Quadratic Equations PDF

A 5 Quadratic Equations AN 5.1 I NT RO DU CT I O N NG Sports committee of Dhannur High School wants to construct a Kho-K...

A 5 Quadratic Equations AN 5.1 I NT RO DU CT I O N NG Sports committee of Dhannur High School wants to construct a Kho-Kho court of dimensions 29 m × 16 m. This is to be x laid in a rectangular plot of area 558 m2. They want to leave space of equal width 16+2x m. all around the court for the spectators. 16 m. What would be the width of the space x LA for spectators? Would it be enough? Suppose the width of the space be 29 m. x meter. So, from the figure the length of 29+2x m. the plot would be (29 + 2x) meter. TE And, breadth of the rectangular plot would be = (16 + 2x) m. Therefore, area of the rectangular plot = length × breadth = (29 + 2x) (16 + 2x) Since the area of the plot is = 558 m2 \ (29 + 2x) (16 + 2x) = 558 T 2 \ 4x + 90x + 464 = 558 ER 2 4x + 90x - 94 =0 (dividing by 2) 2 2x + 45x - 47 =0 2x2 + 45 x - 47 = 0..... (1) In previous class we solved the linear equations of the form ax + b = c to find the value of ‘x’. Similarly, the value of x from the above equation will give the possible width of the space for SC spectators. Can you think of more such examples where we have to find the quantities, like in the above example and get such equations. Let us consider another example: Rani has a square metal sheet. She removed squares of side 9 cm from each corner of this sheet. Of the remaining sheet, she turned up the sides to form an open box as shown. The capacity of the box is 144 cm3. Can we find out the dimensions of the metal sheet?  106 Class-X Mathematics Suppose the side of the square piece of metal sheet A be ‘x’ cm. Then, the dimensions of the box are 9 cm. 9 cm. AN 9 cm × (x-18) cm × (x-18) cm x cm. Since volume of the box is 144 cm3 9 (x-18) (x-18) = 144 9 cm. 9 cm. (x-18)2 = 16 NG x2 - 36x + 308 = 0 x cm. So, the side ‘x’ of the metal sheet have to satisfy the equation. x2 - 36x + 308 = 0..... (2) 9 cm. Let us observe the L.H.S of equation (1) and (2) LA 8 Are they quadratic polynomials? -1 x - 18 x 2 We studied quadratic polynomials of the form ax + bx + c, a ¹ 0 in the previous chapter. Since, the LHS of the above equations are quadratic polynomials and the RHS is 0 they TE are called quadratic equations. In this chapter we will study quadratic equations and methods to find their roots. 5.2 Q UADRATIC E QUATIONS A quadratic equation in the variable x is an equation of the form ax2 + bx + c = 0, where a, b, c are real numbers and a ¹ 0. For example, 2x2 + x - 300 = 0 is a quadratic equation, T 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 ER quadratic equation. 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 ¹ 0 is called the standard form of a quadratic equation and y = ax2 + bx + c is called a quadratic function. TRY THIS SC Check whether the following equations are quadratic or not ? (i) x2 - 6x - 4 = 0 (ii) x3 - 6x2 + 2x - 1 = 0 1 (iii) 7x = 2x2 (iv) x 2 + = 2 (x ¹ 0) x2 (v) (2x + 1) (3x + 1) = b(x - 1) (x - 2) (vi) 3y2 = 192 Free Distribution by T.S. Government 2019-20  Quadratic Equations 107 There are various situations described by quadratic functions. Some of them are:- A 1. When a rocket is fired upward, then the path of the rocket is defined by a ‘quadratic function.’ AN 2. Shapes of the satellite dish, reflecting mirror in a telescope, lens of the eye glasses and orbits of the celestial objects are defined by the quadratic functions. NG Satellite Dish Reflecting Mirror Lens of Spectacles LA Artificial Satellite Earth Earth Sun TE 3. The path of a projectile is defined by a quadratic function. T ER 4. When the brakes are applied to a vehicle, the stopping distance is calculated by using a quadratic equation. Example-1. Represent the following situations with suitable mathematical equations. i. Sridhar and Rajendar together have 45 marbles. Both of them lost 5 marbles each, SC and the product of the number of marbles