Algebra IMP Textbook PDF - Standard Ten - Maharashtra

 The Constitution of India Chapter IV A Fundamental Duties ARTICLE 51A Fundamental Duties- It shall be the duty of every citizen of India- (a) to abide by the Constitution and respect its ideals and institutions, the National Flag and th...

 The Constitution of India Chapter IV A Fundamental Duties ARTICLE 51A Fundamental Duties- It shall be the duty of every citizen of India- (a) to abide by the Constitution and respect its ideals and institutions, the National Flag and the National Anthem; (b) to cherish and follow the noble ideals which inspired our national struggle for freedom; (c) to uphold and protect the sovereignty, unity and integrity of India; (d) to defend the country and render national service when called upon to do so; (e) to promote harmony and the spirit of common brotherhood amongst all the people of India transcending religious, linguistic and regional or sectional diversities, to renounce practices derogatory to the dignity of women; (f) to value and preserve the rich heritage of our composite culture; (g) to protect and improve the natural environment including forests, lakes, rivers and wild life and to have compassion for living creatures; (h) to develop the scientific temper, humanism and the spirit of inquiry and reform; (i) to safeguard public property and to abjure violence; (j) to strive towards excellence in all spheres of individual and collective activity so that the nation constantly rises to higher levels of endeavour and achievement; (k) who is a parent or guardian to provide opportunities for education to his child or, as the case may be, ward between the age of six and fourteen years. The Coordination Committee formed by GR No. Abhyas - 2116/(Pra.Kra.43/16) SD - 4 Dated 25.4.2016 has given approval to prescribe this textbook in its meeting held on 29.12.2017 and it has been decided to implement it from the educational year 2018-19. Mathematics Part I STANDARD TEN Maharashtra State Bureau of Textbook Production and Curriculum Research, Pune - 411 004  Sachin Mehta Asst. Production Officer N/PB/2022-23/1,20,000 IMPRESSIONS,KOLHAPUR  Preface Dear Students, Welcome to the tenth standard! This year you will study two text books - Mathematics Part-I and Mathematics Part-II The main areas in the book Mathematics Part-I are Algebra, Graph, Financial planning and Statistics. All of these topics were introduced in the ninth standard. This year you will study some more details of the same. The new tax system, GST is introduced in Financial planning. Wherever a new unit, formula or application is introduced, its lucid explanation is given. Each chapter contains illustrative solved examples and sets of questions for practice. In addition, some questions in practice sets are star-marked, indicating that they are challenging for talented students. After tenth standard, some students do not opt for Mathematics. This book gives them basic concepts and mathematics needed to work in other fields. The matter under the head ‘For more information’ is useful for those students who wish to study mathematics after tenth standard and achieve proficiency in it. So they are earnestly advised to study it. Read the book thoroughly at least once and understand the concepts. Additional useful audio-visual material regarding each lesson will be available to you by using Q.R. code through ‘App’. It will definitely be useful to you for your studies. Much importance is given to the tenth standard examination. You are advised not to take the stress and study to the best of your ability to achieve expected success. Best wishes for it ! (Dr. Sunil Magar) Pune Director Date : 18 March 2018, Gudhipadva Maharashtra State Bureau of Textbook Indian Solar Year : 27 Falgun 1939 Production and Curriculum Research, Pune.  It is expected that students will develop the following competencies after studying Mathematics- Part I syllabus in standard X Area Topic Competency Statements 1. Knowledge 1.1 Arithmetic The students will be able to- of numbers Progression · solve examples using Arithmetic Progression. · plan steps to achieve a goal in future. 2. Algebra 2.1 Quadratic · solve day to day problems which can be Equations expressed in the form of quadratic equations. · decide the number of variables required to 2.2 Linear find solutions of word problems. equations in · convert a word problem into an equation in two variables two variables and find its solution. 3. Commercial 3.1 Financial · understand the concepts of savings and Mathematics planning investments. · get familiar with financial transactions in business, profession etc. 4. Statistics 4.1 Probability · use the concept of probaility in games, and Probability 4.2 Graph voting etc. · present the collected data in the form of and measures graphs or pictures deciding the suitable form of central of presentation. tendencies · find the mean, median and mode of a provided classified data. Instructions for Teachers Read the book in detail and understand the content thoroughly.Take the help of activities to explain different topics, to verify the formulae etc. Practicals is also a means of evaluation. Activities given can also be used for this purpose. Encourage the students to think independently.Compliment a student if he solves an example by a different and logically correct method.  List of some practicals (Specimen) 1. On a graph paper, draw a line parallel to the X- axis or Y- axis. Write coordinates of any four points on the line. Write how the equation of the line can be obtained from the coordinates. [Instead of parallel lines, lines passing through the origin or intersecting the X or Y- axis can also be considered] 2. Bear a two- digit number in mind. Without disclosing it, cunstruct a puzzle. Create two algebraic relations between the two digits of the number and solve the puzzle. [The above practical can be extended to a three-digit number also.] 3. Read the information about contents on a food packet. Show the information by a pie diagram. For example, see the chart of contents like carbohydrates, proteins, vitamines etc. per given weight on a buiscuit packet. Show the pro- portion of the contents by a pie diagram. The contents can be divided into four classes as carbohydrates, proteins, fats and others. 4. Prepare a frequncy distribution table given by the teacher in Excel sheet on a computer. From the table draw a frequency polygon and a histogram in Excel. 5. Roll a die ten times and record the outcomes in the form of a table. 6. Observe the tax invoice given by your teacher. Record all of its contents. Re- calculate the taxes and verify their correctness. 