Class 10 Quadratic Equations

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1 ID : in-10-quadratic-equations [1] Class 10 Quadratic Equations For more such worksheets visit Answer t he quest ions (1) Had Sunil scored 8 more marks in his mathematics test out of 30 marks, 4 times these marks would have been the square of his actual marks. How many marks did he get in the test? (2) The sum of the n consecutive natural even numbers starting f rom 4 is 70. Find the value of n. (3) In a class, one-third of students had gone to watch drama. Twice the square root of the students had gone to watch magic show, and remaining 12 students had gone to watch f ootball match. Find total number of students in the class. (4) Shilpa buys number of books f or Rs.60. If she had bought 3 more books f or the same amount, each book would have cost Rs.1 less. How many books did she buy? (5) How many integers satisf y the inequality x 2-6 x < - 5 (6) The product of Sneha's age (in years) 5 years ago with her age (in years) 3 years later is 20. Find her current age. (7) Find the sum of the roots of the quadratic equation y 2 + 3y + 2 = 0 Choose correct answer(s) f rom given choice (8) Which of the f ollowing quadratic equations have no real roots? a. -4x 2 + 8x - 4 = 0 b. x 2 + 2x + 1 = 0 c. 2x 2 + 4x + 3 = 0 d. -2x 2 + 4x - 1 = 0 (9) A two digit number is such that the product of its digits is 24. If 18 is subtracted f rom the number, the digits interchange their places. Find the number. a. 59 b. 64 c. 46 d. Cannot be determined (10) Solve quadratic equation = 0 using f actorization. a. x = a/(a + b) or, x = -(a + b)/a b. x = -b/(a + b) or, x = -(a + b)/a c. x = -a/(a + b) or, x = (a + b)/a d. x = -a/(a + b) or, x = -(a + b)/a (11) Equation -x 2-8x + 2 = 0 has a. more than two roots b. two distinct real roots c. two equal real roots d. no real roots

2 ID : in-10-quadratic-equations [2] (12) A piece of copper wire costs Rs.192. If it was 4 meter longer and price of each meter of copper wire costs Rs.4 less, the total cost of the piece would remain unchanged. Find the length of copper wire. a. 11 meter b. 12 meter c. 14 meter d. 10 meter (13) A natural number, when increased by 1, equals 240 times its reciprocal. Find the number. a. 12 b. 16 c. 18 d. 15 (14) Solve quadratic equation p 2 q 2 x 2 + q 2 x - p 2 x - 1 = 0 using f actorization. a. x = -1/p 2, x = 1/q 2 b. x = p 2, x = -q 2 c. x = -p 2, x = q 2 d. x = 1/p 2, x = -1/q 2 (15) Which of the f ollowing equations has -5 as one of its roots? a. x 2-3x - 4 = 0 b. x 2 + 3x - 4 = 0 c. x 2 + 9x + 20 = 0 d. x 2-5x + 4 = Edugain ( All Rights Reserved Many more such worksheets can be generated at

3 Answers ID : in-10-quadratic-equations [3] (1) 8 marks Lets assume Sunil scored x marks out of 30 marks It is given that, 4 (x + 8) = x 2 4x = x 2 x 2-4x - 32 = 0 x 2 + 4x - 8x + (-8)(4) = 0 x (x + 4) - 8 (x + 4)) = 0 (x - 8) (x + 4) = 0 Theref ore x = 8 or x = -4. But since marks should be positive, his actual marks are 8 (2) n terms = 70 n( (n - 1))/2 = 70 [Using AP summation Sn = n/2(2a + d(n - 1) )] n 2 + 3n = 70 n 2 + 3n - 70 = 0 (n - 7) (n + 10) = 0 n = 7 or -10. Since n cannot be negative, n = 7

4 (3) 36 ID : in-10-quadratic-equations [4] Let total number of students = T T/3 + 2 T + 12 = T 2T - 6 T - 36 = 0 Lets assume T = x 2 2x 2-6x - 36 = 0 One of the legal solution of this quadratic equation is x=6. Hence T = x 2 = 36

5 (4) 12 books ID : in-10-quadratic-equations [5] Let the number of books bought = x Price of one book = Rs. 60/x Price of one book, if 3 more items were bought = Rs. 60/(x + 3) Since dif f erence in price is Rs = x 2 + 3x Step 7 x 2 + 3x = 0 Step 8 x x - 12x = 0 Step 9 x (x + 15) - 12(x + 15) = 0 0 (x - 12) (x + 15) = 0 1 x = 12 or -15. Since number of books cannot be negative, x = 12 (5) 3 First lets f ind the value of x f or which x 2-6 x = - 5, or x 2-6 x + 5 = 0 x 2-6 x + 5 = 0 (x - 1) (x - 5) = 0 x = 1 or x = 5 Now you can notice that inequality is satisf ied f or x > 1 and x < 5. Theref ore there 3 such integers

6 (6) 7 years ID : in-10-quadratic-equations [6] Lets Sneha's current age is x (x - 5) (x + 3) = 20 x 2 + (3-5) x - 15 = 20 x 2-2 x - 35 = 0 x 2 + 5x - 7x - 35 = 0 x (x + 5) - 7 ( x + 5) = 0 Step 7 (x + 5) (x - 7) = 0 Step 8 x = 7 or -5. Since age cannot be negative x = 7 years (7) -3 For quadratic equation ay 2 + by + c = 0, sum of roots is -b/a For given equation y 2 + 3y + 2 = 0, a = 1, b = 3 and c = 2 Theref ore sum of roots, S = (-1 b)/(a) S = (-1 3)/(1) S = -3

