 4 + 16 16 4 or, 0 1 Unit-5 Page-9
Bangladesh Open University or, 5 3 Therefore, 5 3 is the solution of the given equation. Eample : The sum of two numers is 45 and their ratio is 7:8. Find the numers. Let, one of the numers e The other numer is (45 ) Using the given information, we get, 45 7 8 By cross multiplication, we get 8 7 (45 ) or, 8 315 7 Transposing the term 7, we have 8 + 7 315 or, 15 315 315 or, 1 15 Hence, the one numer is 1 and the other numer is (45 1) 4. Eample 3: The ages of a mother and a daughter are 31 and 7 years respectively. In 3 how many years will the mother s age e times that of the daughter? Let the required numer of years e. Mother s age after years 31 + Daughter s age after years 7 + Using the given information, we get 3 31 + (7 + ) 1 + 3 or, 31 + By cross multiplication, we have (31 + ) 1 + 3 or, 6 + 1 + 3 Business Mathematics Page-93
School of Business Transposing the terms 3 and 6 3 1 6 or, 41 or, 41 Hence, in 41 years, the mother s age will e daughter s. 3 times age of the Eample 4: The distance etween two stations is 340 km. Two trains start at the same time from these two stations on parallel tracks to cross each other. The speed of one train is greater than that of other y 5 km / hr. If the distance etween the two trains after hours of their start is 30 km, find the speed of each trains? Let the speed of the first train e km / hr. Then the speed of the second train e ( + 5) km / hr. Distance covered y the first train in hrs km. Distance covered y the second train in hrs ( + 5) km ( + 10) km. Since, oth the trains are in opposite directions. Total distance etween the two stations 340 km. Distance etween the two trains after hours 30 km. Therefore, distance covered y two trains in two hours from opposite directions (340 30) 310 km + ( + 10) 310 or, 4 300 75 Hence, speed of the first train 75 km / hr and the speed of the second train 80 km / hr. Unit-5 Page-94
Bangladesh Open University Questions For Review: These questions are designed to help you assess how far you have understood and can apply the learning you have accomplished y answering (in written form) the following questions: 1. What is an equation and inequalities? Mention the characteristics of equation.. What is the difference etween identity and equation? Give eamples. 3. Solve the following equations: (i) ( + 1) + 7 / (+1) 18 (ii) 6 + 9 4 6 + 6 (iii) + 6 + + 6 5 1 4. Wasifa s mother is four times as old as Wasifa. After five years, her mother will e three times as old as she will e then, what are their present ages? 5. A steamer goes downstream and covers the distance etween two parties in 4 hours while it covers the same distance upstream in 5 hours. If the speed of the stream is 1 cm/per hour, find the speed of the steamer in still water. 6. Three prizes are to e distriuted in a quiz contest. The value of the nd prize is five-siths the value of the first prize and the value of the third prize is four fifths that of the second prize. If the total value of the three prizes is Tk.15,000, find the value of each prize. 7. The price of two cows and five horses is Tk.68,000. If the price of horse eceeds that of a cow y Tk.800, find the price of each. Multiple choice questions ( the appropriate answers) 1. The difference etween two numers is 9 and the difference etween the squres is 07. The numers are: a) 17, 8 ) 16, 7 c) 3, 14. The ratio of two numers is 3 : 5. If each is increased y 10 the ratio ecomes 5 : 7, the numers are: a) 6, 10 ) 15, 5 c) 9, 15 3. The sum of three numers is 10. If the ratio etween the first and the second e : 3 and that of etween the second and the third e 5 : 3, the second numer is a) 30 ) 45 c) 7 4. Taka 49 were divided among 150 children. Each girl got 50 paisa and each oy got 5 paisa. How much oys were there? a) 104 ) 105 c) 10 5. If + 3 5 and, then the value of y is: a) 1/3 ) 1 c) 3 Business Mathematics Page-95