7. Calculate the sum of first n natural numbers given by your teacher through the following activity. For example to find the sum of first four natural numbers, take a square-grid piece of paper of 4 ´ 5 squares. Then cut it as shown in the n(n +1) figure. Hence verify the formula Sn = 2 (Here n = 4) 5 n(n +1) 4 (4 + 1) 4´5 20 Sn = \ S4 = 2 = 2 = = 10 1 2 2 2 3 [Note: Here a = 1 and d = 1. The activity can be 4 4 5 6 done taking different values of a and d. Similarly, 7 8 9 10 you can find the sum of even or odd numbers, cubes of natural numbers etc.] 8. Write a = 6 on one side of a card sheet and a = -6 on its backside. Similarly, write b = -3 on one side of another card sheet and b = 7 on its backside. From these values, form different values of (a + b) and (ab); using these values form quadratic equations.  INDEX Chapters Pages 1. Linear Equations in Two Variables..................... 1 to 29 2. Quadratic Equations........................................ 30 to 54 3. Arithmetic Progression.................................... 55 to 80 4. Financial Planning........................................... 81 to 112 5. Probability..................................................... 113 to 128 6. Statistics........................................................ 129 to 168 Answers........................................................ 169 to 176  1 Linear Equations in Two Variables Let’s study. · Methods of solving linear equations in two variables - graphical method, Cramer’s method · Equations that can be transformed in linear equation in two variables · Application of simultaneous equations Let’s recall. Linear equation in two variables An equation which contains two variables and the degree of each term containing variable is one, is called a linear equation in two variables. ax + by + c = 0 is the general form of a linear equation in two variables; a, b, c are real numbers and a, b are not equal to zero at the same time. Ex. 3x - 4y + 12 = 0 is the general form of equation 3x = 4y - 12 Activity : Complete the following table No. Equation Is the equation a linear equation in 2 variables ? 1 4m + 3n = 12 Yes 2 3x2 - 7y = 13 3 2x- 5 y = 16 4 0x + 6y - 3 = 0 5 0.3x + 0y -36 = 0 6 4 5  4 x y 7 4xy - 5y - 8 = 0 1  Simultaneous linear equations When we think about two linear equations in two variables at the same time, they are called simultaneous equations. Last year we learnt to solve simultaneous equations by eliminating one variable. Let us revise it. Ex. (1) Solve the following simultaneous equations. (1) 5x - 3y = 8; 3x + y = 2 Solution : Method (II) Method I : 5x - 3y = 8... (I) 5x - 3y = 8... (I) 3x + y = 2... (II) 3x + y = 2... (II) Multiplying both sides of Let us write value of y in terms equation (II) by 3. of x from equation (II) as 9x + 3y = 6... (III) y = 2 - 3x... (III) 5x - 3y = 8... (I) Substituting this value of y in Now let us add equations (I) equation (I). and (III) 5x - 3y = 8 5x - 3y = 8 \ 5x - 3(2 - 3x) = 8 + 9x + 3y = 6 \ 5x - 6 + 9x = 8 14x = 14 \ 14x - 6 = 8 \ x = 1 \ 14x = 8 + 6 substituting x = 1 in equation (II) \ 14x = 14 3x + y = 2 \ x = 1 \ 3´1 + y = 2 Substituting x = 1 in equation \ 3 + y = 2 (III). \ y = -1 y = 2 - 3x solution is x = 1, y = -1; it is also \ y = 2 - 3´1 written as (x, y) = (1, -1) \ y = 2 - 3 \ y = -1 x = 1, y = -1 is the solution. 2 Ex. (2) Solve : 3x + 2y = 29; 5x - y = 18 Solution : 3x + 2y = 29... (I) and 5x - y = 18... (II) Let’s solve the equations by eliminating ’y’. Fill suitably the boxes below. Multiplying equation (II) by 2. \ 5x ´ - y ´ = 18 ´ \ 10x - 2y =... (III) Let’s add equations (I) and (III) 3x + 2y = 29 + -   = = \ x = Substituting x = 5 in equation (I) 3x + 2y = 29 \ 3 ´ + 2y = 29 \ + 2y = 29 \ 2y = 29 - \ 2y = \ y = (x, y) = ( , ) is the solution. Ex. (3) Solve : 15x + 17y = 21; 17x + 15y = 11 Solution : 15x + 17y = 21... (I) 17x + 15y = 11... (II) In the two equations above, the coefficients of x and y are interchanged. While solving such equations we get two simple equations by adding and subtracting the given equations. After solving these equations, we can easily find the solution. Let’s add the two given equations. 15x + 17y = 21 + 17x + 15y = 11 32x + 32y = 32 3  Dividing both sides of the equation by 32. x + y = 1... (III) Now, let’s subtract equation (II) from (I) 15x + 17y = 21 - -17x -+ 15y =-11 -2x + 2y = 10 dividing the equation by 2. -x + y = 5... (IV) Now let’s add equations (III) and (V). x + y = 1 + -x + y = 5 \ 2y = 6 \ y = 3 Place this value in equation (III). x + y = 1 \ x + 3 = 1 \ x = 1 - 3 \ x = -2 (x, y) = (-2, 3) is the solution. Practice Set 1.1 (1) Complete the following activity to solve the simultaneous equations. 5x + 3y = 9 -----(I) 2x - 3y = 12 ----- (II) Place x = 3 in equation (I). Let’s add eqations (I) and (II). 5 ´ + 3y = 9 5x + 3y = 9 3y = 9 - + 2x - 3y = 12 3y = x = y = 3 x = x = y = \ Solution is (x, y) = ( , ). 4 2. Solve the following simultaneous equations. (1) 3a + 5b = 26; a + 5b = 22 (2) x + 7y = 10; 3x - 2y = 7 (3) 2x - 3y = 9; 2x + y = 13 (4) 5m - 3n = 19; m - 6n = -7 (5) 5x + 2y = -3; x + 5y = 4 (6) 1 x  y  10 ; 2 x  1 y  11 3 3 4 4 (7) 99x + 101y = 499; 101x + 99y = 501 (8) 49x - 57y = 172; 57x - 49y = 252 Let’s recall. Graph of a linear equation in two variables In the 9th standard we learnt that the graph of a linear equation in two variables is a straight line. The ordered pair which satisfies the equation is a solution of that equation. The ordered pair represents a point on that line. Ex. Draw graph of 2x - y = 4. Solution : To draw a graph of the equation let’s write 4 ordered pairs. x 0 2 3 -1 To obtain ordered pair by y -4 0 2 -6 simple way let’s take x = 0 (x, y) (0, -4) (2, 0) (3, 2) (-1, -6) and then y = 0. Y Scale : on both axes 2 (3, 2) 1 cm = 1 unit. 1 (2, 0) X' -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7 8 9 X -1 -2 -3 -4 (0, -4) -5 (-1, -6) -6 Y' 5  Steps to follow for drawing a graph Two points are sufficient to of linear equation in two variables. represent a line, but if co-ordinates of one of the two points are Find at least 4 ordered wrong then you will not get a pairs for given equation correct line. If you plot three points and if they are non collinear then it is Draw X- axis, Y-axis on understood that one of the points graph paper and plot the is wrongly plotted. But it is not points easy to identify the incorrect point. If we plot four points, it is almost See that all 4 points lie certain that three of them will be on one line collinear. A linear equation y = 2 is also written as 0x + y = 2. The graph of this line