Grade 10 Quadratic Equations

Transcription

1 ID : ae-10-quadratic-equations [1] Grade 10 Quadratic Equations For more such worksheets visit Answer t he quest ions (1) A two digit number is such that the product of its digits is 15. If 18 is added to the number, the digits interchange their places. Find the number. (2) In a class, one-f ourth of students had gone to watch museum. Twice the square root of the students had gone to watch f ootball match, and remaining 32 students had gone to watch drama. Find total number of students in the class. (3) Solve quadratic equation x x - (a 2 + 3a - 10) = 0 using f actorization. (4) The product of Ambra's age (in years) 3 years ago with her age (in years) 5 years later is 48. Find her current age. (5) Anan buys number of notebooks f or Dhs84. If she had bought 2 more notebooks f or the same amount, each notebook would have cost Dhs1 less. How many notebooks did she buy? (6) Adiba goes f or a vacation with Dhs1320 f or her expanses. If she extends her vacation by 3 days, she has to cut down her daily expanses by Dhs4. Find original duration of her vacation. (7) Find the sum of the roots of the quadratic equation y 2 + y - 30 = 0 (8) The sum of two numbers is 36. If sum of their reciprocals is 9/80, f ind the numbers. Choose correct answer(s) f rom given choice (9) Find solution of quadratic equation 2a a + 24 = 0 a. -4,-5 b. -4,-3 c. -5, -5 d. -5, -3 (10) Had Ambra scored 3 more marks in her mathematics test out of 30 marks, 4 times these marks would have been the square of her actual marks. How many marks did she get in the test? a. 6 b. 2 c. 5 d. 8 (11) Which of the f ollowing equations has 2 as one of its roots? a. a 2 + 5a - 6 = 0 b. a 2 - a - 12 = 0 c. a 2 + 3a - 10 = 0 d. a 2 + 4a - 5 = 0 (12) Find the roots of the quadratic equation = 0. a. and b. 3 and c. and 1 d. and

2 (13) Which of the f ollowing quadratic equations have no real roots? ID : ae-10-quadratic-equations [2] a. -3x 2 + 7x - 3 = 0 b. -2x 2 + 4x - 1 = 0 c. -x 2 + 4x - 2 = 0 d. -3x 2 + 6x - 4 = 0 (14) Which of the f ollowing is a solution to inequality x 2-6 x < - 5 a. 1 x < 5 b. 1 < x < 5 c. 1 x 5 d. 1 < x 5 (15) Find values of k f or which the quadratic equation 8x 2 kx + 3k = 0 has equal roots. a. 0 and 48 b. 0 and 96 c. 96 only d. 0 only 2016 Edugain ( All Rights Reserved Many more such worksheets can be generated at

3 Answers ID : ae-10-quadratic-equations [3] (1) 35 Let the tens digit be x, hence unit digit = 15/x. Number = 10x + 15/x Number af ter digits interchanged = (10 15/x) + x 9x x = 0 x 2 + 2x - 15 = 0 Step 7 x 2 + 5x - 3x - 15 = 0 Step 8 x (x + 5) - 3 (x + 5) = 0 Step 9 (x + 5) (x - 3) = 0 0 x = -5, or 3. But it cannot be negative hence x = 3, and other digit = 15/3 = 5. Theref ore number is 35

4 (2) 64 ID : ae-10-quadratic-equations [4] Let total number of students = T T/4 + 2 T + 32 = T 3T - 8 T = 0 Lets assume T = x 2 3x 2-8x = 0 One of the legal solution of this quadratic equation is x=8. Hence T = x 2 = 64 (3) x = -(a + 5),or x = a - 2 x x - (a 2 + 3a - 10) = 0 x x - (a + 5) (a - 2) = 0 x 2 + [ (a + 5) - (a - 2) ] x - (a + 5) (a - 2) = 0 [ x + (a + 5)] [ x - (a - 2) ] = 0 Theref ore x = -(a + 5),or x = a - 2

5 (4) 7 years ID : ae-10-quadratic-equations [5] Lets Ambra's current age is x (x - 3) (x + 5) = 48 x 2 + (5-3) x - 15 = 48 x x - 63 = 0 x 2 + 9x - 7x - 63 = 0 x (x + 9) - 7 ( x + 9) = 0 Step 7 (x + 9) (x - 7) = 0 Step 8 x = 7 or -9. Since age cannot be negative x = 7 years

6 (5) 12 notebooks ID : ae-10-quadratic-equations [6] Let the number of notebooks bought = x Price of one notebook = Dhs 84/x Price of one notebook, if 2 more items were bought = Dhs 84/(x + 2) Since dif f erence in price is Dhs = x 2 + 2x Step 7 x 2 + 2x = 0 Step 8 x x - 12x = 0 Step 9 x (x + 14) - 12(x + 14) = 0 0 (x - 12) (x + 14) = 0 1 x = 12 or -14. Since number of notebooks cannot be negative, x = 12