now they have is 124. We would like to find out how many marbles each of them had previously. ii. The hypotenuse of a right triangle is 25 cm. We know that the difference in lengths of the other two sides is 5 cm. We would like to find out the length of the two sides? Solution : i. Let the number of marbles Sridhar had be x. Free Distribution by T.S. Government 2019-20  108 Class-X Mathematics Then the number of marbles Rajendar had = 45 – x (Why?). A The number of marbles left with Sridhar, when he lost 5 marbles = x – 5 AN The number of marbles left with Rajendar, when he lost 5 marbles = (45 – x) – 5 = 40 – x Therefore, their product = (x – 5) (40 – x) = 40x – x2 – 200 + 5x = – x2 + 45x – 200 NG So, – x2 + 45x – 200 = 124 (Given that product = 124) i.e., – x2 + 45x – 324 = 0 i.e., x2 – 45x + 324 = 0 LA (Multiply by -1) Therefore, the number of marbles Sridhar had ‘x’, should satisfy the quadratic equation x2 – 45x + 324 = 0 which is the required representation of the problem. ii. Let the length of smaller side be x cm TE Then length of larger side = (x + 5) cm Given length of hypotenuse = 25 cm We know that in a right angle triangle (hypotenuse)2 = (side)2 + (side)2 So, x2 + (x + 5)2 = (25)2 T x2 + x2 + 10x + 25 = 625 25 x+5 cm. cm. 2 2x + 10x - 600 =0 ER x2 + 5x - 300 =0 x Value of x from the above equation will give the possible value of length of sides of the given right angled triangle. SC Example-2. Check whether the following are quadratic equations: i. (x – 2)2 + 1 = 2x – 3 ii. x(x + 1) + 8 = (x + 2) (x – 2) iii. x (2x + 3) = x2 + 1 iv. (x + 2)3 = x3 – 4 Solution : i. LHS = (x – 2)2 + 1 = x2 – 4x + 4 + 1 = x2 – 4x + 5 Therefore, (x – 2)2 + 1 = 2x – 3 can be written as x2 – 4x + 5 = 2x – 3 Free Distribution by T.S. Government 2019-20  Quadratic Equations 109 i.e., x2 – 6x + 8 = 0 A It is in the form of ax2 + bx + c = 0. AN Therefore, the given equation is a quadratic equation. ii. Here LHS = x(x + 1) + 8 = x2 + x + 8 and RHS = (x + 2)(x – 2) = x2 – 4 Therefore, x2 + x + 8 = x2 – 4 NG x2 + x + 8 - x2 + 4 = 0 i.e., x + 12 = 0 It is not in the form of ax2 + bx + c = 0, (a ¹ 0) Therefore, the given equation is not a quadratic equation. LA iii. Here, LHS = x (2x + 3) = 2x2 + 3x So, x (2x + 3) = x2 + 1 can be rewritten as 2x2 + 3x = x2 + 1 Therefore, we get x2 + 3x – 1 = 0 TE It is in the form of ax2 + bx + c = 0. So, the given equation is a quadratic equation. iv. Here, LHS = (x + 2)3 = (x + 2)2 (x + 2) = (x2 + 4x + 4) (x + 2) T = x3 + 2x2 + 4x2 + 8x + 4x + 8 = x3 + 6x2 + 12x + 8 ER Therefore, (x + 2)3 = x3 – 4 can be rewritten as x3 + 6x2 + 12x + 8 = x3 – 4 i.e., 6x2 + 12x + 12 = 0 or, x2 + 2x + 2 = 0 It is in the form of ax2 + bx + c = 0. SC So, the given equation is a quadratic equation. Remark : In (ii) above, the given equation appears to be a quadratic equation, but it is not a quadratic equation. In (iv) above, the given equation appears to be a cubic equation (an equation of degree 3) and not a quadratic equation. But it turns out to be a quadratic equation. As you can see, often we need to simplify the given equation before deciding whether it is quadratic or not. Free Distribution by T.S. Government 2019-20  110 Class-X Mathematics A EXERCISE - 5.1 1. Check whether the following are quadratic equations : AN i. (x + 1)2 = 2(x – 3) ii. x2 – 2x = (–2) (3 – x) iii. (x – 2)(x + 1) = (x – 1)(x + 3) iv. (x – 3)(2x +1) = x(x + 5) v. (2x – 1)(x – 3) = (x + 5)(x – 1) vi. x2 + 3x + 1 = (x – 2)2 vii. (x + 2)3 = 2x (x2 – 1) viii. x3 – 4x2 – x + 1 = (x – 2)3 NG 2. Represent the following situations in the form of quadratic equations : i. The area of a rectangular plot is 528 m2. The length of the plot is one metre more than twice its breadth. We need to find the length and breadth of the plot. ii. The product of two consecutive positive integers is 306. We need to find the integers. LA iii. Rohan’s mother is 26 years older than him. The product of their ages after 3 years will be 360 years. We need 