is parallel to X- axis; as for any value of x, y is always 2. x 1 4 -3 y 2 2 2 (x, y) (1, 2) (4, 2) (-3, 2) Similarly equation x = 2 is written as x + 0y = 2 and its graph is parallel to Y-axis. Y Scale on both axes 4 1 cm = 1 Unit 3 (2,3) (-3,2) (1,2) (4,2) y = 2 2 1 (2,1) X' -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7 X -1 (2,-1) -2 -3 (2,-3) x = 2 -4 Y' 6  Let’s learn. Graphical method Ex. Let’s draw graphs of x + y = 4, 2x - y = 2 and observe them. x + y = 4 2x - y = 2 x -1 4 1 6 x 0 1 3 -1 y 5 0 3 -2 y -2 0 4 -4 (x, y) (-1, 5) (4, 0) (1, 3) (6,-2) (x, y) (0, -2) (1, 0) (3, 4) (-1,-4) Y Scale Each point on the graph on both axes (-1,5) 5 satisfies the equation. The =2 1 cm = 1 Unit two lines intersect each 4 (3,4) -y other at (2, 2). (1,3) Hence ordered pair (2, 2) 2x 3 (2,2) i.e. x = 2, y = 2 satisfies 2 the equations x + y = 4 1 and 2x - y = 2. (1,0) (4,0) The values of variables that X' -2 -1 0 1 2 3 4 5 6 X satisfy the given equations, -1 give the solution of given (6,-2) x -2 (0,-2) equations. + y \ the solution of given = -3 4 equations x + y = 4, (-1,-4) -4 2x - y = 2 is x = 2, -5 y = 2. Y' Let’s solve these equations by method of elimination. x + y = 4... (I) substituting this value in equation (I) 2x - y = 2... (II) x + y = 4 Adding equations (I) and (II) we \ 2 + y = 4 get, \ y = 2 3x = 6 \ x = 2 7  Activity (I) : Solve the following simultaneous equations by graphical method. · Complete the following tables to get ordered pairs. x - y = 1 5x - 3y = 1 x 0 3 x 2 -4 y 0 -3 y 8 -2 (x, y) (x, y) · Plot the above ordered pairs on the same co-ordinate plane. · Draw graphs of the equations. · Note the co-ordinates of the point of intersection of the two graphs. Write solution of these equations. Activity II : Solve the above equations by method of elimination. Check your solution with the solution obtained by graphical method. Let’s think. The following table contains the values of x and y co-ordinates for ordered pairs to draw the graph of 5x - 3y = 1 x 0 1 1 -2 5 y 1 0 4 11 -3 3 -3 (x, y) (0, - 13 ) ( 15 , 0) (1, 43 ) (-2, - 113 ) · Is it easy to plot these points ? · Which precaution is to be taken to find ordered pairs so that plotting of points becomes easy ? Practice Set 1.2 1. Complete the following table to draw graph of the equations - (I) x + y = 3 (II) x - y = 4 x + y = 3 x - y = 4 x 3 x -1 0 y 5 3 y 0 -4 (x, y) (3, 0) (0, 3) (x, y) (0, -4) 2. Solve the following simultaneous equations graphically. (1) x + y = 6 ; x - y = 4 (2) x + y = 5 ; x - y = 3 (3) x + y = 0 ; 2x - y = 9 (4) 3x - y = 2 ; 2x - y = 3 ê (5) 3x - 4y = -7 ; 5x - 2y = 0 (6) 2x - 3y = 4 ; 3y - x = 4 8  Let’s discuss. To solve simultaneous equations x + 2y = 4 ; 3x + 6y = 12 graphically, following are the ordered pairs. x + 2y = 4 3x + 6y = 12 x -2 0 2 x -4 1 8 y 3 2 1 y 4 1.5 -2 (x, y) (-2, 3) (0, 2) (2, 1) (x, y) (-4, 4) (1, 1.5) (8, -2) Plotting the above ordered pairs, graph is drawn. Observe it and find answers of the following questions. Y Scale on both axes 1 cm = 1 Unit 6 5 (-4,4) 4 x+ 2y (-2,3) =4 3 2 (0,2) (1,1.5) 1 (2,1) X' -4 -3 -2 -1 0 1 2 3 4 5 6 7 8 X -1 3x +6 y =1 (8,-2) -2 2 Y' (1) Are the graphs of both the equations different or same ? (2) What are the solutions of the two equations x + 2y = 4 and 3x + 6y = 12 ? How many solutions are possible ? (3) What are the relations between coefficients of x, coefficients of y and constant terms in both the equations ? (4) What conclusion can you draw when two equations are given but the graph is only one line ? 9 Now let us consider another example. Draw graphs of x - 2y = 4, 2x - 4y = 12 on the same co-ordinate plane. Observe it. Think of the realation between the coefficients of x, coefficients of y and the constant terms and draw the inference. ICT Tools or Links. Use Geogebra software, draw X- axis, Y-axis. Draw graphs of simultaneous equations. Let’s learn. Determinant a b a b  c d is a determinant. (a, b), (c, d) are rows and  ,   c  d  are columns. Degree of this determinant is 2, because there are 2 elements in each column and 2 elements in each row. Determinant represents a number which is (ad-bc). a b i.e. = ad-bc c d a b ad-bc is the value of determinant c d Determinants, usually, are represented with capital letters as A, B, C, D,..... etc. ÒÒÒ Solved Examples ÒÒÒ Ex. Find the values of the following determinants. 5 3 -8 -3 2 3 9 (1) A = 7 9 (2) N = 2 4 (3) B = 2 3 3 10 Solution : 5 3 (1) A = 7 9 = (5 ´ 9) - (3 ´ 7) = 45 - 21 = 24 -8 -3 (2) N = 2 4 = [(-8) ´ (4)] - [(-3 ) ´ 2)] = -32 - (-6) = -32 + 6 = -26 2 3 9 (3) B = = [2 3 ´ 3 3 )] - [2 ´ 9)] = 18 - 18 = 0 2 3 3 Let’s learn. Determinant method (Cramer’s Rule) Using determinants, simultaneous equaions can be solved easily and in less space. This method is known as determinant method. This method was first given by a Swiss mathematician Gabriel Cramer, so it is also known as Cram- er’s method. To use Cramer’s method, the equations are written as a1x + b1y = c1 and a2x + b2y = c2. a1x + b1 y = c1... (I) a2x + b2 y = c2... (II) Here x and y are variables, a1, b1, c1 and a2, b2, c2 are real numbers, a1b2 - a2b1 ¹ 0 Now let us solve these equations. Multiplying equation (I) by b2. a1 b2 x + b1 b2 y = c1 b2... (III) Multiplying equation (II) by b1. a2 b1 x + b2 b1 y = c2 b1... (IV) 11  Subtracting equation (IV) from (III) a1 b2 x + b1 b2 y = c1 b2 - a b x b b y = c2 b1 - 2 1 - 2 1 - (a1 b2 - a2 b1) x = c1 b2- c2 b1 c1 b2- c2 b1 x =... (V) a1 b2 - a2 b1 a1 c2- a2 c1 Similarly y =... (VI) a1 b2 - a2 b1 To remember and write the expressions c1 b2- c2 b1, a1 b2 - a2 b1, a1 c2- a2 c1 we use the determinants. Now a1 x + b1 y = c1  a1   b1   c1  We can write 3 columns.  a  ,  b  ,  c        2 2 2 and a2 x + b2 y = c2 The values x, y in equation (V), (VI) are written using determinants as follows c1 b1 c1 b2- c2 b1 c2 b2 x = = , a1 b2 - a2 b1 a1 b1 a2 b2 a1 c1 a1 c2- a2 c1 a2 c2 y = = a1 b2 - a2 b1 a1 b1 a2 b2 a1 b1 c1 b1 a c To remember let us denote D = a b , Dx = c b , Dy = a1 c1 2 2 2 2 2 2 D Dy \ x = Dx , y = D  a1   b1   c1  For writting D, Dx, Dy remember the order of columns  ,  ,  .  a 2   b2   c2  From the equations, 12  a1 x + b1 y = c1  a1   b1   c1  and a2 x + b2 y = c2 we get the columns  ,  ,  .  a2   b2   c2   c1  In D the column of constants   is omitted.  c2   a1   c1  In Dx the column of the coefficients of x,   is replaced by  .  a2   c2   b1   c1  In Dy the column of the coefficients of y,   is replaced by  .  