maths class 10 textbook, Study notes of Mathematics

Download maths class 10 textbook and more Study notes Mathematics in PDF only on Docsity! 70 MATHEMATICS 4 4.1 Introduction In Chapter 2, you have studied different types of polynomials. One type was the quadratic polynomial of the form ax2 + bx + c, a ≠ 0. When we equate this polynomial to zero, we get a quadratic equation. Quadratic equations come up when we deal with many real-life situations. For instance, suppose a charity trust decides to build a prayer hall having a carpet area of 300 square metres with its length one metre more than twice its breadth. What should be the length and breadth of the hall? Suppose the breadth of the hall is x metres. Then, its length should be (2x + 1) metres. We can depict this information pictorially as shown in Fig. 4.1. Now, area of the hall = (2x + 1). x m2 = (2x2 + x) m2 So, 2x2 + x = 300 (Given) Therefore, 2x2 + x – 300 = 0 So, the breadth of the hall should satisfy the equation 2x2 + x – 300 = 0 which is a quadratic equation. Many people believe that Babylonians were the first to solve quadratic equations. For instance, they knew how to find two positive numbers with a given positive sum and a given positive product, and this problem is equivalent to solving a quadratic equation of the form x2 – px + q = 0. Greek mathematician Euclid developed a geometrical approach for finding out lengths which, in our present day terminology, are solutions of quadratic equations. Solving of quadratic equations, in general form, is often credited to ancient Indian mathematicians. In fact, Brahmagupta (C.E.598–665) gave an explicit formula to solve a quadratic equation of the form ax2 + bx = c. Later, QUADRATIC EQUATIONS Fig. 4.1 2019-20 QUADRATIC EQUATIONS 71 Sridharacharya (C.E. 1025) derived a formula, now known as the quadratic formula, (as quoted by Bhaskara II) for solving a quadratic equation by the method of completing the square. An Arab mathematician Al-Khwarizmi (about C.E. 800) also studied quadratic equations of different types. Abraham bar Hiyya Ha-Nasi, in his book ‘Liber embadorum’ published in Europe in C.E. 1145 gave complete solutions of different quadratic equations. In this chapter, you will study quadratic equations, and various ways of finding their roots. You will also see some applications of quadratic equations in daily life situations. 4.2 Quadratic Equations A quadratic equation in the variable x is an equation of the form ax2 + bx + c = 0, where a, b, c are real numbers, a ≠ 0. For example, 2x2 + x – 300 = 0 is a quadratic equation. Similarly, 2x2 – 3x + 1 = 0, 4x – 3x2 + 2 = 0 and 1 – x2 + 300 = 0 are also quadratic equations. In fact, any equation of the form p(x) = 0, where p(x) is a polynomial of degree 2, is a quadratic equation. But when we write the terms of p(x) in descending order of their degrees, then we get the standard form of the equation. That is, ax2 + bx + c = 0, a ≠ 0 is called the standard form of a quadratic equation. Quadratic equations arise in several situations in the world around us and in different fields of mathematics. Let us consider a few examples. Example 1 : Represent the following situations mathematically: (i) John and Jivanti together have 45 marbles. Both of them lost 5 marbles each, and the product of the number of marbles they now have is 124. We