School of Business Lesson-: Inequality After studying this lesson, you should e ale to: Descrie the nature of inequalities; Eplain the properties of inequality; Solve the inequalities. Nature of Inequality Relationship of two epressions with an inequality sign etween them is called inequality. Relationship of two epressions with an inequality sign ( or, < or >) etween them is called inequality. For eample, > y is greater than y < y is smaller than y > y is not greater than y < y is not smaller than y y is smaller than or equal to y y is greater than or equal to y Properties of Inequalities The fundamental properties of inequalities are as follows: (a) Order Aioms: If and are only elements, then (i) One and only one of the following is true:, < y and > y (ii) If < y and y < z, then y < c (iii) If < y and < z, then z < yz Since, > y and y < are the same statements, the aove aioms can e replaced in terms of > y. As shown earlier sometimes equality signs are comined with inequality signs < y means y or < y. Again < y means is not less than y and that means either > y. So, < y means y. We also say that is positive when 0 and is negative, when < 0. All equals may e add or sutracted from oth sides of inequalities and the inequality is preserved. () Operation Aioms: (i) All equals may e add or sutracted from oth sides of inequalities and the inequality is preserved. For eample, if 5 9 < 1 We may add 5 to oth sides and we get 5 9 + 5 < 1 + 5 or, 5 4 < 17 Unit-5 Page-96
Bangladesh Open University (ii) Any term in an inequality can e moved from one side to the other provided that its sign is changed. For eample, if 5 4 < 17. or, 5 < 17 + 4 And, again if z > y or, > y + z Both sides of an inequality may e multiplied or divided y a positive numer and the inequality is preserved. For eample, if 1 < 36. After Multiplying the oth sides of inequality y 5, we get 1 5 < 36 5 or, 60 < 180 Again, after dividing oth sides of inequality y 3, we get 1 3 < 36 3 or, 4 < 1 (iii) Both sides on an inequality are multiplied or divided y a negative numer and the direction of the inequality is reversed. For eample, if 7 < 40, (where 4). (iv) (v) By multiplying oth sides of the inequality y 5, we have 7 ( 5) > 40 ( 5) < to >] or, 35 > 00 [Note that inequality sign has een changed from This is ecause when 4, the inequality 35 140 is greater than 00. An inequality can y converted into an equation: If > y then y + p Where p is the positive real numer (i.e. p > 0) If z > m, then we write z m + q, where q > 0 Hence,, z (y + p) (m + q) ym + yq + pm + pq Now p and q are positive. If in addition y and m are positive, then every term on the right-hand side is also positive so that. z > y. m If y z m >, then < z m y Both sides of an inequality may e multiplied or divided y a positive numer and the inequality is preserved. Both sides on an inequality are multiplied or divided y a negative numer and the direction of the inequality is reversed. (vi) If < y, then > y (vii) Now if 1 > y 1, > y, 3 > y 3 ----------- n > y n, then 1 + + 3 + ----------- + n > y 1 + y + y 3 + --------- +y n and 1.. 3 --------- n > y 1. y. y 3 ----------- y n Business Mathematics Page-97
School of Business (viii) If > y and n > 0 then n > y n and 1 n < The following section of this lesson contains some applications of inequality. Eample-1: Solve: 3[4 5( 3)] 7 [ + 3 (4 )] 3[4 5( 3)] 7 [ + 3 (4 )] or, 3[4 10 15)] 7 [ + 1 3)] or, 1 30 45 7 + 4 + 6 Transposing oth sides we have or, 1 30 + 6 4 45 or, 1 6 Multiplying oth sides y 1, we get or, 6 31 or, 11 1 y n Eample-: Solve: 5 (3 4) > 4[ 3 (1 3)] 5 (3 4) > 4[ 3 (1 3)] or, 5 6 8) > 8 1 +36 or, 5 6 8 36 > 1 8 or, 45 > 0, Multiplying oth side we get 1, or, 45 < 0 0 or, < 45 i.e. < 9 4 Unit-5 Page-98