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. TE 5.3 S OLUTION OF A Q UADRATIC E QUATION BY F ACTORISATION We have learned to represent some of the daily life situations in the form of quadratic equation with an unknown variable ‘x’. T Now we need to find the value of x. Consider the quadratic equation 2x2 – 3x + 1 = 0. If we replace x by 1. Then, we get (2 × 12) – (3 × 1) + 1 = 0 = RHS of the equation. Since 1 satisfies the equation , we say that 1 ER is a root of the quadratic equation 2x2 – 3x + 1 = 0. \ x = 1 is a solution of the quadratic equation. This also means that 1 is a zero of the quadratic polynomial 2x2 – 3x + 1. In general, a real number a is called a root of the quadratic equation ax2 + bx + c = 0, SC if aa2 + b a + c = 0. We also say that x = a is a solution of the quadratic equation, or a satisfies the quadratic equation. Note that the zeroes of the quadratic polynomial ax2 + bx + c (a ¹ 0) and the roots of the quadratic equation ax2 + bx + c = 0 (a ¹ 0) are the same. We have observed, in Chapter 3, that a quadratic polynomial can have at most two zeroes. So, any quadratic equation can have at most two roots. (Why?) Free Distribution by T.S. Government 2019-20  Quadratic Equations 111 We have learnt in Class-IX, how to factorise quadratic polynomials by splitting their middle A terms. We shall use this knowledge for finding the roots of a quadratic equation. Let us see. Example-3. Find the roots of the equation 2x2 – 5x + 3 = 0, by factorisation. AN Solution : Let us first split the middle term. Recall that if ax2 + bx + c is a quadratic polynomial then to split the middle term we have to find two numbers p and q such that p + q = b and p × q = a × c. So to split the middle term of 2x2 – 5x + 3, we have to find two numbers p and q such that p + q = –5 and p × q = 2 × 3 = 6. NG For this we have to list out all possible pairs of factors of 6. They are (1, 6), (–1, –6); (2, 3); (–2, –3). From the list it is clear that the pair (–2, –3) will satisfy our condition p + q = –5 and p × q = 6. The middle term ‘–5x’ can be written as ‘–2x – 3x’. So, 2x2 – 5x + 3 = 2x2 – 2x – 3x + 3 = 2x (x – 1) –3(x – 1) = (2x – 3)(x – 1) LA Now, 2x2 – 5x + 3 = 0 can be rewritten as (2x – 3)(x – 1) = 0. So, the values of x for 2x2 – 5x + 3 = 0 are the same for (2x – 3)(x – 1) = 0, i.e., either 2x – 3 = 0 or x – 1 = 0. TE 3 Now, 2x – 3 = 0 gives x = and x – 1 = 0 gives x = 1. 2 3 So, x = and x = 1 are the solutions of the equation. 2 3 In other words, 1 and are the roots of the equation 2x2 – 5x + 3 = 0. T 2 Do This ER Find the roots of the following equations using factorisation method. (i) x2 + 5x + 6 = 0 (ii) x2 - 5x + 6= 0 (iii) x2 + 5x - 6 = 0 (iv) x2 - 5x – 6= 0 SC TRY THIS 3 Verify whether 1 and are the roots of the equation 2x2 – 5x + 3 = 0. 2 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. Free Distribution by T.S. Government 2019-20  112 Class-X Mathematics A 1 1 Example 4 : Find the roots of the equation x- = (x ¹ 0) 3x 6 AN 1 1 Solution : We have x- = Þ 6x2 - x - 2 = 0 3x 6 6x – x – 2 = 6x2 + 3x – 4x – 2 2 = 3x (2x + 1) – 2 (2x + 1) = (3x – 2)(2x + 1) 2 The roots of 6x – x – 2 = 0 are the values of x for which (3x – 2)(2x + 1) = 0 NG Therefore, 3x – 2 = 0 or 2x + 1 = 0, 2 1 i.e., x = or x = - 3 2 2 1 Therefore, the roots of 6x2 – x – 2 = 0 are and -. 3 2 LA 2 1 We verify the roots, by checking that and - satisfy 6x2 – x – 2 = 0. 3 2 Example-5. Find the width of the space for spectators discussed in section 5.1. Solution : In Section 5.1, we found that if the width of the space for spectators is x m, then x satisfies the equation 2x2 + 45x - 47 = 0. Applying the factorisation method we write this equation TE as:- 2x2 - 2x + 47x - 47 = 0 2x (x - 1) + 47 (x - 1) = 0 i.e., (x - 1) (2x + 47) = 0 T -47 So, the roots of the given equation are x = 1 or x =. Since ‘x’ is the width of space 2 ER of the spectators it cannot be negative. Thus, the width is 1 m. So it is not enough for spectators. E XERCISE - 5.2 SC 1. Find the roots of the following quadratic equations by factorisation: i. x2 – 3x – 10 = 0 ii. 2x2 + x – 6 = 0 iii. 2 x2 + 7 x + 5 2 = 0 2 1 iv. 2 x - x + =0 v. 100x2 – 20x + 1 = 0 vi. x(x + 4) = 12 8 3 vii. 3x2 – 5x + 2 = 0 viii. x - = 2 (x ¹ 0) ix. 3(x – 4)2 – 5(x – 4) = 12 x Free Distribution by T.S. Government 2019-20  Quadratic Equations 113 A 2. Find two numbers whose sum is 27 and product is 182. 