would like to find out how many marbles they had to start with. (ii) A cottage industry produces a certain number of toys in a day. The cost of production of each toy (in rupees) was found to be 55 minus the number of toys produced in a day. On a particular day, the total cost of production was ` 750. We would like to find out the number of toys produced on that day. Solution : (i) Let the number of marbles John had be x. Then the number of marbles Jivanti had = 45 – x (Why?). The number of marbles left with John, when he lost 5 marbles = x – 5 The number of marbles left with Jivanti, when she lost 5 marbles = 45 – x – 5 = 40 – x 2019-20 74 MATHEMATICS (ii) The product of two consecutive positive integers is 306. We need to find the integers. (iii) Rohan’s mother is 26 years older than him. The product of their ages (in years) 3 years from now will be 360. We would like to find Rohan’s present age. (iv) A train travels a distance of 480 km at a uniform speed. If the speed had been 8 km/h less, then it would have taken 3 hours more to cover the same distance. We need to find the speed of the train. 4.3 Solution of a Quadratic Equation by Factorisation Consider the quadratic equation 2x2 – 3x + 1 = 0. If we replace x by 1 on the LHS of this equation, we get (2 × 12) – (3 × 1) + 1 = 0 = RHS of the equation. We say that 1 is a root of the quadratic equation 2x2 – 3x + 1 = 0. This also means that 1 is a zero of the quadratic polynomial 2x2 – 3x + 1. In general, a real number α is called a root of the quadratic equation ax2 + bx + c = 0, a ≠ 0 if a α2 + bα + c = 0. We also say that x = α is a solution of the quadratic equation, or that α satisfies the quadratic equation. Note that the zeroes of the quadratic polynomial ax2 + bx + c and the roots of the quadratic equation ax2 + bx + c = 0 are the same. You have observed, in Chapter 2, that a quadratic polynomial can have at most two zeroes. So, any quadratic equation can have atmost two roots. You have learnt in Class IX, how to factorise quadratic polynomials by splitting their middle terms. We shall use this knowledge for finding the roots of a quadratic equation. Let us see how. Example 3 : Find the roots of the equation 2x2 – 5x + 3 = 0, by factorisation. Solution : Let us first split the middle term – 5x as –2x –3x [because (–2x) × (–3x) = 6x2 = (2x2) × 3]. So, 2x2 – 5x + 3 = 2x2 – 2x – 3x + 3 = 2x (x – 1) –3(x – 1) = (2x – 3)(x – 1) Now, 2x2 – 5x + 3 = 0 can be rewritten as (2x – 3)(x – 1) = 0. So, the values of x for which 2x2 – 5x + 3 = 0 are the same for which (2x – 3)(x – 1) = 0, i.e., either 2x – 3 = 0 or x – 1 = 0. Now, 2x – 3 = 0 gives 3 2 x = and x – 1 = 0 gives x = 1. So, 3 2 x = and x = 1 are the solutions of the equation. In other words, 1 and 3 2 are the roots of the equation 2x2 – 5x + 3 = 0. Verify that these are the roots of the given equation. 