Questions for review Bangladesh Open University These questions are designed to help you assess how far you have understood and can apply the learning you have accomplished y answering (in written form) the following questions: 1. Solve the following inequalities: (i) (ii) 3 < 4 + 6 3(4 ) < 7 4 (1 ) (iii) 3 + 4 (5 3) 4 19 (iv) 3 [5 + 11 (3 + )] 11 3 [ 5 (3 )] (v) [(5 7) / ( 3)] (3 4) < 0. Solve each inequality with a sign graph (i) ( + 3) ( + 4) < 0 (ii) ( 1) ( ) ( 3) < 0 Multiple choice questions ( the appropriate answers) 1. If + y 3, > 0 and y > 0, then one of the solution is a) 1, y ) 1, y 1 c), y 1. The system of linear in equalities + y 0, 0 and y 0, has a) eactly 1 solution ) 3 solutions c) no solution 3. The value of from linear inequality 4 7 > 6 + 5 a) < 6 ) < 8 c) < 4 Business Mathematics Page-99
School of Business Lesson-3: Degree of an Equation After studying this lesson, you should e ale to: Eplain the nature of degree of an equation; Solve the simultaneous linear equation. Nature of Degree of an Equation An ordinary equation involving only the first power of the unknown quantity is called simple or linear equation or equation of the first degree. An equation involving only one unknown quantity is called ordinary equation. An ordinary equation involving only the first power of the unknown quantity is called simple or linear equation or equation of the first degree. When the highest power of the unknown quantity is, it is called quadratic or the second degree equation; when the highest power of the unknown quantity is 3, the equation is termed as cuic or the third degree equation. When the highest power of is 4, the equation is called iquadratic or the fourth degree equation. For eample, + 18 y Linear equation + 5 + 7 0 Quadratic equation 3 + 5 + 3 + 9 0 Cuic equation 4 + 10 3 + 5 + + 10 35 Biquadratic equation If an equation in is unaltered y changing to 1, it is known as a reciprocal equation. An equation in which the variale occurs as indices or eponents is called an eponential equation. For eample, 3 1, 81 9 +4 etc. are called eponential equations. If more than one unknown quantity are involved, the numer in independent equations required for solution is equal to the numer of the unknown quantities. Such set of linear equations is called simultaneous linear equations. The following section of this lesson contains some model applications. Eample-1: Solve 4 4 16 3 + 3 16 + 4 0 4 4 16 3 + 3 16 + 4 0 Rearranging the terms, we have 4 4 4 16 3 16 + 3 0 Dividing oth sides y we have, 4 + 4 16 16 + 3 0 Unit-5 Page-100
Bangladesh Open University 1 1 or, 4 + 16 + + 3 0 1 1 1 or, 4 +.. 16 + + 3 0 Putting y for + 1, we have 4 (y ) 16 (y) + 3 0 or, 4y 8 16y + 3 0 or, 4y 16y + 15 0 or, 4y 10y 6y + 15 0 or, y (y 5) 3 (y 5) 0 or, (y 5) (y 3) 0 either, y 5 0 or, y 3 0 or, y 5 or, y 3 or, y 5 5 1 5 When, y, then + or, + 1 5 or, + 5 or, 4 + 0 or, ( ) 1 ( ) 0 or, ( 1) ( ) 0 or, y 3 either, 1 0 or, 0 or, 1 1 3 1 3 When, y, then + + 1 3 or, + 3 or, 3 + 0 Business Mathematics Page-101
School of Business We know that ± 4ac (Here, a, 3, c ) a ( Here, 1) 3 ± 9 16 3 ± 7 3 ± 7i i 4 4 4 Hence, Eample-: 4 Solve + 1 y 1, or, 3 ± 7i 4 4 y + 5 4 + 1... (i) y 4 y + 5... (ii) 4 From equation (i) + 1 y or, y + 4 y 1 4 From equation (ii) y + 5 or, y + 4 y... (iii) y + 4 or, 5 or, y + 4 5... (iv) Sutracting the equation (iii) from equation (iv) 0 5 y or, y 5 Putting the value of y in equation (iii), we get (5) + 4 5 or, 5 + 4 5 or, 5 5 + 4 0 or, 5 0 5 + 4 0 or, 5 (5 4) 1 (5 4) 0 or, (5 4) (5 1) 0 either, 5 4 0 or, 5 1 0 or, 5 4 or, 5 1 Unit-5 Page-10