3. Find two consecutive positive integers, sum of whose squares is 613. AN 4. The altitude of a right triangle is 7 cm less than its base. If the hypotenuse is 13 cm, find the other two sides. 5. 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 NG day was Rs 90, find the number of articles produced and the cost of each article. 6. Find the dimensions of a rectangle whose perimeter is 28 meters and whose area is 40 square meters. 7. The base of a triangle is 4cm longer than its altitude. If the area of the triangle is 48 sq.cm LA then find its base and altitude. 8. Two trains leave a railway station at the same time. The first train travels towards west and the second train towards north. The first train travels 5 km/hr faster than the second train. If after two hours they are 50 km. apart, find the average speed of each train. TE 9. In a class of 60 students, each boy contributed rupees equal to the number of girls and each girl contributed rupees equal to the number of boys. If the total money then collected was D1600. How many boys were there in the class? 10. A motor boat heads upstream a distance of 24 km in a river whose current is running at 3 km per hour. The trip up and back takes 6 hours. Assuming that the motor boat maintained T a constant speed, what was its speed in still water? ER 5.4 S OLUTION OF A Q UADRATIC E QUATION BY C OMPLETING THE S QUARE In the previous section, we have learnt method of factorisation for obtaining the roots of a quadratic equation. Is method of factorization applicable to all types of quadratic equation? Let us try to solve x2 + 4x - 4 = 0 by factorisation method To solve the given equation x2 + 4x - 4 = 0 by factorization method. SC We have to find ‘p’ and ‘q’ such that p + q = 4 and p × q = -4 We have no integers p, q satisfying above equation. So by factorization method it is difficult to solve the given equation. Therefore, we shall try another method. Free Distribution by T.S. Government 2019-20  114 Class-X Mathematics Consider the following situation A The product of Sunita’s age (in years) two years ago and her age four years hence is one more than twice her present age. What is her present age? AN To answer this, let her present age be x years. Their age two years ago would be x – 2 and the age after four years will be x + 4. So, the product of both the ages is (x – 2)(x + 4). Therefore, (x – 2)(x + 4) = 2x + 1 i.e., x2 + 2x – 8 = 2x + 1 NG i.e., x2 – 9 = 0 So, Sunita’s present age satisfies the quadratic equation x2 – 9 = 0. We can write this as x2 = 9. Taking square roots, we get x = 3 or x = – 3. Since the age is a positive number, x = 3. LA So, Sunita’s present age is 3 years. Now consider another quadratic equation (x + 2)2 – 9 = 0. To solve it, we can write it as (x + 2)2 = 9. Taking square roots, we get x + 2 = 3 or x + 2 = – 3. Therefore, x = 1 or x = –5 TE So, the roots of the equation (x + 2)2 – 9 = 0 are 1 and – 5. In both the examples above, the term containing x is inside a square, and we found the roots easily by taking the square roots. But, what happens if we are asked to solve the equation x2 + 4x – 4 = 0, which cannot be solved by factorisation also. So, we now introduce the method of completing the square. The idea behind this T method is to adjust the left side of the quadratic equation so that it becomes a perfect square of the first degree polynomial and the RHS without x term. ER The process is as follows: x2 + 4x – 4 = 0 Þ x2 + 4x = 4 x2 + 2. x. 2 = 4 SC Now, the LHS is in the form of a2 + 2ab. If we add b2 it becomes