Class 10 Quadratic Equations - PDF Free Download

ID : in-10-quadratic-equations [1] Class 10 Quadratic Equations For more such worksheets visit www.edugain.com Answer the questions (1) The sum of square of two positive numbers is 832. If square of the larger number is 36 times the smaller number, find the numbers. (2) Had Neha scored 10 more marks in her mathematics test out of 50 marks, 9 times these marks would have been the square of her actual marks. How many marks did she get in the test? (3) Find a natural number whose square diminished by 37 is equal to 4 times of 2 more than the given number. (4) A natural number, when decreased by 4, equals 320 times its reciprocal. Find the number. (5) Solve quadratic equation p 2 q 2 x 2 + q 2 x - p 2 x - 1 = 0 using factorization. (6) Solve quadratic equation x 2 + 5 x - (a 2 + 3a - 4) = 0 using factorization. (7) Find solution of quadratic equation b 2-9b + 20 = 0 (8) The sum of the n consecutive natural odd numbers starting from 3 is 35. Find the value of n. (9) A train, traveling at a uniform speed for 6900 km, would have taken 9 hours less to travel the same distance if its speed were 15 km/h more. Find the original speed of the train. (10) The sum of the first n even natural numbers is 30. Find the value of n. (11) The speed of a boat in still water is 20 km/hour. It goes 96 km upstream and return downstream to starting point in 10 hours. Find the speed of the stream. (12) Find the roots of the quadratic equation -2x 2-6x - 3 = 0. (13) If the price of a sweet is reduced by Rs. 5, Radha can buy 15 more sweets for Rs. 900. Find the original price of sweet. Choose correct answer(s) from the given choices (14) Which of the following quadratic equations have no real roots? a. -3x 2 + 4x - 1 = 0 b. -x 2 + 3x - 4 = 0 c. -2x 2 + 7x - 4 = 0 d. x 2 + 2x + 1 = 0 (15) Which of the following is a solution to inequality x 2-12 x < - 32 a. 4 < x < 8 b. -8 < x < -4 c. 4 x 8 d. 4 x < 8

Answers ID : in-10-quadratic-equations [3] (1) 24 and 16 Let smaller number be x. Therefore square of larger number = 36x. x 2 + 36x = 832 x 2 + 36x - 832 = 0 x 2 + 52x - 16x - 832 = 0 x (x + 52) - 16 (x + 52) = 0 Step 6 (x + 52) (x - 16) = 0 Step 7 x = 16 or -52. Since numbers are positive, smaller number x = 16. Step 8 Larger number = (36 x 16) = 24 (2) 15 marks Lets assume Neha scored x marks out of 50 marks It is given that, 9 (x + 10) = x 2 9x + 9 10 = x 2 x 2-9x - 90 = 0 x 2 + 6x - 15x + (-15)(6) = 0 x (x + 6) - 15 (x + 6)) = 0 (x - 15) (x + 6) = 0 Therefore x = 15 or x = -6. But since marks should be positive, her actual marks are 15

(3) 9 ID : in-10-quadratic-equations [4] Let's assume that the natural number is 'x'. It is given that the square of the natural number when diminished by 37 is equal to 4 times of 2 more than the given number. We can write this fact as an equation and solve for x as: x 2-37 = 4(x + 2) x 2-37 = 4x + 8 x 2-37 - 4x - 8 = 0 x 2-4x - 45 = 0 Let us now solve the quadratic equation x 2-4x - 45 = 0 by the factorization method: x 2-4x - 45 = 0 x 2-9x + 5x - 45 = 0 x(x - 9) + 5(x - 9) = 0 (x - 9)(x + 5) = 0 either, or, (x - 9) = 0 (x + 5) = 0 x = 9 x = -5 x -5, since x is a natural number. Therefore, x = 9 Thus, the natural number is 9.

(4) 20 ID : in-10-quadratic-equations [5] Let's assume the natural number is 'x'. The reciprocal of 'x' = 1/x. It is given that when the natural number 'x' is decreased by 4, it equals 320 times its reciprocal. We can write this fact as an equation and solve for x as: x - 4 = 320(1/x) x - 4 = 320/x x 2-4x = 320 x 2-4x - 320 = 0 Let us now solve the quadratic equation x 2-4x - 320 = 0 by the factorization method: x 2-4x - 320 = 0 x 2-20x + 16x - 320 = 0 x(x - 20) + 16(x - 20) = 0 (x - 20)(x + 16) = 0 either, or, (x - 20) = 0 (x + 16) = 0 x = 20 x = -16 x -16, since we have been told that x is a natural number. Therefore, x = 20 Thus, the natural number is 20. (5) 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