Bangladesh Open University so, 5 4 so, 5 1 Now 5 4, then y 5 5 4 0 5 1, then y 5 5 1 5 Thus, 5 4, 5 1 y 0, y 5 Eample-3: Find the solution of the system of equations. 4 3y + z 1... (1) y + z 6... () 3 + 4y 4z 1... (3) Let us first eliminate z: We rewrite: (1): 8 6y + z (): y + z 6 Sutraction: 6 5y 4... (4) Now we rewrite: (3) : 3 + 4y 4z 1 (): 4 y + 4z 1 Addition: 7 + y 11... (5) Let us now eliminate y from (4) and (5); i.e. 5 (5) : 35 10y 55 (4) : 1 10y 8 Addition: 47 47 So, 1. Now, sustituting the value of in (5), we get the value of y, i.e. 7 (1) + y 11 or, y 11 7 or, y 4 so, y Now, sustituting the value of and y in (1), we have: 4 (1) 3 () + z 1 or, z 1 4 + 6 or, z 3 Therefore the solution is: 1, y and z 3. Business Mathematics Page-103
School of Business Questions for Review: These questions are designed to help you assess how far you have understood and apply the learning you have accomplished y answering (in written form) the following questions: 1. Eplain the nature of degree of an equation.. Solve the following equations: (i) 4 3 3 3 + 1 0 (ii) y + y 3 y 3 (iii) 3 + 7 5z 0 3. Solve 7 3y 9z 0 + y + 3z 3 (i) 3 + y 3 4914 (ii) + y 18 7 a y 81 y 43.3 Multiple choice questions ( the appropriate answers) 1. The solution of the simultaneous linear equation and y 3 3 1 y 3 6 a), y 3 ), y 3 c), y 3. The solution of the system of simultaneous linear equation 4 3y 7 and 7 + 5y is: a) 1, y 1 ) 1, y 1 c) 1, y 1 3. If 7 + 9y 85 and 4 + 5y 48, then a) 7, y 4 ) 7 40, y 5 c) 37, y 8 3 6 4. The solution of the pair of equations + 1 and y 4 1 + 44 is: y a) 1, y 3 ), y 3 c) 1 1, y 3 Unit-5 Page-104
Lesson-4: Graphical Equation After studying this lesson, you should e ale to: State the nature of graph; Solve the equation with the help of graph. Graphical Equation Bangladesh Open University In this lesson, we will discuss aout graphical equations with the help of following practical eamples. Eample 1: Solve the equation graphically, +5y 1 and y 1 +5y 1...(i) y 1...(ii) From equation (i), 5y 1. 1 So, y 5 So for 0, 1, and 3; y.4,, 1.6 and 1. respectively. Plotting the points (0,, 4), (1, ), (, 1.6), (, 1.6), (3, 1.) and jointing them, we otain graph of + 5y 1, which is AB in the following figure - 5.1. Again, from equation (ii), y 1 So, y 1+ So, for 0, 1, and 3; y 1,, 3 and 4. Plotting the points (0, 1), (1,), (3, 4) and jointing them, we otain graph of y 1, which is CD in the following figure-5.1 Y A D X X O M C Y Figure-5.1 B Business Mathematics Page-105
School of Business From the aove graph it is clear that AB is the line of equation (i) and CD is the line of equation (ii) they are intersecting each other at the point M. The co-ordinates of M are (1, ). Hence the required equation is 1 and y. This gives the solution of 1, y for the pair of simultaneous equations + 5y 1 and y 1. Eample : Draw the graph and solve the following equations; 4y y + 11 and 4 6y + 0 The given equations are 4 y + 11 0 or, y 11 4 or, y (11 + 4) so, y 11 + 4... (i) and 4 6y + 0 or, 6y 4 or, 6y (4+) or, y 4 + 6...(ii) We put the values 0, 1, and 3 to, and find corresponding values of y then putting down the values in the following tale: 0 1 3 y 11 + 4 11 15 19 3 y (4 + )/6 0.33 4.67 8.67 1.67 The system of equations has no solution, ecause the two lines of the equations are parallel and distinct. The system of equations has no solution, ecause the two lines of the equations are parallel and distinct. Hence the given equations are inconsistent. Y X X O Y Figure-5. Unit-5 Page-106
Bangladesh Open University The system of equations has no solution, ecause the two lines of the equations are parallel and distinct. Hence the given equations are inconsistent. Eample 3: Solve the following quadratic equation graphically; 4 + 3 0 Let y 4 +3 We give values 0, 1,, 3, 4... to and find corresponding values of y put down in the following tale: 0 1 3 4 0 1 4 9 16 4 0 4 8 1 16 +3 3 3 3 +3 3 y 3 0 1 0 3 Plotting the points (0, 3), (1, 0), (, 1), (3, 0), (4, 3) and jointing them, we get the graph y 4 + 3 on shown in the following figure 5.3. Y P N X O X O M Y Figure-5.3 The curve MNOP is the graph of 4 + 3 0. The graph of the function y 4 +3 is the curve MNOP which crosses ais at the points N and O where 1 and 3. Hence 1 and 3 are the solutions of the quadratic equation 4 + 3 0. Business Mathematics Page-107