as a2 + 2ab + b2 which is perfect square. So, by adding b2 = 22 = 4 to both sides we get, x2 + 2.x.2 + 22 = 4 + 4 Þ (x + 2)2 = 8 Þ x + 2 = ± 8 Þ x = –2 ± 2 2 Free Distribution by T.S. Government 2019-20  Quadratic Equations 115 Now consider the equation 3x2 – 5x + 2 = 0. Note that the coefficient of x2 is not 1. So A we divide the entire equation by 3 so that the coefficient of x2 is 1 5 2 AN \ x2 - x + = 0 3 3 5 -2 Þ x2 - x = 3 3 5 -2 Þ x 2 - 2.x. = NG 6 3 5 æ5ö 2 -2 æ 5 ö 2 æ æ5ö 2 ö Þ 2 x - 2.x. + ç ÷ = +ç ÷ çç add ç ÷ both side ÷÷ 6 è6ø 3 è6ø è è6ø ø 2 æ 5ö -2 25 çx- ÷ = + è 6ø 3 36 LA æ 5ö 2 (12 ´ -2) + ( 25 ´ 1) çè x - ÷ø = 6 36 TE 2 æ 5ö -24 + 25 çè x - ÷ø = 6 36 2 æ 5ö 1 çx- ÷ = (take square root both sides) è 6ø 36 5 1 T x- = ± 6 6 5 1 5 1 ER So, x = + or x = - 6 6 6 6 4 Therefore, x = 1 or x = 6 2 i.e., x = 1 or x = SC 3 2 Therefore, the roots of the given equation are 1 and. 3 From the above examples we can deduce the following algorithm for completing the square. Algorithm : Let the quadratic equation be ax2 + bx + c = 0 (a ¹ 0) Step-1 : Divide each side by ‘a’ Free Distribution by T.S. Government 2019-20  116 Class-X Mathematics Step-2 : Rearrange the equation so that constant term c/a is on the right side. (RHS) A 2 é 1 æ b öù Step-3 : Add ê ç ÷ ú to both sides to make LHS, a perfect square. AN ë 2 è a øû Step-4 : Write the LHS as a square and simplify the RHS. Step-5 : Solve it. Example-6. Find the roots of the equation 5x2 – 6x – 2 = 0 by the method of completing the NG square. Solution : Given : 5x2 – 6x – 2 = 0 Now we follow the Alogarithm 6 2 2 LA Step-1 : x - x - = 0 (Dividing both sides by 5) 5 5 2 6 2 Step-2 : x - x = 5 5 TE 6 æ3ö 2 2 æ 3ö 2 æ æ3ö 2 ö 2 Step-3 : x - x + ç ÷ = + ç ÷ çç Adding ç ÷ to both sides ÷÷ 5 è5ø 5 è5ø è è5ø ø 2 æ 3ö 2 9 Step-4 : ç x - ÷ = + è 5ø 5 25 T 2 æ 3 ö 19 Step-5 : ç x - ÷ = ER è 5ø 25 3 19 x- =± 5 25 3 19 3 19 SC x= + or x = - 5 5 5 5 3 + 19 3 - 19 \ x= or x = 5 5 Free Distribution by T.S. Government 2019-20  Quadratic Equations 117 Example-7. Find the roots of 4x2 + 3x + 5 = 0 by the method of completing the square. A Solution : Given 4x2 + 3x + 5 = 0 AN 3 5 x2 + x+ = 0 4 4 3 -5 x2 + x= 4 4 NG 2 2 2 3 æ 3ö -5 æ 3 ö x + x+ç ÷ = +ç ÷ 4 è8ø 4 è8ø 2 æ 3ö -5 9 çx+ ÷ = + è 8ø 4 64 LA -71 2 æ 3ö çx+ ÷ = 0, then by taking the square roots, we get b ± b 2 - 4ac x+ = 2a 2a -b ± b 2 - 4ac Therefore, LA x= 2a -b + b 2 - 4ac -b - b 2 - 4ac So, the roots of ax 2 + bx + c = 0 are and , 2a 2a if b2 – 4ac > 0. TE If b2 – 4ac < 0, the equation will have no real roots. (Why?) Thus, if b2 – 4ac > 0, then the roots of the quadratic equation ax2 + bx + c = 0 are - b ± b 2 - 4ac given by. T 2a This formula for finding the roots of a quadratic equation is known as the quadratic formula. ER Let us consider some examples by using quadratic formula. Example-8. Solve Q. 2(i) of Exercise 5.1 by using the quadratic formula. Solution : Let the breadth of the plot be x metres. SC Then the length is (2x + 1) metres. Since area of rectangular plot is 528 m2 , we can write x(2x + 1) = 528, i.e., 2x2 + x – 528 = 0. This is in the form of ax2 + bx + c = 0, where a = 2, b = 1, c = – 528. So, the quadratic formula gives us the solution as Free Distribution by T.S. Government 2019-20  Quadratic Equations 119 A -1 ± 1 + 4(2)(528) -1 ± 4225 -1 ± 65 x= = = 4 4 4 AN 64 -66 i.e., x= or x = 4 4 33 i.e., x = 16 or x = - 2 Since x cannot be negative. So, the breadth of the plot is 16 metres and hence, the length NG of the plot is (2x + 1) = 33m. You should verify that these values satisfy the conditions of the problem. T HINK - D ISCUSS LA We have three methods to solve a quadratic equation. Among these three, which method would you like to use? Why? Example-9. Find two consecutive positive odd integers, sum of whose squares is 290. TE Solution : Let the first positive odd integer 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 T i.e., 2x2 + 4x – 286 = 0 i.e., x2 + 2x – 143 = 0 ER which is a quadratic equation in x. -b ± b 2 - 4ac Using the quadratic formula x = 2 SC -2 ± 4 + 572 -2 ± 576 -2 ± 24 we get, x = = = 2 2 2 i.e., x = 11 or x = – 13 But x is given to be positive odd integer. Therefore, x ¹ – 13. Thus, the two consecutive odd integers are 11 and (x + 2) = 11 + 2 = 13. Check : 112 + 132 = 121 + 169 = 290. Free Distribution by T.S. Government 2019-20  120 Class-X Mathematics Example-10. A rectangular park is to be designed whose breadth is 3 m less than its length. Its A 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 AN 12 m. 