Grade 10 Quadratic Equations - PDF Free Download

ID : ae-10-quadratic-equations [1] Grade 10 Quadratic Equations For more such worksheets visit www.edugain.com Answer the questions (1) Solve quadratic equation (x 0 and x -1). (2) The age of a mother is equal to the square of her son's age. One year ago, the mother's age was 9 times the son's age. Find their current ages. (3) Find a natural number whose square diminished by 12 is equal to 4 times of 12 more than the given number. (4) Ghada goes for a vacation with Dhs936 for her expanses. If she extends her vacation by 2 days, she has to cut down her daily expanses by Dhs3. Find original duration of her vacation. (5) In a class, one-third of students had gone to watch magic show. Twice the square root of the students had gone to watch drama, and remaining 72 students had gone to watch museum. Find total number of students in the class. (6) The sum of a number and its reciprocal is 13/6. Find the number. (7) Find solution of quadratic equation 2x 2 + 2x - 4 = 0 (8) The speed of a boat in still water is 12 km/hour. It goes 45 km upstream and return downstream to starting point in 8 hours. Find the speed of the stream. (9) The sum of the first n even natural numbers is 42. Find the value of n. (10) A family has 480 kg of rice for x number of weeks. If they need to use same amount of rice for 4 more weeks, they need to cut down their weekly consumption of rice by 4 kgs. Find value of x. (11) A piece of copper wire costs Dhs180. If it was 3 meter longer and price of each meter of copper wire costs Dhs3 less, the total cost of the piece would remain unchanged. Find the length of copper wire. (12) Solve quadratic equation x 2 + 8 x - (a 2 + 2a - 15) = 0 using factorization. (13) The sum of the squares of two consecutive natural numbers is 421. Find the numbers. (14) A two digit number is such that the product of its digits is 18. If 27 is added to the number, the digits interchange their places. Find the number. Choose correct answer(s) from the given choices (15) Equation -3x 2-5x + 2 = 0 has a. two distinct real roots b. more than two roots c. two equal real roots d. no real roots

Answers ID : ae-10-quadratic-equations [3] (1) x = 4 or, x = -5 On adding two fractions on LHS, 20 (x 2 + x 2 + x + 1) = 41 (x 2 + x) x 2 + x - 20 = 0 x 2 + 5 x - 4 x - 20 = 0 x(x + 5 x) - 4 (x + 5) = 0 (x - 4) (x + 5) = 0 Hence x = 4, or x = -5

(2) 64 years and 8 years ID : ae-10-quadratic-equations [4] Let the age of the son = x, hence age of father = x 2 One year ago, x 2-1 = 9(x - 1) x 2-9x + 8 = 0 x 2-8x - x + 8 = 0 x (x - 8) - (x + 8) = 0 (x - 1) (x - 8) = 0 x = 1 or 8. Age cannot be 1 (that would make age of mother and son to be same) Therefore age of son = x = 8 years. Age of mother = x 2 = 64 years

(3) 10 ID : ae-10-quadratic-equations [5] Let's assume that the natural number is 'x'. It is given that the square of the natural number when diminished by 12 is equal to 4 times of 12 more than the given number. We can write this fact as an equation and solve for x as: x 2-12 = 4(x + 12) x 2-12 = 4x + 48 x 2-12 - 4x - 48 = 0 x 2-4x - 60 = 0 Let us now solve the quadratic equation x 2-4x - 60 = 0 by the factorization method: x 2-4x - 60 = 0 x 2-10x + 6x - 60 = 0 x(x - 10) + 6(x - 10) = 0 (x - 10)(x + 6) = 0 either, or, (x - 10) = 0 (x + 6) = 0 x = 10 x = -6 x -6, since x is a natural number. Therefore, x = 10 Thus, the natural number is 10.