2019-20 QUADRATIC EQUATIONS 75 Note that we have found the roots of 2x2 – 5x + 3 = 0 by factorising 2x2 – 5x + 3 into two linear factors and equating each factor to zero. Example 4 : Find the roots of the quadratic equation 6x2 – x – 2 = 0. Solution : We have 6x2 – x – 2 = 6x2 + 3x – 4x – 2 = 3x (2x + 1) – 2 (2x + 1) = (3x – 2)(2x + 1) The roots of 6x2 – x – 2 = 0 are the values of x for which (3x – 2)(2x + 1) = 0 Therefore, 3x – 2 = 0 or 2x + 1 = 0, i.e., x = 2 3 or x = 1 2 − Therefore, the roots of 6x2 – x – 2 = 0 are 2 1 .and – 3 2 We verify the roots, by checking that 2 1 and 3 2 − satisfy 6x2 – x – 2 = 0. Example 5 : Find the roots of the quadratic equation 23 2 6 2 0x x− + = . Solution : 23 2 6 2x x− + = 23 6 6 2x x x− − + = ( ) ( )3 3 2 2 3 2x x x− − − = ( )( )3 2 3 2x x− − So, the roots of the equation are the values of x for which ( )( )3 2 3 2 0x x− − = Now, 3 2 0x − = for 2 3 x = . So, this root is repeated twice, one for each repeated factor 3 2x − . Therefore, the roots of 23 2 6 2 0x x− + = are 2 3 , 2 3 . 2019-20 76 MATHEMATICS Example 6 : Find the dimensions of the prayer hall discussed in Section 4.1. Solution : In Section 4.1, we found that if the breadth of the hall is x m, then x satisfies the equation 2x2 + x – 300 = 0. Applying the factorisation method, we write this equation as 2x2 – 24x + 25x – 300 = 0 2x (x – 12) + 25 (x – 12) = 0 i.e., (x – 12)(2x + 25) = 0 So, the roots of the given equation are x = 12 or x = – 12.5. Since x is the breadth of the hall, it cannot be negative. Thus, the breadth of the hall is 12 m. Its length = 2x + 1 = 25 m. EXERCISE 4.2 1. Find the roots of the following quadratic equations by factorisation: (i) x2 – 3x – 10 = 0 (ii) 2x2 + x – 6 = 0 (iii) 2 2 7 5 2 0x x+ + = (iv) 2x2 – x + 1 8 = 0 (v) 100 x2 – 20x + 1 = 0 2. Solve the problems given in Example 1. 3. Find two numbers whose sum is 27 and product is 182. 4. Find two consecutive positive integers, sum of whose squares is 365. 5. The altitude of a right triangle is 7 cm less than its base. If the hypotenuse is 13 cm, find the other two sides. 6. A cottage industry produces a certain number of pottery articles in a day. It was observed on a particular day that the cost of production of each article (in rupees) was 3 more than twice the number of articles produced on that day. If the total cost of production on that day was ` 90, find the number of articles produced and the cost of each article. 4.4 Solution of a Quadratic Equation by Completing the Square In the previous section, you have learnt one method of obtaining the roots of a quadratic equation. In this section, we shall study another method. Consider the following situation: The product of Sunita’s age (in years) two years ago and her age four years from now is one more than twice her present age. What is her present age? To answer this, let her present age (in years) be x. Then the product of her ages two years ago and four years from now is (x – 2)(x + 4). 2019-20 QUADRATIC EQUATIONS 79 So, 9x2 – 15x + 6 = 0 can be written as 2 5 1 3 2 4 x   − −    = 0 i.e., 2 5 3 2 x   −    = 1 4 So, the solutions of 9x2 – 15x + 6 = 0 are the same as those of 2 5 1 3 2 4 x   − =    . i.e., 3x – 5 2 = 1 2 or 5 3 2 x− = 1 2 − (We can also write this as 5 1 3 2 2 x − = ± , where ‘±’ denotes ‘plus minus’.) Thus, 3x = 5 1 2 2 + or 5 1 3 2 2 x = − So, x = 5 1 6 6 + or 5 1 6 6 x = − Therefore, x = 1 or x = 4 6 i.e., x = 1 or x = 2 3 Therefore, the roots of the given equation are 1 and 2 . 3 Remark : Another way of showing this process is as follows : The equation 3x2 – 5x + 2 = 0 is the same as 2 5 2 3 3 x x− + = 0 Now, x2 – 5 2 3 3 x + = 2 2 1 5 1 5 2 2 3 2 3 3 x        − − +              2019-20 80 MATHEMATICS = 2 5 2 25 6 3 36 x   − + −    = 2 2 2 5 1 5 1 6 36 6 6 x x       − − = − −            So, the solutions of 3x2 – 5x + 2 = 0 are the same as those of 2 2 5 1 0 6 6 x     − − =        , which are x – 5 6 = ± 1 6 , i.e., x = 5 1 6 6 + = 1 and x = 5 1 6 6 − = 2 3 . Let us consider some examples to illustrate the above process. Example 7 : Solve the equation given in Example 3 by the method of completing the square. Solution : The equation 2x2 – 5x + 3 = 0 is the same as 2 5 3 0. 