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. NG Now, base of the isosceles triangle = x m. 1 Therefore, its area = × x × 12 = 6 x m2. 2 According to our requirements, 12 LA x2 + 3x = 6x + 4 i.e., x2 – 3x – 4 = 0 Using the quadratic formula, we get x+3 TE 3 ± 25 3 ± 5 x= = = = 4 or – 1 2 2 x But x ¹ – 1 (Why?). Therefore, x = 4. So, the breadth of the park = 4m and its length will be x + 3 = 4 + 3 = 7m. Verification : Area of rectangular park = 28 m2, T area of triangular park = (28 – 4)m2 = 24m2. ER Example-11. Find the roots of the following quadratic equations, if they exist. (i) x2 + 4x + 5 = 0 (ii) 2x2 – 2 2 x + 1 = 0 Solution : SC (i) 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 b2 - 4ac will not have any real value. So, there are no real roots for the given equation. (ii) 2x2 – 2 2 x + 1 = 0. Here, a = 2, b = -2 2 , c = 1. Free Distribution by T.S. Government 2019-20  Quadratic Equations 121 So, b2 – 4ac = 8 – 8 = 0 A 2 2± 0 2 1 Therefore, x = = ± 0 i.e., x =. AN 4 2 2 1 1 So, the roots are ,. 2 2 Example-12. Find the roots of the following equations: NG 1 1 1 (i) x + = 3, x ¹ 0 (ii) - = 3, x ¹ 0, 2 x x x-2 Solution : 1 LA (i) x + = 3. Multiplying both sides of equation by x, we get x x2 + 1 = 3x i.e., x2 – 3x + 1 = 0, which is a quadratic equation. TE Here, a = 1, b = – 3, c = 1 So, b2 – 4ac = 9 – 4 = 5 > 0 3± 5 Therefore, x= (why ?) 2 T 3+ 5 3- 5 So, the roots are and. 2 2 ER 1 1 (ii) - = 3, x ¹ 0, 2. x x-2 As x ¹ 0, 2, multiplying the equation by x (x – 2), we get (x – 2) – x = 3x (x – 2) SC = 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 6 ± 12 6± 2 3 3± 3 Therefore, x= = =. 6 6 3 Free Distribution by T.S. Government 2019-20  122 Class-X Mathematics A 3+ 3 3- 3 So, the roots are and. 3 3 AN Example-13. A motor boat whose speed is 18 km/h in still water. It takes 1 hour more to go 24 km upstream than to return downstream to the same spot. Find the speed of the stream. Solution : Let the speed of the stream be x km/h. Therefore, the speed of the boat upstream = (18 – x) km/h and the speed of the boat NG downstream = (18 + x) km/h. distance 24 The time taken to go upstream = = hours. speed 18 - x 24 Similarly, the time taken to go downstream = hours. 18 + x LA According to the question, 24 24 - =1 18 - x 18 + x TE i.e., 24(18 + x) – 24(18 – x) = (18 – x) (18 + x) i.e., x2 + 48x – 324 = 0 Using the quadratic formula, we get -48 ± 482 + 1296 -48 ± 3600 x= = T 2 2 -48 ± 60 = = 6 or -54 ER 2 Since x is the speed of the stream, it cannot be negative. So, we ignore the root x = – 54. Therefore, x = 6 gives the speed of the stream as 6 km/h. SC E XERCISE - 5.3 1. Find the roots of the following quadratic equations, if they exist. i. 2x2 + x – 4 = 0 ii. 4 x 2 + 4 3x + 3 = 0 iii. 5x2 - 7x - 6 = 0 iv. x2 + 5 = -6x Free Distribution by T.S. Government 2019-20  Quadratic Equations 123 2. Find the roots of the quadratic equations given in Q.1 by applying the quadratic formula. A 3. Find the roots of the following equations: 1 1 1 11 AN (i) x - = 3, x ¹ 0 (ii) - = , x ¹ -4, 7 x x + 4 x - 7 30 4. The sum of the reciprocals of Rehman’s ages, (in years) 3 years ago and 5 years from now 1 is. Find his present age. 3 5. In a class test, the sum of Moulika’s marks in Mathematics and English is 30. If she got 2 NG marks more in Mathematics and 3 marks less in English, the product of her marks would have been 210. Find her marks in the two subjects. 