(4) 24 days ID : ae-10-quadratic-equations [6] Lets assume original duration of vacation = x days Per day expense = 936/x Per day expense if vacation is extended = 936/(x + 2) Reduction in daily expense = Dhs 3 312 2 = x 2 + 2x x 2 + 2x - 624 = 0 Step 9 x 2 + 26x - 24x - 624 = 0 0 x (x + 26) - 24 (x + 26) = 0 1 (x + 26) (x - 24) = 0 2 x = -26 or 24. Since days cannot be negative, x = 24 days

(5) 144 ID : ae-10-quadratic-equations [7] Let total number of students = T T/3 + 2 T + 72 = T 2T - 6 T - 216 = 0 Lets assume T = x 2 2x 2-6x - 216 = 0 One of the legal solution of this quadratic equation is x=12. Hence T = x 2 = 144 (6) 2/3 or 3/2 Let the number is x. Then 6 x 2 + 6 = 13 x 6 x 2-13 x + 6 = 0 6 x 2-4 x - 9 x + 6 = 0 2 x(3 x - 2) - 3 (3 x - 2) = 0 (2 x - 3) (3 x - 2) = 0 Hence x = 2/3, or 3/2 (7) -2, 1

(8) 3 km/hour ID : ae-10-quadratic-equations [8] Let the speed of stream = x km/hour Downstream speed = (12 + x) km/hour Upstream speed = (12 - x) km/hour Its is given that x 2 = 144-135 = 9 x = 3 km/hour

(9) 6 ID : ae-10-quadratic-equations [9] 2 + 4 + 6 +... n terms = 42 n(2 2 + 2(n - 1))/2 = 42 [Using AP summation Sn = n/2(2 2 + d(n - 1) )] n 2 + n = 42 n 2 + n - 42 = 0 (n - 6) (n + 7) = 0 n = 6 or -7. Since n cannot be negative, n = 6

(10) 20 ID : ae-10-quadratic-equations [10] Original weekly consumption = 480/x New weekly consumption = 480/(x + 4) Reduction in weekly consumption = 4 kg 120 4 = x 2 + 4x x 2 + 4x - 480 = 0 x 2 + 24x - 20x - 480 = 0 Step 9 x (x + 24) - 20 (x + 24) = 0 0 (x + 24) (x - 20) = 0 1 x = -24 or 20. Since number of weeks cannot be negative, x = 20 weeks

(11) 12 meter ID : ae-10-quadratic-equations [11] Lets the length of copper wire = x meters Original price for per meter copper wire = 180/x New prie = 180/(x + 3) Reduction in price = Dhs3 60 3 = x 2 + 3x x 2 + 3x - 180 = 0 Step 9 x 2 + 15x - 12x - 180 = 0 0 x (x + 15) - 12 (x + 15) = 0 1 (x + 15) (x - 12) = 0 2 x = -15 or 12. Since length cannot be negative, x = 12 meter

(12) x = -(a + 5),or x = a - 3 ID : ae-10-quadratic-equations [12] x 2 + 8 x - (a 2 + 2a - 15) = 0 x 2 + 8 x - (a + 5) (a - 3) = 0 x 2 + [ (a + 5) - (a - 3) ] x - (a + 5) (a - 3) = 0 [ x + (a + 5)] [ x - (a - 3) ] = 0 Therefore x = -(a + 5),or x = a - 3 (13) 14 and 15 Let the numbers be x and x+1 x 2 + (x+1) 2 = 421 2x 2 + 2x + 1 = 421 2x 2 + 2x - 420 = 0 x 2 + x - 210 = 0 x 2 + 15x - 14x - 210 = 0 x(x + 15) - 14(x + 15) = 0 (x - 14) (x + 15) = 0 Step 9 x = 14 or -15. Since x cannot be negative, x = 14. Hence numbers are 14 and 15.

(14) 36 ID : ae-10-quadratic-equations [13] Let the tens digit be x, hence unit digit = 18/x. Number = 10x + 18/x Number after digits interchanged = (10 18/x) + x 9x 2 + 27x - 162 = 0 x 2 + 3x - 18 = 0 x 2 + 6x - 3x - 18 = 0 x (x + 6) - 3 (x + 6) = 0 Step 9 (x + 6) (x - 3) = 0 0 x = -6, or 3. But it cannot be negative hence x = 3, and other digit = 18/3 = 6. Therefore number is 36

(15) a. two distinct real roots ID : ae-10-quadratic-equations [14] Let's compare the equation -3x 2-5x + 2 = 0 with the quadratic equation ax 2 + bx + c = 0, we get, a = -3, b = - 5, c = 2. Now, D = b 2-4ac = (- 5) 2-4(-3)(2) = 5 + 24 = 29 Since, D > 0, the equation -3x 2-5x + 2 = 0 has two distinct real roots.