b2   c2  Let’s remember! Cramer’s method to solve simultaneous equations. Write given equations in the form ax + by = c. Find the values of determinants D, Dx and Dy Dx Dy Using, x = and y = D D find values of x, y. Gabriel Cramer (31 July, 1704 to 4 January, 1752) This Swiss mathematician was born in Geneva. He was very well versed in mathematics, since childhood. At the age of eighteen, he got a doctorate. He was a professor in Geneva. 13  ÒÒÒ Solved ExampleÒÒÒ Ex. (1) Solve the following simultaneous equations using Cramer’s Rule. 5x + 3y = -11 ; 2x + 4y = -10 Solution : Given equations 5x + 3y = -11 2x + 4y = -10 5 3    D = 2 4 = (5 ´ 4) - (2 ´ 3) = 20 - 6 = 14 -11 3 Dx = -10 4 = (-11) ´ 4 - (-10) ´ 3 = -44 -(-30) = -44 + 30 = -14 5 -11 Dy = 2 -10 = 5 ´ (-10) - 2 ´ (-11) = -50 -(-22) = -50 + 22 = -28 D Dy x = Dx = −14 −28 = -1 and y = D = = -2 14 14 \ (x, y) = (-1, -2) is the solution. Activity 1 : To solve the simultaneous equations by determinant method, fill in the blanks y + 2x - 19 = 0 ; 2x - 3y + 3 = 0 Solution : Write the given equations in the form ax + by = c 2x + y = 19 2x - 3y = -3 D = 2 -3 = [ ´ (-3)] - [2 ´ ( )] = - ( ) = - = 19 Dx = = [19 ´ ( )] - [( ) ´ ( )] = - -3   = 14  19 Dy = 2 = [( ) ´ ( )] - [( ) ´ ( )] = - = By Cramer’s Rule - D Dy x = x y = D D \ x = = y = = \ (x, y) = ( , ) is the solution of the given equations. Activity 2 : Complete the following activity - 3x-2y=3 2x+y=16 Find the values of determinants in the given equations D = = Dx = = Dy = = Values according to Cramer’s Rule x = = y = = \ (x, y) = ( , ) is the solution. 15  Let’s think. What is the nature of solution if D = 0 ? What can you say about lines if common solution is not possible? Practice Set 1.3 1. Fill in the blanks with correct number 3 2 4 5 = 3 ´ - ´ 4 = - 8 = 2. Find the values of following determinants. 7 5 3 3 -1 7 5 3 (1) (2) (3) 3 1 2 4 -7 0 2 2 3. Solve the following simultaneous equations using Cramer’s rule. (1) 3x - 4y = 10 ; 4x + 3y = 5 (2) 4x + 3y - 4 = 0 ; 6x = 8 - 5y (3) x + 2y = -1 ; 2x - 3y = 12 (4) 6x - 4y = -12 ; 8x - 3y = -2 y 1 (5) 4m + 6n = 54 ; 3m + 2n = 28 (6) 2x + 3y = 2 ; x - 2 = 2 Let’s learn. Equations reducible to a pair of linear equations in two variables Activity : Complete the following table. Equation No. of variables whether linear or not 3 x - 4y = 8 2 Not linear 6 3 x −1 + y−2 = 0 7 13 2x +1 + y+2 =0 14 3 x+ y + x− y = 5 16  Let’s think. In the above table the equations are not linear. Can you convert the equations into linear equations ? Let’s remember! We can create new variables making a proper change in the given variables. Substituting the new variables in the given non-linear equations, we can convert them in linear equations. m Also remember that the denominator of any fraction of the form n cannot be zero. ÒÒÒ Solved examples ÒÒÒ Solve: 4 5 3 4 Ex. (1) x + y = 7; x + y = 5 4 5 3 4 Solution : x + y = 7; x + y = 5 1 1 4  + 5  = 7... (I)  x  y 1 1 3  + 4  = 5... (II) x  y 1 1 Replacing   by m and   by n in equations (I) and (II), we get x  y 4m + 5n = 7... (III) 3m + 4n = 5... (IV) On solving these equations we get m = 3, n = -1 1 1 1 Now, m = x \ 3 = x \ x = 3 1 1 n = y \ -1 = y \ y = -1 1 \ Solution of given simultaneous equations is (x, y) = ( 3 , -1) 17  4 1 2 3 Ex.(2) x − y + x + y = 3 ; x − y - x + y = 5 4 1 2 3 Solution : x − y + x + y = 3 ; x − y - x + y = 5  1   1  4  x  y  + 1  x  y  = 3... (I)    1   1  2 x  y  - 3  x  y  = 5... (II)    1   1  Replacing  x  y  by a and  x  y  by b we get   4a + b = 3... (III) 2a - 3b = 5... (IV) On solving these equations we get, a = 1 b = -1  1   1  But a =   , b =    x y  x y  1   1  \   = 1,  x  y  = -1  x y   \ x - y = 1... (V) x + y = -1... (VI) Solving equation (V) and (VI) we get x = 0, y = -1 \ Solution of the given equations is (x, y) = (0, -1) Let’s think. In the above examples the simultaneous equations obtained by transformation are solved by elimination method. If you solve these equations by graphical method and by Cramer’s rule will you get the same answers ? Solve and check it. 18 Activity : To solve given equations fill the boxes below suitably. 5 1 6 3 + y−2 = 2 ; x −1 - y−2 = 1 x −1  1   1  Replacing   by m ,  y − 2  by n  x −1  New equations 6m - 3n = 1 On solving m = , n = Replacing m, n by their original values.  1  1   =  x −1  3 On solving x = , y = \ (x, y) = ( , ) is the solution of the given simultaneous equations. Practice Set 1.4 1. Solve the following simultaneous equations. (1) 2  3  15 ; 8  5  77 x y x y 10 2 15 5 (2)  4 ;   2 x y x y x y x y 27 31 31 27 (3)   85 ;   89 x2 y3 x2 y3 1 1 3 1 1 1 (4)   ;   3x  y 3x  y 4 2(3 x  y ) 2(3 x  y ) 8 19  Let’s learn. Application of Simultaneous equations Activity : There are some instructions given below. Frame the equations from the information and write them in the blank boxes shown by arrows. Sarthak’s age is less by 8 than double the age of Sakshi I am My Present Sarthak age is x years. 4 years ago The sum of present Sakshi’s age was My Present I am ages of Sarthak 3 years less than age is y years. Sakshi and Sakshi is 25. Sarthak’s age at that time. Ex. (1) The perimeter of a rectangle is 40 cm. The length of the rectangle is more than double its breadth by 2. Find length and breadth. Solution : Let length of rectangle be x cm and breadth be y cm. From first condition - 2(x + y) = 40 x + y = 20... (I) From 2nd condition - x = 2y + 2 \ x - 2y = 2... (II) Let’s solve eq. (I), (II) by determinant method x + y = 20 x - 2y = 2 20  1 1 D = 1 -2 = [1 ´ (-2)] - (1 ´ 1) = - 2 - 1 = - 3 20 1 Dx = 2 -2 = [20 ´ (-2)] - (1 ´ 2) = - 40 - 2 = - 42 1 20 Dy = 1 2 = (1 ´ 2) - (20 ´ 1) = 2 - 20 = -18 Dx Dy x = and y = D D - 42 −18 \ x = and y = -3 −3 \ x = 14, y = 6 \ Length of the rectangle is 14 cm and breadth is 6 cm. Ex. (2) Sale ! Sale !! Sale !!! only for 2 days I have some analogue wrist watches and some digital wrist watches. I am going to sell them at a discount Sale of 1st day Sale of the 2nd day Analogue watch = 11 Analogue watch = 22 Digital watch = 6 Digital watch = 5 Received amount = ` 4330 Received amount = ` 7330 Find selling price of wrist watch of each type. 21 Solution : Let selling price of each analogue watch be ` x Selling price of each digital watch be ` y From first condition - 11x + 6y = 4330... (I) from 2nd condition - 22x + 5y = 7330... (II) multiplying equation (I) by 2 we get, 22x + 12y = 8660... (III) subtract equation (III) from equation (II). 22x + 5y = 7330... (II) - 22x + 12y = 8660... (III) - - - -7y = -1330 \ y = 190 Substitute this value of y in equation (I) 11x + 6y = 4330 \ 11x + 6(190) = 4330 \ 11x + 1140 = 4330 \ 11x = 3190 \ x = 290 \ selling price of each analogue watch is ` 290 and of each digital watch is ` 190. 