2 2 x x− + = Now, 2 5 3 2 2 x x− + = 2 2 5 5 3 4 4 2 x     − − +        = 2 5 1 4 16 x   − −    Therefore, 2x2 – 5x + 3 = 0 can be written as 2 5 1 0 4 16 x   − − =    . So, the roots of the equation 2x2 – 5x + 3 = 0 are exactly the same as those of 2 5 1 0 4 16 x   − − =    . Now, 2 5 1 4 16 x   − −    =0 is the same as 2 5 1 4 16 x   − =    Therefore, 5 4 x − = 1 4 ± i.e., x = 5 1 4 4 ± i.e., x = 5 1 5 1 or 4 4 4 4 x+ = − i.e., x = 3 2 or x = 1 2019-20 QUADRATIC EQUATIONS 81 Therefore, the solutions of the equations are 3 2 x = and 1. Let us verify our solutions. Putting 3 2 x = in 2x2 – 5x + 3 = 0, we get 2 3 3 2 – 5 3 0 2 2     + =        , which is correct. Similarly, you can verify that x = 1 also satisfies the given equation. In Example 7, we divided the equation 2x2 – 5x + 3 = 0 throughout by 2 to get x2 – 5 3 2 2 x + = 0 to make the first term a perfect square and then completed the square. Instead, we can multiply throughout by 2 to make the first term as 4x2 = (2x)2 and then complete the square. This method is illustrated in the next example. Example 8 : Find the roots of the equation 5x2 – 6x – 2 = 0 by the method of completing the square. Solution : Multiplying the equation throughout by 5, we get 25x2 – 30x – 10 = 0 This is the same as (5x)2 – 2 × (5x) × 3 + 32 – 32 – 10 = 0 i.e., (5x – 3)2 – 9 – 10 = 0 i.e., (5x – 3)2 – 19 = 0 i.e., (5x – 3)2 = 19 i.e., 5x – 3 = 19± i.e., 5x = 3 19± So, x = 3 19 5 ± Therefore, the roots are 3 19 5 + and 3 19 5 − . Verify that the roots are 3 19 5 + and 3 19 5 − . 2019-20 84 MATHEMATICS Example 11 : Find two consecutive odd positive integers, sum of whose squares is 290. Solution : Let the smaller of the two consecutive odd positive integers be x. Then, the second integer will be x + 2. According to the question, x2 + (x + 2)2 = 290 i.e., x2 + x2 + 4x + 4 = 290 i.e., 2x2 + 4x – 286 = 0 i.e., x2 + 2x – 143 = 0 which is a quadratic equation in x. Using the quadratic formula, we get x = 2 4 572 2 576 2 24 2 2 2 − ± + − ± − ± = = i.e., x = 11 or x = – 13 But x is given to be an odd positive integer. Therefore, x ≠ – 13, x = 11. Thus, the two consecutive odd integers are 11 and 13. Check : 112 + 132 = 121 + 169 = 290. Example 12 : A rectangular park is to be designed whose breadth is 3 m less than its length. Its area is to be 4 square metres more than the area of a park that has already been made in the shape of an isosceles triangle with its base as the breadth of the rectangular park and of altitude 12 m (see Fig. 4.3). Find its length and breadth. Solution : Let the breadth of the rectangular park be x m. So, its length = (x + 3) m. Therefore, the area of the rectangular park = x(x + 3) m2 = (x2 + 3x) m2. Now, base of the isosceles triangle = x m. Therefore, its area = 1 2 × x × 12 = 6 x m2. According to our requirements, x2 + 3x = 6x + 4 i.e., x2 – 3x – 4 = 0 Using the quadratic formula, we get Fig. 4.3 2019-20 QUADRATIC EQUATIONS 85 x = 3 25 2 ± = 3 5 2 ± = 4 or – 1 But x ≠ – 1 (Why?). Therefore, x = 