School of Business Question for Review: These questions are designed to help you assess how far you have understood and can apply the learning you have accomplished y answering (in written form) the following questions: 1. Draw the graphs and solve the following questions: (a) 1 7y 6 () 5 + y 6 5 7y 7 y 9 + 16. Draw the graph of the solution set for the system of liner inequalities 0 y 0 + y 45 + y 60 3. Draw the graph of the solution set for the system of liner inequalities + y 30 y 0 4 + y 39 4. Draw the graphs and solve the equations (i) 5 + 6 0 (ii) 3 y + 5 0 + y + 7 0 Unit-5 Page-108
Lesson 5: Quadratic Equation After studying this lesson, you should e ale to: State the nature of quadratic equation; Bangladesh Open University Eplain the relationship etween roots and coefficient of quadratic equation; Eplain the formation of quadratic equation; and Solve the quadratic equation. Nature of Quadratic Equation Generally an equation contains the square of unknown variale is called a quadratic equation. The general method of solving a quadratic equation of the form a + + c 0 is given. Since a + + c 0; y transposition a + c. Hence dividing oth sides y a, the co-efficient of we have, Generally an equation contains the square of unknown variale is called a quadratic equation. + a c a Adding to oth sides + a a + a c a a + or, or, +.. + a a 4ac + a 4a 4a c a or, + a ± a 4ac or, or, a ± ± 4ac a 4ac a Thus the required roots of a + + c 0 are + 4ac a and 4ac a Business Mathematics Page-109
School of Business Relationship etween Roots and the Co-efficient of Quadratic Equation A quadratic equation has eactly two roots. A quadratic equation has thus eactly two roots. The relationship etween the roots and the co-efficient of the quadratic equations as as follows: Let the quadratic equation is a + + c 0, then if α and β denote the roots of this quadratic equation, we have + α 4ac a β 4ac a Therefore y addition, α + β + 4ac a + 4ac a + 4ac a 4ac a a And y multiplication, αβ + 4ac a 4ac a ( ) 4ac 4a + 4ac 4ac 4a 4a c a Thus, we have shown that Sum of the roots (α+β) a Coefficient of Coefficient of Product of the roots (αβ) c a Coefficient of Formation of Quadratic Equation Constant term The formation of the quadratic equation whose roots are given can e eplained as follows: Unit-5 Page-110
Bangladesh Open University Let the general form of quadratic equation is a + + c 0 and α and β denote the two given roots, where a, and c are constants whose values we have to find out. The sum of the roots α + β, a So, a (α+β) And product of the roots, αβ a c so, c a αβ The aove relations imply that a + + c 0 or, a +[ a(α+β)] + aαβ 0 or, a aα aβ + aαβ 0 or, (α+β) + αβ 0 [Dividing oth sides y a ] So, the required quadratic equation will e (sum of the roots) + (product of the roots) 0 The following section of this lesson contains some model applications of quadratic equation. Eample 1: Solve 4 + 13 0 4 + 13 0 We know that ± 4ac a Here a 1, 4, c 13 Sustituting the given values we have ( 4) ± ( 4) 4. ( 1)( 13).1 4 ± 16 5 Business Mathematics Page-111
School of Business 4 ± 36 4 ± 6i [ Since, i 1] Therefore, + 3i or, 3i Eample : Solve 3 + 10 6 4 We have 3 + 10 6 4 or, 3 +.5 (3 ) ( 3 + )( 3 ) 5 or, ( 3) ( ) 5 or, or, 9 4 5 0 We know that ± 4ac a Here a 9; 4; c 5 Sustituting the given values we have ( 4) ± ( 4) 4( 9)( 5) 4 ± 16 + 180 18 4 ± 196 18 4 ±14 18.9 18 So, or, 18 10 18 Unit-5 Page-11