22 Ex. (3) A boat travels 16 km upstream The same boat travels 36 km and 24 km downstream in 6 upstream and 48 km downstream hours. in 13 hours. Find the speed of water current and speed of boat in still water. Solution : Let the speed of the boat in still water be x km/hr and the speed of water current be y km/hr \ speed of boat in downstream = (x + y) km/hr. and that in upstream = (x - y) km/hr. distance Now distance = speed ´ time \ time = speed 16 Time taken by the boat to travel 16 km upstream = hours and it x− y 24 takes x + y hours to travel 24 km downstream. from first condition - 16 24  = 6... (I) x y x y from 2nd condition 36 48  = 13... (II) x y x y 1 1 By replacing by m and x + y by n we get x− y 16m + 24n = 6... (III) 36m + 48n = 13... (IV) 23  1 1 Solving equations (III) and (IV) m = 4 , n = 12 Repalcing m, n by their original values we get x - y = 4... (V) x + y = 12... (VI) Solving equations (V), (VI) we get x = 8, y = 4 \ speed of the boat in still water is 8 km/hr. and speed of water current is 4 km/hr. Ex. (4) A certain amount is equally distributed among certain number of students. Each would get ` 2 less if 10 students were more and each would get ` 6 more if 15 students were less. Find the number of students and the amount distributed. Solution : Let the number of students be x and amount given to each student be ` y. \ Total amount distributed is xy From the first condition we get, (x + 10) (y - 2) = xy \ xy - 2x + 10y - 20 = xy \- 2x + 10y = 20 \- x + 5y = 10... (I) From the 2nd condition we get, (x - 15) (y + 6) = xy \ xy + 6x - 15y - 90 = xy \ 6x - 15y = 90 \ 2x - 5y = 30... (II) Adding equations (I) and (II) - x + 5y = 10 +   2x - 5y = 30 x = 40 Substitute this value of x in equation (I) - x + 5y = 10 \- 40 + 5y = 10 \ 5y = 50 24  \ y = 10 Total amount distributed is = xy = 40 ´ 10 = ` 400. \ ` 400 distributed equally among 40 students. Ex. (5) A three digit number is equal to 17 times the sum of its digits; If the digits are reversed, the new number is 198 more than the old number ; also the sum of extreme digits is less than the middle digit by unity. Find the original number. Solution : Let the digit in hundreds place be x and that in unit place be y. H T unit x x + y + 1 y \ the three digit number is 100x + 10(x + y + 1) + y = 100x + 10x + 10y + 10 + y = 110x + 11y + 10 the sum of the digits in the given number = x + (x + y + 1) + y = 2x + 2y + 1 \ From first condition Given number = 17 ´ (sum of the digits) \ 110x + 11y + 10 = 17 ´ (2x + 2y + 1) \ 110x + 11y + 10 = 34x + 34y + 17 \ 76x - 23y = 7... (I) The number obtained by reversing the digits = 100y + 10(x + y + 1) + x = 110y + 11x + 10 Given number = 110x + 11y + 10 From 2nd condition, Given number + 198 = new number. 110x + 11y + 10 + 198 = 110y + 11x + 10 99x - 99y = -198 x - y = -2 \ x = y - 2... (II) Substitute this value of x in equation (I). \ 76(y - 2) - 23y = 7 \ 76y - 152 - 23y = 7 53y = 159 25  \ y = 3 \ the digit in units place is = 3 Substitute this value in equation (II) x = y - 2 \ x = 3 - 2 = 1 \ x = 1 \ The digit in hundred’s place is 1 the digit in ten’s place is 3 + 1 + 1 = 5 \ the number is 153. Practice Set 1.5 (1) Two numbers differ by 3. The sum of twice the smaller number and thrice the greater number is 19. Find the numbers. (2) Complete the following. 2x + y + 8 x + 4 Find the values of x 2y I am a rectangle. and y. 4x - y Find my perimeter and area. (3) The sum of father’s age and twice the age of his son is 70. If we double the age of the father and add it to the age of his son the sum is 95. Find their present ages. (4) The denominator of a fraction is 4 more than twice its numerator. Denominator becomes 12 times the numerator, if both the numerator and the denominator are reduced by 6. Find the fraction. (5) Two types of boxes A, B are to be placed in a truck having capacity of 10 tons. When 150 boxes of type A and 100 boxes of type B are loaded in the truck, it weighes 10 tons. But when 260 boxes of type A are loaded in the truck, it can still accommodate 40 boxes of type B, so that it is fully loaded. Find the weight of each type of box. ê (6) Out of 1900 km, Vishal travelled some distance by bus and some by aeroplane. Bus travels with average speed 60 km/hr and the average speed of aeroplane is 700 km/hr. It takes 5 hours to complete the journey. Find the distance, Vishal travelled by bus. 26  Problem Set - 1 1. Choose correct alternative for each of the following questions (1) To draw graph of 4x + 5y = 19, Find y when x = 1. (A) 4 (B) 3 (C) 2 (D) -3 (2) For simultaneous equations in variables x and y, Dx = 49, Dy = - 63, D = 7 then what is x ? 1 −1 (A) 7 (B) -7 (C) 7 (D) 7 5 3 (3) Find the value of -7 -4 (A) -1 (B) -41 (C) 41 (D) 1 (4) To solve x + y = 3 ; 3x - 2y - 4 = 0 by determinant method find D. (A) 5 (B) 1 (C) -5 (D) -1 (5) ax + by = c and mx + ny = d and an ¹ bm then these simultaneous equations have - (A) Only one common solution. (B) No solution. (C) Infinite number of solutions. (D) Only two solutions. 2. Complete the following table to draw the graph of 2x - 6y = 3 x -5 y 0 (x, y) 3. Solve the following simultaneous equations graphically. (1) 2x + 3y = 12 ; x - y = 1 (2) x - 3y = 1 ; 3x - 2y + 4 = 0 (3) 5x - 6y + 30 = 0 ; 5x + 4y - 20 = 0 (4) 3x - y - 2 = 0 ; 2x + y = 8 (5) 3x + y = 10 ; x - y = 2 4. Find the values of each of the following determinants. (1) 4 3 (2) 5 -2 (3) 3 -1 2 7 -3 1 1 4 27 5. Solve the following equations by Cramer’s method. (1) 6x - 3y = -10 ; 3x + 5y - 8 = 0 (2) 4m - 2n = -4 ; 4m + 3n = 16 5 1 4 (3) 3x - 2y = 2 ; x + 3y = - 3 3 (4) 7x + 3y = 15 ; 12y - 5x = 39 (5) x  y  8  x  2 y  14  3 x  y 2 3 4 6. Solve the following simultaneous equations. 2 2 1 3 2 7 13 13 7 (1) x  3 y  6 ; x  0 y (2) 2 x  1  y  2  27 ; 2 x  1  y  2  33 148 231 527 231 148 610 7x  2 y 8x  7 y (3)   ;   (4) 5 ;  15 x y xy x y xy xy xy (5) 1 1 1 5 2 3   ;   2( 3 x  4 y ) 5( 2 x  3 y ) 4 (3x  4 y ) (2x  3y) 2 7. Solve the following word problems. (1) A two digit number and the number with digits interchanged add up to 143. In the given number the digit in unit’s place is 3 more than the digit in the ten’s place. Find the original number. Let the digit in unit’s place is x and that in the ten’s place is y \ the number = y + x The number obtained by interchanging the digits is x + y According to first condition two digit number + the number obtained by interchanging the digits = 143 \ 10y + x + = 143 \ x + y = 143 x + y =..... (I) From the second condition, digit in unit’s place = digit in the ten’s place + 3 \ x = + 3 \ x - y = 3..... (II) Adding equations (I) and (II) 28  2x = x = 8 Putting this value of x in equation (I) x + y = 13 8 + = 13 \ y = The original number is 10 y + x = + 8 = 58 1 (2) Kantabai bought 1 2 kg tea and 5 kg sugar from a shop. She paid ` 50 as return fare for rickshaw. Total expense was ` 700. Then she realised that by ordering online the goods can be bought with free home delivery at the same price. So next month she placed the order online for 2 kg tea and 7 kg sugar. She paid ` 880 for that. Find the rate of sugar and tea per kg. (3) To find number of notes that Anushka had, complete the following activity. Suppose that Anushka had x notes of ` 100 and y notes of ` 50 each Anushka got ` 2500/- from Anand If Anand would have given her the as denominations mentioned above amount by interchanging number of \............. equation I notes, Anushka would have received ` 500 less than the previous amount \ The No. of notes ( , ) \.............. equation II (4) Sum of the present ages of Manish and Savita is 31. Manish’s age 3 years ago was 4 times the age of Savita. Find their present ages.  (5) In a factory the ratio of salary of skilled and unskilled workers is 5 : 3. Total salary of one day of both of them is ` 720. Find daily wages of skilled and unskilled workers.  (6) Places A and B are 30 km apart and they are on a straight road. Hamid travels from A to B on bike. At the same time Joseph starts from B on bike, travels towards A. They meet each other after 20 minutes. If Joseph would have started from B at the same time but in the opposite direction (instead of towards A) Hamid would have caught him after 3 hours. Find the speed of Hamid and Joseph.  29  2 Quadratic Equations Let’s study. · Quadratic equation : Introduction · Methods of solving quadratic equation · Nature of roots of quadratic equation · Relation between roots and coefficients · Applications of quadratic equations Let’s recall. You have studied polynomials last year. You know types of polynomials according to their degree. When the degree of polynomial is 1 it is called a linear polynomial and if degree of a polynomial is 2 it is called a quadratic polynomial. Activity : Classify the following polynomials as linear and quadratic. 5x + 9, x2 + 3x -5, 3x - 7, 3x2 - 5x, 5x2 Linear polynomials Quadratic polynomials Now equate the quadratic polynomial to 0 and study the equation we get. Such type of equation is known as quadratic equation. In practical life we may use quadratic equations many times. Ex. Sanket purchased a rectangular plot having area 200 m2. Length of the plot was 10 m more than its breadth. Find the length and the breadth of the plot. Let the breadth of the plot be x metre. \ Length = (x + 10) metre Area of rectangle = length ´ breadth \ 200 = (x + 10) ´ x \ 200 = x2 + 10 x That is x2 + 10x = 200 \ x2 + 10x - 200 = 0 30  Now, solving equation x2 + 10x - 200 = 0, we will decide the dimensions of the plot. Let us study how to solve the quadratic equation. Let’s recall. Activity : x2 + 3x -5, 3x2 - 5x, 5x2; Write the polynomials in the index form. Observe the coefficients and fill in the boxes. x2 + 3x -5 , 3x2 - 5x + 0 , 5x2 + 0x + 0 Coefficients of x2 are 1 , 3 and 5 these coefficients are non zero. Coefficients of x are 3, and respectively. Constants terms are , and respectively. Here constant term of second and third polynomial is zero. Let’s learn. Standard form of quadratic equation The equation involving one variable with all indices as whole numbers and having 2 as the maximum index of the variable is called the quadratic equation. General form is ax2 + bx + c = 0 In ax2 + bx + c = 0, a, b, c are real numbers and a ¹ 0. ax2 + bx + c = 0 is the general form of quadratic equation. Activity : Complete the following table Quadratic General form a b c Equation x2 - 4 = 0 x2 + 0x - 4 = 0 1 0 -4 y2 = 2y - 7.................. x2 + 2x = 0.................. ÒÒÒ Solved Examples ÒÒÒ Ex. (1) Decide which of the following are quadratic equations ? (1) 3x2 - 5x + 3 = 0 (2) 9y2 + 5= 0 (3) m3 - 5m2 + 4 = 0 (4) (l + 2) (l - 5) = 0 Solution : (1) In the equation 3x2 - 5x + 3 = 0, x is the only variable and maximum index of the variable is 2 \ It is a quadratic equation. 31  (2) In the equation 9y2 + 5= 0, is the only variable and maximum index of the variable is \It a quadratic equation. (3) In the equation m3 - 5m2 + 4 = 0, is the only variable but maximum index of the variable is not 2. \It a quadratic equation. (4) (l + 2) (l - 5) = 0 \ l (l - 5) + 2 (l - 5) = 0 \ l2 - 5l + 2l - 10 = 0 \ l2 - 3l - 10 = 0, In this equation is the only variable and maximum index of the variable is. \It a quadratic equation. Let’s learn. Roots of a quadratic equation In the previous class you have studied that if value of the polynomial is zero for x = a then (x - a) is a factor of that polynomial. That is if p(x) is a polynomial and p(a) = 0 then (x - a) is a factor of p(x). In this case ’a’ is the root or solution of p(x) = 0 For Example , Let x = -6 in the polynomial x2 + 5x - 6 Let x = 2 in polynomial x2 + 5x - 6 x2 + 5x - 6 = (-6)2 + 5 ´ (-6) - 6 x2 + 5x - 6 = 22 + 5 ´ 2 - 6 = 36 - 30 - 6 =0 = 4 + 10 - 6 \ x = -6 is a solution of the equation. = 8¹0 Hence -6 is one root of the equation \ x = 2 is not a solution of the x2 + 5x - 6 = 0 equation x2 + 5x - 6 = 0 ÒÒÒ Solved Example ÒÒÒ 3 Ex. 2x2 - 7x + 6 = 0 check whether (i) x = 2 , (ii) x = -2 are solutions of the equations. 3 Solution : (i) Put x = 2 in the polynomial 2x2 - 7x + 6 3 2 3 2x - 7x + 6 = 2   - 7  2  + 6 2 2 32  9 21 =2 ´ - +6 4 2 9 21 12 = 2 - + 2 =0 2 3 \ x= 2 is a solution of the equation. (ii) Let x = -2 in 2x2 - 7x + 6 2x2 - 7x + 6 = 2(-2)2 - 7(-2) + 6 = 2 ´ 4 + 14 + 6 = 28 ¹ 0 \ x = -2 is not a solution of the equation. Activity : If x = 5 is a root of equation kx2 - 14x - 5 = 0 then find the value of k by completing the following activity. Solution : One of the roots of equation kx2 - 14x - 5 = 0 is. \ Now Let x = in the equation. 2 k - 14 -5=0 \ 25k - 70 - 5 = 0 25k - =0 25k = \ k= =3 Let’s remember! (1) ax2 + bx + c = 0 is the general form of equation where a, b, c are real numbers and ’a ’ is non zero. (2) The values of variable which satisfy the equation [or the value for which both the sides of equation are equal] are called solutions or roots of the equation. 33  Practice Set 2.1 1. Write any two quadratic equations. 2. Decide which of the following are quadratic equations. 1 (1) x2 + 5 x - 2 = 0 (2) y2 = 5 y - 10 (3) y2 + =2 y 1 (4) x + = -2 (5) (m + 2) (m - 5) = 0 (6) m + 3 m2 -2 = 3 m3 3 x 3. Write the following equations in the form ax2 + bx + c = 0, then write the values of a, b, c for each equation. (1) 2y =10 - y2 (2) (x - 1)2 = 2 x + 3 (3) x2 + 5x = -(3 - x) (4) 3m2 = 2 m2 - 9 (5) P (3 + 6p) = -5 (6) x2 - 9 = 13 4. Determine whether the values given against each of the quadratic equation are the roots of the equation. 