6. The diagonal of a rectangular field is 60 metres more than the shorter side. If the longer side is 30 metres more than the shorter side, find the sides of the field. 7. The difference of squares of two numbers is 180. The square of the smaller number is 8 LA times the larger number. Find the two numbers. 8. A train travels 360 km at a uniform speed. If the speed had been 5 km/h more, it would have taken 1 hour less for the same journey. Find the speed of the train. 3 TE 9. Two water taps together can fill a tank in 9 hours. The tap of larger diameter takes 10 8 hours less than the smaller one to fill the tank separately. Find the time in which each tap can separately fill the tank. 10. An express train takes 1 hour less than a passenger train to travel 132 km between Mysore and Bangaluru (without taking into consideration the time they stop at intermediate stations). If the average speed of the express train is 11km/h more than that of the passenger train, T find the average speed of the two trains. 11. Sum of the areas of two squares is 468 m2. If the difference of their perimeters is 24 m, find ER the sides of the two squares. 12. An object is thrown upwards with an initial velocity of 17 m/sec from a building with 12 m height. It is at a height of S = 12 + 17t – 5t2 from the ground after a flight of ‘t’ seconds. Find the time taken by the object to touch the ground. SC 1 13. If a polygon of ‘n’ sides has n (n-3) diagonals. How many sides are there in a polygon 2 with 65 diagonals? Is there a polygon with 50 diagonals? Free Distribution by T.S. Government 2019-20  124 Class-X Mathematics 5.5 N ATURE ROOTS A OF In the previous section, we have seen that the roots of the equation ax2 + bx + c = 0 are AN given by -b ± b 2 - 4ac x= 2a Now let us try to study the nature of roots. Remember that zeroes are those points where value of polynomial becomes zero or we NG can say that the curve of quadratic polynomial cuts the X-axis. Similarly, roots of a quadratic equation are those points where the curve cuts the X-axis. Case-1 : If b2 - 4ac > 0; -b + b 2 - 4ac -b - b 2 - 4ac We get two distinct real roots , LA 2a 2a In such case if we draw corresponding graph for the given quadratic equation we get the following types of figures. TE Figure shows that the corresponding curve of the quadratic equation cuts the X-axis at two distinct points Case-2 : If b2 - 4ac = 0 T -b + 0 x= 2a ER -b -b So, x = , 2a 2a Figure shows that the graph of the quadratic equation touches X-axis at one point. Case-3 : b2 - 4ac < 0 SC There are no real roots. Roots are imaginary. Free Distribution by T.S. Government 2019-20  Quadratic Equations 125 In this case, the graph neither intersects nor touches the X-axis at all. So, there are no real A roots. Since b2 – 4ac determines whether the quadratic equation ax2 + bx + c = 0 (a ¹ 0) has AN real roots or not, b2 – 4ac is called the discriminant of the quadratic equation. So, a quadratic equation ax2 + bx + c = 0 (a ¹ 0) has i. two distinct real roots, if b2 – 4ac > 0, ii. two equal real roots, if b2 – 4ac = 0, NG iii. no real roots, if b2 – 4ac < 0. Let us consider some examples. Example-14. Find the discriminant of the quadratic equation 2x2 – 4x + 3 = 0, and hence find the nature of its roots. LA Solution : The given equation is in the form of 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 TE So, the given equation has no real roots. Example-15. 