Bangladesh Open University Eample 3: If α and β are the roots of the equation p + q 0, find the values of (i) α β (ii) α + β (iii) α 3 β 3. Since α and β are the roots of the equation p + q 0 So, sum of the roots, α + β Coefficient of Coefficient of P P 1 Product of the roots, αβ Constant term Coefficient of q q 1 (i) (α β) α + β αβ α + β + αβ 4αβ (α+β) 4αβ p 4q so, α β p 4q (ii) α + β (α + β) αβ p q (iii) α 3 β 3 (α + β) (α αβ + β ) (α + β) [(α β) + αβ] p (p 4q + q) p(p 3q). Eample 4: The roots of + 1 0 are α and β; from a quadratic equation whose roots are α 4 + β 4 and α + β 4 Since α and β are the roots of the equation + 1 0 Coefficient of 1 Sum of the roots, α + β 1 Coefficient of 1 Constant term 1 And product of the roots, αβ 1 Coefficient of 1 Now, the sum of the roots of the required equation, α + β + α + β 4 (α + β ) + [(α + β ) α β ] Business Mathematics Page-113
School of Business [(α+β) αβ] + [(α+β) αβ] (αβ) (1.1) + [(1).1] (1) 1 + 1 3 + 1. And the product of the roots of the required equation. (α 4 + β ) (α + β 4 ) α 6 + α 4 β 4 + α β + β 6 (α 6 +β 6 ) (αβ) 4 + (αβ) (α +β ) (α 4 α β + β 4 ) + (αβ) 4 + (αβ) [(α+β) αβ] [(α +β ) 3α β ] + (αβ) + (αβ) [(α+β) αβ] [{(α+β) αβ} 3(αβ) ] + (αβ) 4 + (αβ) [(1) (1)] [{(1) (1)} 3(1) ] + (1) + (1) ( 1) (1 3) + 1 + 1 + 1 + 1 4. Hence the required quadratic equation is, (sum of the roots) + (product of the roots) 0 or, ( ) + 4 0 so, + + 4 0 Unit-5 Page-114
Bangladesh Open University Questions for Review: These questions are designed to help you assess how far you have understood and can apply the learning you have accomplished y answering (in written form) the following questions: 1. What is a quadratic equation? Give an eample. State the characteristics of a quadratic equation. 3. If α and β are the roots of p + q 0, form a quadratic equation whose roots are (αβ α β). 4. If α and β are the roots of + 3 + 7 0, find the values of (1) α + β (ii) α/β (iii) β/α (iv) α 3 + β 3 5. If the equation a + + c 0 and + c + a 0 have a common root, then prove that a + + c 0 and a c. 6. Solve: (i) 3 5 0 (ii) + 4 + 4 y Multiple Choice Questions ( the appropriate answers) 1. The roots of a quadratic equation are 5 and. The equation is: (a) 3 10 0 () 3 + 10 0 (c) + 3 10 0. The quadratic equation whose one root is 3 + 3, is (a) + 6 10 0 () 6 3 0 (c) + 6 + 3 0 3. If α and β are the roots of the question 8 + P 0 and α +β 40 then P is equal to: (a) 8 () 1 (c) 1 4. If α and β are the roots of the equation 4 + 3 0 the value of the α 3 + β 3 is: (a) 1 () 5 (c) 5. If α and β are the roots of the equation 3 + 1 0, then the equation whose roots are α/β and β/α is: (a) + 5 + 0 () 5 + 0 (c) +5 0 6. If one root of the equation 3 10 + 3 0 is 1/3, then the other root is: (a) 3 () 1/3 (c) 3 Business Mathematics Page-115
School of Business Lesson-6: Application of Equation in Business Prolems After studying this lesson, you should e ale to: Apply the principles of equations to solve the usiness prolems Introduction The application of equations in the solution of usiness prolems is of great use. The application of equations in the solution of usiness prolems is of great use. It is usual to take the following steps in the solution of usiness prolems. (i) Read the prolem very carefully and try to understand the relation etween the quantities stated therein. (ii) Denote the unknown quantity y the letter, y, z, etc. (iii) Translate the statements of the prolem step y step into mathematical statements; to the etent it is possile. (iv) Solve the equation for the unknown (v) check whether the otained solution satisfies the condition given in the prolem. The following section of this lesson contains some model applications of equations to solve usiness prolems. Eample-1: A man says to his son, Seven years ago I was seven times as old as you were, and three years hence, I shall e three times as old as you will e. Find their present ages. Let the present age of the son e years. His age seven years ago 7 The father s age 7 years ago 7 ( 7) Father s present age 7 ( 7) + 7. Son s age 3 years hence + 3 Father s age 3 years hence 7 ( 7) + 7 + 3 7 ( 7) + 10 Using the given information we can write 7 ( 7) + 10 3 ( + 3) or, 7 49 + 10 3 + 9 or, 7 3 49 + 9 10 or, 4 48 so, 1 So, son s present age 1 years. Father s present age 7 (1 7) + 7 35 + 7 4 years. Unit-5 Page-116