5 (1) x2 + 4x - 5 = 0 , x = 1, -1 (2) 2m2 - 5m = 0 , m = 2, 2 5. Find k if x = 3 is a root of equation kx2 - 10x + 3 = 0. −7 6. One of the roots of equation 5m2 + 2m + k = 0 is. Complete the following activity 5 to find the value of ’k’. Solution : is a root of quadratic equation 5m2 + 2m + k = 0 \ Put m = in the equation. 2 5´ +2´ +k=0 + +k=0 + k=0 k= Let’s recall. Last year you have studied the methods to find the factors of quadratic polynomials like x2 - 4x - 5, 2m2 - 5m, a2 - 25. Try the following activity and revise the same. Activity : Find the factors of the following polynomials. (1) x2 - 4 x - 5 (2) 2m2 - 5 m (3) a2 - 25 = x2 - 5 x + 1x - 5 =...... = a2 - 52 = x (....) +1(....) = (....) (....) = (....) (....) 34  Let’s learn. Solutions of a quadratic equation by factorisation By substituting arbitrary values for the variable and deciding the roots of quadratic equation is a time consuming process. Let us learn to use factorisation method to find the roots of the given quadratic equation. x2 - 4 x - 5 = (x - 5) (x + 1) (x - 5) and (x + 1) are two linear factors of quadratic polynomial x2 - 4 x - 5. So the quadratic equation obtained from x2 - 4 x - 5 can be written as (x - 5) (x + 1) = 0 If product of two numbers is zero then at least one of them is zero. \ x - 5 = 0 or x + 1 = 0 \ x = 5 or x = -1 \ 5 and the -1 are the roots of the given quadratic equation. While solving the equation first we obtained the linear factors. So we call this method as ’factorization method’ of solving quadratic equation. ÒÒÒ Solved Examples ÒÒÒ Ex. Solve the following quadratic equations by factorisation. (1) m2 - 14 m + 13 = 0 (2) 3x2 - x - 10 = 0 (3) 3y2 = 15 y (4) x2 = 3 (5) 6 3 x2 + 7x = 3 (1) m2 - 14 m + 13 = 0 (2) 3x2 - x - 10 = 0 \ m2 - 13 m - 1m + 13 = 0 \ 3x2 - 6x + 5x - 10 = 0 \ m (m - 13) -1 (m - 13) = 0 \ 3x (x - 2) +5 (x - 2) = 0 \ (m - 13) (m - 1) = 0 \ (3x + 5) (x - 2) = 0 \ m - 13 = 0 or m - 1 = 0 \ (3x + 5) = 0 or (x - 2) = 0 5 \ m = 13 or m = 1 \ x = - 3 or x = 2 5 \ 13 and 1 are the roots of the given \ - , and 2 are the roots of the given 3 quadratic equation. quadratic equation. 35  (3) 3y2 = 15 y (4) x2 = 3 \ 3y2 - 15 y = 0 \ x2 - 3 = 0 \ 3y (y - 5) = 0 \ x2 - ( 3 )2 = 0 \ 3y = 0 or (y - 5) = 0 \ (x + 3 ) (x - 3 ) = 0 \ y = 0 or y = 5 \(x + 3 ) = 0 or (x - 3 ) = 0 \ 0 and 5 are the roots of quadratic \ x = - 3 or x = 3 equation. \ - 3 and 3 are the roots of given quadratic equation. (5) 6 3 x2 + 7x = 3 \6 3 x2 + 7x - 3= 0 । 6 3 ´- 3 = -18 । -18 \6 3 x2 + 9x - 2x - 3= 0 । । 9 -2 \3 3 x(2x + 3 ) -1 (2x + 3) = 0 । । 9=3 3 ´ 3 । \ (2x + 3 ) (3 3 x - 1) = 0 \2x + 3 = 0 or 3 3x -1=0 \ 2x = - 3 or 3 3x = 1 3 1 \ x=- or x = 3 3 2 3 1 \ - and 3 3 are the roots of the given quadratic equation. 2 Practice Set 2.2 1. Solve the following quadratic equations by factorisation. (1) x2 - 15 x + 54 = 0 (2) x2 + x - 20 = 0 (3) 2y2 + 27 y + 13 = 0 1 2 (4) 5m2 = 22 m + 15 (5) 2x2 - 2 x + 2 =0 (6) 6x - x =1 (7) 2x +7x+5 2 2 = 0 to solve this quadratic equation by factorisation, complete the following activity. Solution : 2x +7x+5 2 2 =0 2x 2 + + +5 2 =0 x(.....) + 2 (.....) = 0 36  (.....)(x + 2 ) = 0 (.....) = 0 or (x + 2 ) = 0 \x= or x = - 2 \ and - 2 are roots of the equation. ê (8) 3x2 - 2 6 x + 2 = 0 (9) 2m (m - 24) = 50 (10) 25m2 = 9 (11) 7m2 = 21m (12) m2 - 11 = 0 Let’s learn. Solution of a quadratic equation by completing the square Teacher : Is x2 + 10 x + 2 = 0 a quadratic equation or not ? Yogesh : Yes Sir, because it is in the form ax2 + bx + c = 0, maximum index of the variable x is 2 and ’a’ is non zero. Teacher : Can you solve this equation ? Yogesh : No Sir, because it is not possible to find the factors of 2 whose sum is 10. Teacher : Right, so we have to use another method to solve such equations. Let us learn the method. Let us add a suitable term to x2 + 10 x so that the new expression would be a complete square. If x2 + 10 x + k = (x + a)2 then x2 + 10 x + k = x2 + 2ax + a2 \ 10 = 2a and k = a2 by equating the coefficients for the variable x and constant term \ a = 5 \ k = a2 = (5)2 = 25 \ x2 + 10 x + 2 = (x + 5)2 - 25 + 2 = (x + 5)2 - 23 Now can you solve the equation x2 + 10 x + 2 = 0 ? Rehana : Yes Sir, left side of the equation is now difference of two squares and we can factorise it. (x + 5)2 - ( 23 ) =02 \ (x + 5 + 23 )(x + 5 - 23 ) = 0 \x+5+ 23 = 0 or x + 5 - 23 = 0 \ x = - 5 - 23 or x = - 5 + 23 37 Hameed : Sir, May I suggest another way ? (x + 5)2 - ( 23 )2= 0 \ (x + 5)2 = ( 23 )2 \ x + 5 = 23 or x + 5 = - 23 \ x = - 5 + 23 or x = - 5 - 23 ÒÒÒ Solved Examples ÒÒÒ Ex. (1) Solve : 5x2 - 4x - 3 = 0 Solution : It is convenient to make coefficient of x2 as 1 and then convert the equation as the of difference of two squares, so dividing the equation by 5, 4 3 we get, x2 - x- =0 55 4 4 now if x - x + k = (x - a)2 then x2 - x + k = x2 - 2ax + a2. 2 5 5 4 compare the terms in x - x and x2 - 2ax. 2 5 4 1 4 2 - 2 ax = - x \a= 2 ´ = 5 5 2 5 When equation is in the form 2 4 \ k = a =  5  = 2 x2 + b x + c = 0, it can be written as   25 4 3 2 Now, x2 - 5 x - = 0 b 2 b 5 x + b x +   -  2  + c = 0 that is, 2 2 4 4 4 3 \ x2 - 5 x+ - - =0  b 2 b 2 25 25 5  x   =  2  - c 2  2  2 4 3 \ x  -( + 5) = 0  5 25 2  2  19  \ x  -  25  =0  5   2  2  19  \ x  =  25   5   2 19 2 19 \ x - 5 = or x- 5 =- 5 5 2 19 2 19 \ x = 5 + or x= 5 - 5 5 2 + 19 2 − 19 \ x = 5 or x= 5 2 + 19 2 − 19 \ 5 and are roots of the equation. 5 38 Ex. (2) Solve : x2 + 8x - 48 = 0 Method I : Completing the square. Method II : Factorisation x2 + 8x - 48 = 0 x2 + 8x - 48 = 0 \ x2 + 8x + 16 - 16 - 48 = 0 \ x2 + 12x - 4x - 48 = 0 \ (x + 4)2 - 64 = 0 \ x (x + 12) - 4(x + 12) = 0 \ (x + 4)2 = 64 \ (x + 12) (x - 4) = 0 \ x + 4 = 8 or x + 4 = -8 \ x + 12 = 0 or x - 4 = 0 \ x = 4 or x = - 12 \ x = -12 or x = 4 Practice Set 2.3 Solve the following quadratic equations by completing the square method. (1) x2 + x - 20 = 0 (2) x2 + 2 x - 5 = 0 (3) m2 - 5 m = -3 (4) 9y2 - 12 y + 2 = 0 (5) 2y2 + 9y + 10 = 0 (6) 5x2 = 4x +7 Let’s learn. Formula for solving a quadratic equation b c ax2 + bx + c, Divide the polynomial by a ( a ¹ 0) to get x2 + a x + a. \ b c Let us write the polynomial x2 + a x + a in the form of difference of two square b c numbers. Now we can obtain roots or solutions of equation x2 + a x + a = 0 which is equivalent to ax2 + bx + c = 0. ax2 + bx + c = 0... (I) b c x2 + a x + a = 0..... dividing both sides by a 2 2 b  b   b  c \x + 2 a x+    2a  -    2a  + a =0 2 b2 c \  x   b - + a =0  2a  4a 2 39  2 2  b  b 2 − 4ac  b  b 2 − 4ac \  x  2a  - =0 \  x  2a  = 4a 2 4a 2 b b 2 − 4ac b b 2 − 4ac \ x+ 2a = 4a 2 or x + 2a = - 4a 2 b b 2 − 4ac b b 2 − 4ac \ x = - 2a + 4a 2 or x = - 2a - 4a 2 b  b 2  4ac −b − b 2 − 4ac \ x = 2a or x = 2a b  b 2  4ac In short the solution is written as x = and these values are denoted 2a by a, b. b  b 2  4ac −b − b 2 − 4ac \a = 2a , b=.......... (I) 2a The values of a, b, c from equation ax2 + bx + c = 0 are substituted in b  b 2  4ac and further simplified to obtain the roots of the equation. So 2a b  b 2  4ac x = is the formula used to solve quadratic equation. Out of the two 2a roots any one can be represented by a and the other by b. −b − b 2 − 4ac b  b 2  4ac That is, instead (I) we can write a = ,b =.... (II) 2a 2a b  b 2  4ac −b − b 2 − 4ac Note that : If a = then a > b, if a = then a < b 2a 2a ÒÒÒ Solved Examples ÒÒÒ Solve quadratic equations using formula. b  b 2  4ac Ex. (1) m2 - 14 m + 13 = 0 m= 2a Solution : m2 - 14 m + 13 = 0 comparing (14)  144 = with ax2 + bx + c = 0 2 1 14 ± 12 we get a = 1, b = -14, c = 13, = 2 \ b2 - 4 ac = (-14)2 - 4 ´ 1 ´ 13 14 + 12 14 − 12 \m= or m = 2 2 = 196 -52 26 2 \m= 2 or m = = 144 2 \ m = 13 or m = 1 \ 13 and 1 are roots of the equation. 40 Ex. (2) : x2 + 10 x + 2 = 0 Solution : x2 + 10 x + 2 = 0 comparing with ax2 + bx + c = 0 we get a = 1, b = 10, c = 2, \ b2 - 4 ac = (10)2 - 4 ´ 1 ´ 2 = 100 - 8 = 92 b  b 2  4ac x= 2a 10  92 = 2 1 10  4  23 x= 2 10  2 23 = 2 2(5  23 ) = 2 \ x = -5 ± 23 \ x= 5  23 or x= −5 − 23 \ the roots of the given quadratic equation are 5  23 and −5 − 23. Ex. (3) : x2 - 2x - 3 = 0 Solution : comparing with ax2 + bx + c = 0 we get a = 1, b = -2, c = -3, \ b2 - 4 ac = (-2)2 - 4 ´ 1 ´ (-3) = 4 + 12 = 16 ( 2)  16 −( −2) − 16 \x = or x = 2 2 2+4 2−4 = or 2 2 = 3 or -1 41  For more information : Let us understand the solution of equation x2 - 2x - 3 = 0 when solved graphically. x2 - 2x - 3 = 0 \ x2 = 2x + 3 The values which satisfy the equation are the roots of the equation. Let y = x2 = 2x + 3. Let us draw graph of y = x2 and y = 2x + 3 y = x2 x 3 2 1 0 -1 -2 -3 y 9 4 1 0 1 4 9 y = 2x + 3 x -1 0 1 -2 y 1 3 5 -1 Y 10 These graphs intersect (-3, 9) 9 (3, 9) each other at (-1, 1) and (3, 9). 8 \ The solutions of x2 = 2x + 3 7 i.e x2 - 2x - 3 = 0 are x = -1 or 6 x = 3. 5 In the adjacent diagram the (-2, 4) 4 graphs of equations y = x2 (2, 4) and y = 2x + 3 are given. 3 From their points of 2 intersection, observe and (-1, 1) 1 (1, 1) understand how you get the solutions of x2 = 2x + 3 i.e X' -4 -3 -2 -1 0 1 2 3 4 X solutions of x2 - 2x - 3 = 0. -1 -2 Y' 42 Ex. (4) 25x2 + 30x + 9 = 0 Ex. (5) x2 + x + 5 = 0 Solution : 25x2 + 30x + 9 = 0 comparing Solution : x2 + x + 5 = 0 comparing with the equation with ax2 + bx + c = 0 ax2 + bx + c = 0 we get a = 25, b = 30, c = 9, we get a = 1, b = 1, c = 5, \ b2 - 4 ac = (30)2 - 4 ´ 25 ´ 9 \ b2 - 4 ac = (1)2 - 4 ´ 1 ´ 5 = 900 - 900 =0 = 1 - 20 b  b 2  4ac = -19 x= 2a b  b 2  4ac 30  0 x= = 2a 2  25 1  19 30  0 −30 − 0 = 2 1 \ x= 50 or x = 50 1  19 30 30 = 2 \ x = - 50 or x = - 50 But −19 is not a real number. Hence 3 3 \ x=- or x = - roots of the equation are not real. 5 5 that is both the roots are equal. Also note that 25x2 + 30x + 9 = 0 means (5x + 3)2 = 0 Activity : Solve the equation 2x2 + 13x + 15 = 0 by factorisation method, by completing the square method and by using the formula. Verify that you will get the same roots every time. Practice Set 2.4 1. Compare the given quadratic equations to the general form and write values of a, b, c. (1) x2 - 7x + 5 = 0 (2) 2m2 = 5m - 5 (3) y2 = 7y 2. Solve using formula. (1) x2 + 6x + 5 = 0 (2) x2 - 3x - 2 = 0 (3) 3m2 + 2m - 7 = 0 1 (4) 5m2 - 4m - 2 = 0 (5) y2 + y = 2 (6) 5x2 + 13x + 8 = 0 3 43 (3) With the help of the flow chart given below solve the equation x2 + 2 3 x + 3 = 0 using the formula. Solution : compare equations Write formula Substitute Find value x2 + 2 3 x + 3 = 0 and to solve values of of b2 - 4ac ax2 + bx + c = 0 find quadratic a,b,c and the values of a,b,c. find roots. equation. Let’s learn. Nature of roots of a quadratic equation b  b 2  4ac You know that x = are roots of quadratic equation ax2 + bx + c = 0 2a b  0 b  0 −b − 0 (1) If b2 - 4ac = 0 then, x = \ x = 2a or x = 2a 2a \ the roots of the quadratic equation are real and equal. b  b 2  4ac (2) If b2 - 4ac > 0, then x = 2a b  b 2  4ac −b − b 2 − 4ac i.e. x = and x = 2a 2a \ roots of the quadratic equation are real and unequal. b  b 2  4ac (3) If b2 - 4ac < 0 then x = are not real numbers \the roots of 2a quadratic equations are not real. Nature of roots of quadratic equation is determined by the value of b2 - 4ac. b2 - 4ac is called discriminant of a quadratic equation and is denoted by Greek letter ∆ (Delta) Activity - Fill in the blanks. Value of discriminant Nature of roots (1) 50 (2) -30 (3) 0 44  ÒÒÒ Solved examples ÒÒÒ Ex. (1) Find the value of the discriminant of the equation x2 + 10x - 7 = 0 Solution : Comparing x2 + 10x - 7 = 0 with ax2 + bx + c = 0. a = 1, b = 10 , c = -7 , 2 \ b2 - 4 ac = 10 - 4 ´ 1 ´ -7 = 100 + 28 = 128 Ex. (2) Determine nature of roots of the quadratic equations. (i) 2x2 - 5x + 7 = 0 (ii) x2 + 2x - 9 = 0 Solution : Compare 2x2 - 5x + 7 = 0 with Solution : Compare x2 + 2x - 9 = 0 with ax2 + bx + c = 0 ax2 + bx + c = 0. a = 2, b = -5 , c = 7 , a= , b = 2, c = , \ b2 - 4 ac = (-5)2 - 4 ´ 2 ´ 7 \ b2 - 4 ac = 22 - 4 ´ ´ D = 25 - 56 D = 4 - D = -31 D = 40 \ b2 - 4 ac < 0 \ b2 - 4 ac > 0 \ the roots of the equation are not real. \ the roots of the equation are real and unequal. Ex. (3) 3 x2 + 2 3 x + 3 =0 Solution : Compare 3 x2 + 2 3 x + 3 = 0 with ax2 + bx + c = 0 We get a = 3 , b = 2 3 , c = 3 , \ b2 - 4 ac = (2 3 )2 - 4 ´ 3 ´ 3 = 4´3- 4´3 = 12 - 12 = 0 \ b2 - 4 ac = 0 \ Roots of the equation are real and equal. 45  Let’s learn. The relation between roots of the quadratic equation and coefficients a and b are the roots of the equation ax2 + bx + c = 0 then, b  b 2  4 ac −b − b 2 − 4ac a+b= + b  b 2  4 ac −b − b 2 − 4ac 2a 2a ´ a b= ´ 2a 2a b  b  4ac  b  b  4ac  b    2 2 = b 2  4ac  b  b 2  4ac 2a = 2 2b 4a = - 2a b 2   b 2  4ac  = 4a 2 \ a + b = -b a 4ac = 4a 2 c = a c \ ab= a Activity : Fill in the empty boxes below properly For 10x2 + 10x + 1 = 0, a+b= and a ´ b = ÒÒÒ Solved examples ÒÒÒ Ex. (1) If a and b are the roots of the quadratic equation 2x2 + 6x - 5 = 0, then find (a + b) and a ´ b. Solution : Comparing 2x2 + 6x - 5 = 0 with ax2 + bx + c = 0. \ a = 2, b = 6 , c = -5 b 6 \ a + b = - a = - 2 = -3 c −5 and a ´ b = = 2 a 46 Ex. (2) The difference between the roots of the equation x2 - 13x + k = 0 is 7 find k. Solution : Comparing x2 - 13x + k = 0 with ax2 + bx + c = 0 a = 1, b = -13 , c = k Let a and b be the roots