4. So, the breadth of the park = 4m and its length will be 7m. Verification : Area of rectangular park = 28 m2, area of triangular park = 24 m2 = (28 – 4) m2 Example 13 : Find the roots of the following quadratic equations, if they exist, using the quadratic formula: (i) 3x2 – 5x + 2 = 0 (ii) x2 + 4x + 5 = 0 (iii) 2x2 – 2 2 x + 1 = 0 Solution : (i) 3x2 – 5x + 2 = 0. Here, a = 3, b = – 5, c = 2. So, b2 – 4ac = 25 – 24 = 1 > 0. Therefore, x = 5 1 5 1 6 6 ± ± = , i.e., x = 1 or x = 2 3 So, the roots are 2 3 and 1. (ii) x2 + 4x + 5 = 0. Here, a = 1, b = 4, c = 5. So, b2 – 4ac = 16 – 20 = – 4 < 0. Since the square of a real number cannot be negative, therefore 2 4b ac− will not have any real value. So, there are no real roots for the given equation. (iii) 2x2 – 2 2 x + 1 = 0. Here, a = 2, b = 2 2− , c = 1. So, b2 – 4ac = 8 – 8 = 0 Therefore, x = 2 2 0 2 1 0 i.e., 4 2 ± = ± =, 2 2 0 2 1 .0 i.e., 2 x= ± = So, the roots are 1 2 , 1 2 . 2019-20 86 MATHEMATICS Example 14 : Find the roots of the following equations: (i) 1 3, 0x x x + = ≠ (ii) 1 1 3, 0,2 2 x x x − = ≠ − Solution : (i) 1 3x x + = . Multiplying throughout by x, we get x2 + 1 = 3x i.e., x2 – 3x + 1 = 0, which is a quadratic equation. Here, a = 1, b = – 3, c = 1 So, b2 – 4ac = 9 – 4 = 5 > 0 Therefore, x = 3 5 2 ± (Why?) So, the roots are 3 5 3 5 and 2 2 + − . (ii) 1 1 3, 0, 2 2 x x x − = ≠ − . As x ≠ 0, 2, multiplying the equation by x (x – 2), we get (x – 2) – x = 3x (x – 2) = 3x2 – 6x So, the given equation reduces to 3x2 – 6x + 2 = 0, which is a quadratic equation. Here, a = 3, b = – 6, c = 2. So, b2 – 4ac = 36 – 24 = 12 > 0 Therefore, x = 6 12 6 2 3 3 3 . 6 6 3 ± ± ± = = So, the roots are 3 3 3 3 and 3 3 + − . 2019-20 QUADRATIC EQUATIONS 89 If b2 – 4ac = 0, then x = 0 i.e., or – 2 2 2 b b b a a a − ± = − ⋅, 0 i.e., 2 b x a − ± So, the roots of the equation ax2 + bx + c = 0 are both 2 b a − ⋅ Therefore, we say that the quadratic equation ax2 + bx + c = 0 has two equal real roots in this case. If b2 – 4ac < 0, then there is no real number whose square is b2 – 4ac. Therefore, there are no real roots for the given quadratic equation in this case. Since b2 – 4ac determines whether the quadratic equation ax2 + bx + c = 0 has real roots or not, b2 – 4ac is called the discriminant of this quadratic equation. So, a quadratic equation ax2 + bx + c = 0 has (i) two distinct real roots, if b2 – 4ac > 0, (ii) two equal real roots, if b2 – 4ac = 0, (iii) no real roots, if b2 – 4ac < 0. Let us consider some examples. Example 16 : Find the discriminant of the quadratic equation 2x2 – 4x + 3 = 0, and hence find the nature of its roots. Solution : The given equation is of the form ax2 + bx + c = 0, where a = 2, b = – 4 and c = 3. Therefore, the discriminant b2 – 4ac = (– 4)2 – (4 × 2 × 3) = 16 – 24 = – 8 < 0 So, the given equation has no real roots. Example 17 : A pole has to be erected at a point on the boundary of a circular park of diameter 13 metres in such a way that the differences of its distances from two diametrically opposite fixed gates A and B on the boundary is 7 metres. Is it possible to do so? If yes, at what distances from the two gates should the pole be erected? Solution : Let us first draw the diagram (see Fig. 4.4). Let P be the required location of the pole. Let the distance of the pole from the gate B be x m, i.e., BP = x m. Now the difference