(6) x = -(a + 4),or x = a - 1 ID : in-10-quadratic-equations [6] x 2 + 5 x - (a 2 + 3a - 4) = 0 x 2 + 5 x - (a + 4) (a - 1) = 0 x 2 + [ (a + 4) - (a - 1) ] x - (a + 4) (a - 1) = 0 [ x + (a + 4)] [ x - (a - 1) ] = 0 Therefore x = -(a + 4),or x = a - 1 (7) 4, 5 (8) 5 3 + 5 + 7 +... n terms = 35 n(2 3 + 2(n - 1))/2 = 35 [Using AP summation Sn = n/2(2a + d(n - 1) )] n 2 + 2n = 35 n 2 + 2n - 35 = 0 (n - 5) (n + 7) = 0 Step 6 n = 5 or -7. Since n cannot be negative, n = 5

(9) 100 km/hour ID : in-10-quadratic-equations [7] We know that, s = d/t, where, s = Speed of the train, d = Distance traveled by the train, t = time taken by train. Let's assume the speed of the train is 's' km/hour. It is given that, the train is traveling at a uniform speed for 6900 km, therefore, s = 6900/t -----(1) or t = 6900/s ------(2) It is also given that, the train would have taken 9 hours less to travel the same distance if its speed were 15 km/h more. 6900 Therefore, s + 15 = t - 9 By putting the value of 't' from equation (2), we get, 6900 s + 15 = 6900/s - 9 6900s s + 15 = 6900-9s (s + 15)(6900-9s) = 6900s 6900s - 9s 2 + 103500-135s = 6900s - 9s 2 + 103500-135s = 0 9s 2 + 135s - 103500 = 0 On solving the quadratic equation, 9s 2 + 135s - 103500 = 0, we get, s = 100 or -115 s -115, since the speed of the train cannot be negative. Therefore, s = 100 Thus, the speed of the train is 100 km/hour.

(10) 5 ID : in-10-quadratic-equations [8] 2 + 4 + 6 +... n terms = 30 n(2 2 + 2(n - 1))/2 = 30 [Using AP summation Sn = n/2(2 2 + d(n - 1) )] n 2 + n = 30 n 2 + n - 30 = 0 (n - 5) (n + 6) = 0 Step 6 n = 5 or -6. Since n cannot be negative, n = 5

(11) 4 km/hour ID : in-10-quadratic-equations [9] Let the speed of stream = x km/hour Downstream speed = (20 + x) km/hour Upstream speed = (20 - x) km/hour Its is given that Step 6 Step 7 x 2 = 400-384 = 16 Step 8 x = 4 km/hour

ID : in-10-quadratic-equations [10] (12) and On comparing the given equation with the general quadratic equation form a x 2 + b x + c = 0, we can say: a = -2, b = -6 and c = -3 Now the roots of the general quadratic equation a x 2 + b x + c = 0 can be calculated using the quadratic formula as: x = Let us substitute the values of a, b and c: x = x = x = x = x = or

(13) Rs. 20 ID : in-10-quadratic-equations [11] Let the original price = x Number of sweets bought for Rs. 900 = 900/x Number of sweets bought for Rs. 900 after reduction in price = 900/(x - 5) Since difference in price is Rs. Step 6 900 5 = 15( x 2-5x) Step 7 60 5 = x 2-5x Step 8 x 2-5x - 300 = 0 Step 9 x 2 + 15x - 20x - 300 = 0 0 x (x + 15) - 20(x + 15) = 0 1 (x - 20) (x + 15) = 0 2 x = 20 or -15. Since price cannot be negative, x = 20

ID : in-10-quadratic-equations [12] (14) b. -x 2 + 3x - 4 = 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 for real roots. -3x 2 + 4x - 1 = 0 Here, a = -3, b = 4 and c = -1 Now, D = b 2-4ac = (4) 2-4(-3)(-1) = 4 Since, D > 0, the quadratic equation -3x 2 + 4x - 1 = 0 has two distinct real roots. -x 2 + 3x - 4 = 0 Here, a = -1, b = 3 and c = -4 Now, D = b 2-4ac = (3) 2-4(-1)(-4) = -7 Since, D < 0, the quadratic equation -x 2 + 3x - 4 = 0 has no real roots. -2x 2 + 7x - 4 = 0 Here, a = -2, b = 7 and c = -4 Now, D = b 2-4ac = (7) 2-4(-2)(-4) = 17 Since, D > 0, the quadratic equation -2x 2 + 7x - 4 = 0 has two distinct real roots. 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. Step 6 Thus, the quadratic equation -x 2 + 3x - 4 = 0 has no real roots.

(15) a. 4 < x < 8 ID : in-10-quadratic-equations [13] First lets find the value of x for which x 2-12 x = - 32, or x 2-12 x + 32 = 0 x 2-12 x + 32 = 0 (x - 4) (x - 8) = 0 x = 4 or x = 8 Now you can notice that inequality is satisfied for x > 4 and x < 8.