7 (6) 30 days ID : ae-10-quadratic-equations [7] Lets assume original duration of vacation = x days Per day expense = 1320/x Per day expense if vacation is extended = 1320/(x + 3) Reduction in daily expense = Dhs 4 Step = x 2 + 3x Step 8 x 2 + 3x = 0 Step 9 x x - 30x = 0 0 x (x + 33) - 30 (x + 33) = 0 1 (x + 33) (x - 30) = 0 2 x = -33 or 30. Since days cannot be negative, x = 30 days

8 (7) -1 ID : ae-10-quadratic-equations [8] For quadratic equation ay 2 + by + c = 0, sum of roots is -b/a For given equation y 2 + y - 30 = 0, a = 1, b = 1 and c = -30 Theref ore sum of roots, S = (-1 b)/(a) S = (-1 1)/(1) S = -1 (8) 16 and 20 Let the numbers are x and 36-x. Then On adding two f ractions on LHS 2880 = 9x (36 - x) 9 x x = 0 x 2-36 x = 0 x 2-16 x - 20 x = 0 Step 7 x (x - 16) - 20 (x - 16) = 0 Step 8 (x - 16) (x - 20) = 0 Step 9 x = 16 or 20. Hence numbers are 16 and 20. (9) b. -4,-3

9 (10) a. 6 ID : ae-10-quadratic-equations [9] Lets assume Ambra scored x marks out of 30 marks It is given that, 4 (x + 3) = x 2 4x = x 2 x 2-4x - 12 = 0 x 2 + 2x - 6x + (-6)(2) = 0 x (x + 2) - 6 (x + 2)) = 0 (x - 6) (x + 2) = 0 Theref ore x = 6 or x = -2. But since marks should be positive, her actual marks are 6

10 ID : ae-10-quadratic-equations [10] (11) c. a 2 + 3a - 10 = 0 To check if 2 is a valid root of an equation, we need to replace variable in equation by 2 and see if equation is satisf ied or not. Lets try this f or all f our equations. a 2 + 5a - 6 = 0 On putting a = 2 we get, L.H.S = (2) 2 + 5(2) - 6 = 8 Now, L.H.S R.H.S, theref ore 2 is not the root of the equation a 2 + 5a - 6 = 0. a 2 - a - 12 = 0 On putting a = 2 we get, L.H.S = (2) 2 - (2) - 12 = -10 Now, L.H.S R.H.S, theref ore 2 is not the root of the equation a 2 - a - 12 = 0. a 2 + 3a - 10 = 0 On putting a = 2 we get, L.H.S = (2) 2 + 3(2) - 10 = 0 Now, L.H.S = R.H.S, theref ore 2 is the root of the equation a 2 + 3a - 10 = 0. a 2 + 4a - 5 = 0 On putting a = 2 we get, L.H.S = (2) 2 + 4(2) - 5 = 7 Now, L.H.S R.H.S, theref ore 2 is not the root of the equation a 2 + 4a - 5 = 0. Thus, the equation a 2 + 3a - 10 = 0 has 2 as one of the root.

11 (12) d. and ID : ae-10-quadratic-equations [11] On comparing the quadratic equation = 0, with the standard f orm ax 2 + bx + c = 0, we get: a = 1, b = -, c = 9 D = b 2-4ac = (- ) 2-4(1)(9) = = 12 By using the quadratic f ormula, we get: a = -b D 2a = a = 2 1 or Thus, the roots of the quadratic equation = 0, are and.

12 ID : ae-10-quadratic-equations [12] (13) d. -3x 2 + 6x - 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 f or real roots. -3x 2 + 7x - 3 = 0 Here, a = -3, b = 7 and c = -3 Now, D = b 2-4ac = (7) 2-4(-3)(-3) = 13 Since, D > 0, the quadratic equation -3x 2 + 7x - 3 = 0 has two distinct 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. -x 2 + 4x - 2 = 0 Here, a = -1, b = 4 and c = -2 Now, D = b 2-4ac = (4) 2-4(-1)(-2) = 8 Since, D > 0, the quadratic equation -x 2 + 4x - 2 = 0 has two distinct real roots. -3x 2 + 6x - 4 = 0 Here, a = -3, b = 6 and c = -4 Now, D = b 2-4ac = (6) 2-4(-3)(-4) = -12 Since, D < 0, the quadratic equation -3x 2 + 6x - 4 = 0 has no real roots. Thus, the quadratic equation -3x 2 + 6x - 4 = 0 has no real roots.

13 (14) b. 1 < x < 5 ID : ae-10-quadratic-equations [13] 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. (15) b. 0 and 96 On comparing equation with standard quadratic equation ax 2 + bx + c = 0, we get, a = 8, b = -k and c = 3k. It is given that, the quadratic equation 8x 2 kx + 3k = 0 has two equal roots, Theref ore, b 2-4ac = 0 (-k) 2-4(8)(3k) = 0 k 2-96k = 0 k(k - 96) = 0 either, k = 0 or, k - 96 = 0 k = 96 Thus, the value of k could be 0 or 96.