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? T Solution : Let us first draw the diagram. Let P be the required location of the pole. Let the distance of the B ER pole from the gate B be x m, i.e., BP = x m. Now the difference of the m 13 distances of the pole from the two gates = AP – BP (or, BP – AP)= 7 m. Therefore, AP = (x + 7) m. A P Now, AB = 13m, and since AB is a diameter, SC ÐAPB = 900 (Why?) 2 2 2 Therefore, AP + PB = AB (By Pythagoras theorem) i.e., (x + 7)2 + x2 = 132 i.e., x2 + 14x + 49 + x2 = 169 i.e., 2x2 + 14x – 120 = 0 Free Distribution by T.S. Government 2019-20  126 Class-X Mathematics So, the distance ‘x’ of the pole from gate B satisfies the equation A x2 + 7x – 60 = 0 AN 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. NG Solving the quadratic equation x2 + 7x – 60 = 0, by the quadratic formula, we get -7 ± 289 -7 ± 17 x= = 2 2 Therefore, x = 5 or – 12. LA 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. TE TRY THIS 1. Explain the benefits of evaluating the discriminant of a quadratic equation before attempting to solve it. What does its value signifies? T 2. Write three quadratic equations, one having two distinct real solutions, one having no real solution and one having exactly one real solution. ER 1 Example-16. Find the discriminant of the equation 3 x 2 - 2 x + = 0 and hence find the nature 3 of its roots. Find them, if they are real. 1 Solution : Here a = 3, b = – 2 and c = SC 3 1 Therefore, discriminant b 2 - 4ac = (-2) 2 - 4 ´ 3 ´ = 4 - 4 = 0. 3 Hence, the given quadratic equation has two equal real roots. -b -b 2 2 1 1 The roots are , , i.e., , , i.e., ,. 2a 2a 6 6 3 3 Free Distribution by T.S. Government 2019-20  Quadratic Equations 127 A EXERCISE - 5.4 1. Find the nature of the roots of the following quadratic equations. If real roots exist, find AN them: (i) 2x2 – 3x + 5 = 0 (ii) 3x2 - 4 3x + 4 = 0 (iii) 2x2 – 6x + 3 = 0 NG 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 (k ¹ 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. LA The sum of the ages of two friends is 20 years. Four years ago, the product of their ages in years was 48. Is the above situation possible? If so, determine their present ages. 5. Is it possible to design a rectangular park of perimeter 80 m and area 400 m2? If so, find its length and breadth. Comment on you answer. TE O PTIONAL E XERCISE [For extensive Learning] 1. Some points are plotted on a plane such that any three of them are non collinear. Each point is joined with all remaining points by line segments. Find the number of points if the T number of line segments are 10. 2. A two digit number is such that the product of its digits is 8. When 18 is added to the ER number they interchange their places. Determine the number. 3. A piece of wire 8 m. in length is cut into two pieces, and each piece is bent into a square. Where should the cut in the wire be made if the sum of the areas of these squares is to be 2 m2? é 2 æ xö æ y ö 2 2 æ xö æ 8 - xö 2 ù ê Hint : x + y = 8, ç ÷ + ç ÷ = 2 Þ ç ÷ + ç = 2 ú. SC è 4ø è 4 ø è 4 ø è 4 ÷ø ëê úû 4. Vinay and Praveen working together can paint the exterior of a house in 6 days. Vinay by himself can complete the job in 5 days less than Praveen. How long will it take Vinay to complete the job. -b 5. Show that the sum of roots of a quadratic equation ax2 + bx + c = 0 (a ¹ 0) is. a Free Distribution by T.S. Government 2019-20  128 Class-X Mathematics A c 6. Show that the product of the roots of a quadratic equation ax2 + bx + c = 0 (a ¹ 0) is. a 16 AN 7. If the sum of the fraction and its reciprocal is 2 , find the fraction. 21 Suggested Projects Solving quadratic equations by geometrical methods. NG l Take two or three quadratic equations of the form ax2 + bx + c = 0, where a ¹ 0, for different situations like a > 0, a < 0, b = 0 and solve them by graphical methods. WHAT WE H AVE D ISCUSSED LA 1. Standard form of quadratic equation in variable x is ax2 + bx + c = 0, where a, b, c are real numbers and a ¹ 0. 2. A real number a is said to be a root of the quadratic equation ax2 + bx + c = 0, if TE aa2 + ba + 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 ¹ 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. T 5. Quadratic formula: The roots of a quadratic equation ax2 + bx + c = 0 (a ¹ 0) are given ER by -b ± b 2 - 4ac , provided b2 – 4ac > 0. 2a 6. A quadratic equation ax2 + bx + c = 0 (a ¹ 0) has SC (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. Free Distribution by T.S. Government 2019-20