of the distances of the pole from the two gates = AP – BP (or, BP – AP) = 7 m. Therefore, AP = (x + 7) m. Fig. 4.4 2019-20 90 MATHEMATICS Now, AB = 13m, and since AB is a diameter, ∠APB = 90° (Why?) Therefore, AP2 + PB2 = AB2 (By Pythagoras theorem) i.e., (x + 7)2 + x2 = 132 i.e., x2 + 14x + 49 + x2 = 169 i.e., 2x2 + 14x – 120 = 0 So, the distance ‘x’ of the pole from gate B satisfies the equation x2 + 7x – 60 = 0 So, it would be possible to place the pole if this equation has real roots. To see if this is so or not, let us consider its discriminant. The discriminant is b2 – 4ac = 72 – 4 × 1 × (– 60) = 289 > 0. So, the given quadratic equation has two real roots, and it is possible to erect the pole on the boundary of the park. Solving the quadratic equation x2 + 7x – 60 = 0, by the quadratic formula, we get x = 7 289 2 − ± = 7 17 2 − ± Therefore, x = 5 or – 12. Since x is the distance between the pole and the gate B, it must be positive. Therefore, x = – 12 will have to be ignored. So, x = 5. Thus, the pole has to be erected on the boundary of the park at a distance of 5m from the gate B and 12m from the gate A. Example 18 : Find the discriminant of the equation 3x2 – 2x + 1 3 = 0 and hence find the nature of its roots. Find them, if they are real. Solution : Here a = 3, b = – 2 and 1 3 c = . Therefore, discriminant b2 – 4ac = (– 2)2 – 4 × 3 × 1 3 = 4 – 4 = 0. Hence, the given quadratic equation has two equal real roots. The roots are 2 2 1 1, ,, , .i.e., , i.e., 2 2 6 6 3 3 b b a a − − 2019-20 QUADRATIC EQUATIONS 91 EXERCISE 4.4 1. Find the nature of the roots of the following quadratic equations. If the real roots exist, find them: (i) 2x2 – 3x + 5 = 0 (ii) 3x2 – 4 3 x + 4 = 0 (iii) 2x2 – 6x + 3 = 0 2. Find the values of k for each of the following quadratic equations, so that they have two equal roots. (i) 2x2 + kx + 3 = 0 (ii) kx (x – 2) + 6 = 0 3. Is it possible to design a rectangular mango grove whose length is twice its breadth, and the area is 800 m2? If so, find its length and breadth. 4. Is the following situation possible? If so, determine their present ages. The sum of the ages of two friends is 20 years. Four years ago, the product of their ages in years was 48. 5. Is it possible to design a rectangular park of perimeter 80 m and area 400 m2? If so, find its length and breadth. 4.6 Summary In this chapter, you have studied the following points: 1. A quadratic equation in the variable x is of the form ax2 + bx + c = 0, where a, b, c are real numbers and a ≠ 0. 2. A real number α is said to be a root of the quadratic equation ax2 + bx + c = 0, if aα2 + bα + c = 0. The zeroes of the quadratic polynomial ax2 + bx + c and the roots of the quadratic equation ax2 + bx + c = 0 are the same. 3. If we can factorise ax2 + bx + c, a ≠ 0, into a product of two linear factors, then the roots of the quadratic equation ax2 + bx + c = 0 can be found by equating each factor to zero. 4. A quadratic equation can also be solved by the method of completing the square. 5. Quadratic formula: The roots of a quadratic equation ax2 + bx + c = 0 are given by 2 4 , 2 b b ac a − ± − provided b2 – 4ac ≥ 0. 6. A quadratic equation ax2 + bx + c = 0 has (i) two distinct real roots, if b2 – 4ac > 0, (ii) two equal roots (i.e., coincident roots), if b2 – 4ac = 0, and (iii) no real roots, if b2 – 4ac < 0. 2019-20