7 ID : in-10-quadratic-equations [7] (8) c. 2x 2 + 4x + 3 = 0 In quadratic equation, ax 2 + bx + c = 0. D = b 2-4ac. If in a quadratic equation, D < 0, then the quadratic equation has no real roots. If in a quadratic equation, D > 0, then the quadratic equation has two distinct real roots. If in a quadratic equation, D = 0, then the quadratic equation has only one root. Let's check all of the quadratic equations f or real roots. -4x 2 + 8x - 4 = 0 Here, a = -4, b = 8 and c = -4 Now, D = b 2-4ac = (8) 2-4(-4)(-4) = 0 Since, D = 0, the quadratic equation -4x 2 + 8x - 4 = 0 has only one root. x 2 + 2x + 1 = 0 Here, a = 1, b = 2 and c = 1 Now, D = b 2-4ac = (2) 2-4(1)(1) = 0 Since, D = 0, the quadratic equation x 2 + 2x + 1 = 0 has only one root. 2x 2 + 4x + 3 = 0 Here, a = 2, b = 4 and c = 3 Now, D = b 2-4ac = (4) 2-4(2)(3) = -8 Since, D < 0, the quadratic equation 2x 2 + 4x + 3 = 0 has no real roots. -2x 2 + 4x - 1 = 0 Here, a = -2, b = 4 and c = -1 Now, D = b 2-4ac = (4) 2-4(-2)(-1) = 8 Since, D > 0, the quadratic equation -2x 2 + 4x - 1 = 0 has two distinct real roots. Thus, the quadratic equation 2x 2 + 4x + 3 = 0 has no real roots.

8 (9) b. 64 ID : in-10-quadratic-equations [8] Let the tens digit be x, hence unit digit = 24/x. Number = 10x + 24/x Number af ter digits interchanged = (10 24/x) + x 9x 2-18x = 0 x 2-2x - 24 = 0 Step 7 x 2-6x + 4x - 24 = 0 Step 8 x (x - 6) + 4 (x - 6) = 0 Step 9 (x - 6) (x + 4) = 0 0 x = 6, or -4. But it cannot be negative hence x = 6, and other digit = 24/6 = 4. Theref ore number is 64

9 (10) d. x = -a/(a + b) or, x = -(a + b)/a ID : in-10-quadratic-equations [9] LHS = = = Hence x = -a/(a + b) or, x = -(a + b)/a (11) b. two distinct real roots Let's compare the equation -x 2-8x + 2 = 0 with the quadratic equation ax 2 + bx + c = 0, we get, a = -1, b = - 8, c = 2. Now, D = b 2-4ac = (- 8) 2-4(-1)(2) = = 16 Since, D > 0, the equation -x 2-8x + 2 = 0 has two distinct real roots.

10 (12) b. 12 meter ID : in-10-quadratic-equations [10] Lets the length of copper wire = x meters Original price f or per meter copper wire = 192/x New prie = 192/(x + 4) Reduction in price = Rs.4 Step = x 2 + 4x Step 8 x 2 + 4x = 0 Step 9 x x - 12x = 0 0 x (x + 16) - 12 (x + 16) = 0 1 (x + 16) (x - 12) = 0 2 x = -16 or 12. Since length cannot be negative, x = 12 meter

11 (13) d. 15 ID : in-10-quadratic-equations [11] Let's assume that the natural number is x. The reciprocal of x = 1/x. It is given that when the natural number x is increased by 1, it equals 240 times its reciprocal. We can write this f act as an equation and solve f or x as: x + 1 = 240(1/x) x + 1 = 240/x x 2 + 1x = 240 x 2 + 1x = 0 Let us now solve the equation x 2 + 1x = 0 by the f actorization method: x 2 + 1x = 0 x x - 15x = 0 x(x + 16) - 15(x + 16) = 0 (x + 16)(x - 15) = 0 either, or, (x + 16) = 0 (x - 15) = 0 x = -16 x = 15 x -16, x is a natural number. Theref ore, x = 15 Thus, the natural number is 15. (14) a. x = -1/p 2, x = 1/q 2 p 2 q 2 x 2 + q 2 x - p 2 x - 1 = 0 (p 2 q 2 x 2 + q 2 x) - ( p 2 x + 1) = 0 (p 2 x + 1) q 2 x - ( p 2 x + 1) = 0 (p 2 x + 1) (q 2 x - 1) = 0 x = -1/p 2, x = 1/q 2

12 ID : in-10-quadratic-equations [12] (15) c. x 2 + 9x + 20 = 0 To check if -5 is a valid root of an equation, we need to replace variable in equation by -5 and see if equation is satisf ied or not. Lets try this f or all f our equations. x 2-3x - 4 = 0 On putting x = -5 we get, L.H.S = (-5) 2-3(-5) - 4 = 36 Now, L.H.S R.H.S, theref ore -5 is not the root of the equation x 2-3x - 4 = 0. x 2 + 3x - 4 = 0 On putting x = -5 we get, L.H.S = (-5) 2 + 3(-5) - 4 = 6 Now, L.H.S R.H.S, theref ore -5 is not the root of the equation x 2 + 3x - 4 = 0. x 2 + 9x + 20 = 0 On putting x = -5 we get, L.H.S = (-5) 2 + 9(-5) + 20 = 0 Now, L.H.S = R.H.S, theref ore -5 is the root of the equation x 2 + 9x + 20 = 0. x 2-5x + 4 = 0 On putting x = -5 we get, L.H.S = (-5) 2-5(-5) + 4 = 54 Now, L.H.S R.H.S, theref ore -5 is not the root of the equation x 2-5x + 4 = 0. Thus, the equation x 2 + 9x + 20 = 0 has -5 as one of the root.

Grade 10 Quadratic Equations

ID : ae-10-quadratic-equations [1] Grade 10 Quadratic Equations For more such worksheets visit www.edugain.com Answer t he quest ions (1) A two digit number is such that the product of its digits is 15.