Bangladesh Open University Eample-: A man s annual income has increased for Tk.7,500 ut there is no change in the income ta payale for him since the rate of income ta has een reduced from 10% to 7%. Find his present income. Let previous annual income Present income + 7500 His income ta of previous year 10% of 0.10 His income ta for present year 70% of ( + 7500) 0.07 + 55 According to the question we can write 0.10 0.07 + 55 or, 0.10 0.07 + 55 or, 0.03 55 or, (55 0.03) so, 17,500 So, the present income ( + 7500) (17,500 + 7500) Tk.5,000. Eample-3: Tk.0,000 is invested in two shares. The first yield is Tk.1% p.a. and the second yield is 14% p.a. If the total yield at the end of one year is 13% p.a.; how much was invested at each rate? Let in Tk. e invested at 1%. Then (Tk.0000 ) is invested at 14% Interest at 1% 1% of 0.1 Interest at 14% 14% of (0,000 ) 800 0.14 Total Interest on 0,000 at 13% 13% of 0,000,600 According to the given information we can write 60 800 0.14 + 0.1 or, 0.0 800 600 or, (00 0.0) 10,000 Hence Tk.10,000 is invested at 1% and (0,000 10,000) Tk.10,000 is invested at 14%. Eample-4: Demand and supply equations are p + q 11 and p + q 7 respectively. Find the equilirium price and quantity. (Where p stands for price and for quantity). Business Mathematics Page-117
School of Business The demand function is : p + q 11... (1) And supply function is : p + q 7 So, p 7 q... () Putting the value of p in equation (1) we have, (7 q) + q 11 or, (49 8q + 4q ) + q 11 or, 98 56q + 8q + q 11 or, 9q 56q + 98 11 0 or, 9q 56q + 87 0 or, 9q 7q 9q + 87 0 or, 9q (q 3) 9 (q 3) 0 or, (q 3) (9q 9) 0 either, q 3 0 or, 9q 9 9 so, q 3 or, q 9 When q 3; the price is, p 7 q 7 3 7 6 1. 9 9 58 63 58 5 When q, the price is, p 7 7 9 9 9 9 9 Therefore the equilirium price and quantity are: 5 9 (p, q) (1, 3) or, 9 9 Unit-5 Page-118
Questions for Review: Bangladesh Open University These questions are designed to help you assess how far you have understood and apply the learning you have accomplished y answering (in written form) the following questions: 1. Mr. Saji invested Tk.16,000 in two types of deentures of Tk.100 each. 13% deentures were purchased at Tk.150 each and 10% deentures were purchased at Tk.10 each. If he got interest of Tk.970 after 1 year, find the sum invested in each type of deenture.. The ages (in years) of Ramesh and Rahim are in the ratio 5 : 7. If Ramesh were 9 years older and Rahim were 9 years younger, the age of Ramesh would have een were twice the age of Rahim. Find their ages. 3. If the total manufacturing cost y of making units of a toy is: 5 9000 y (i) What is the variale cost per unit? (ii) What is the fied cost? (iii) What is the average cost of manufacturing 5,000 units? (iv) What is the marginal cost of producing 3,000 units? (v) What will e the effect on average cost per unit if volume changes? Multiple choice questions ( the appropriate answers) 1. The total cost of 6 ooks and 4 pencils is Tk.34 and that of 5 ooks and 5 pencils is Tk.30. The costs of each ook and each pencil respectively are: a) Tk.5 and Tk.1 ) Tk.1 and Tk.5 c) Tk.6 and Tk.1. If 3 chairs and tues cost Tk.100 and 5 chairs and 3 tues cost Tk.1900, then the cost of chairs and tues is: a) Tk.1000 ) Tk.900 c) Tk.700 3. The monthly income of A & B are in the ratio 4 : 3, Each of them saves Tk.600. If the ratio of the ependitures is 3 :, then the monthly income of A is: a) Tk.1800 ) Tk.400 c) Tk.000 4. If the laws of demand and supply are respectively given y the q equation 4q + 9p 48 and p 9 +, then the value of the equilirium price and quantity respectively are: a) 8 / 3 and 6 ) 3 / 8 and 5 c) 8 and 6 Business Mathematics Page-119