Quadratic Equations 6 QUESTIONS. Relatively Easy: Questions 1 to 2 Moderately Difficult: Questions 3 to 4 Difficult: Questions 5 to 6

Transcription

1 Quadratic Equations 6 QUESTIONS Relatively Easy: Questions 1 to 2 Moderately Difficult: Questions 3 to 4 Difficult: Questions 5 to 6

2 Questions Page 2 of 11

3 Q1. The factors of 2x² - 7x - 39 are a) (2x - 13) and (x + 3) b) (2x + 13) and (x + 3) c) (2x + 13) and (x - 3) d) (2x - 13) and (x - 3) Q2. {-2, 3} is the solution of the quadratic equation a) x² + x 6 = 0 b) x² - x + 6 = 0 c) x² - x 6 = 0 d) x² + x + 6 = 0 Q3. The values of y satisfying the equation 5(3y + 1)² + 6(3y + 1) 8 = 0 are 1 a) (-1, ) 15 b) (-1, 1 15 ) c) (1, 1 15 ) 1 d) (1, ) 15 Q4. The nature of the roots of the equation 2x² + 6x + 3 = 0 are a) Imaginary b) Real and equal c) Real and unequal d) Irrational and unequal Page 3 of 11

4 Q5. Two natural numbers differ by 3. The sum of their squares is 117. The numbers are a) (3, 6) b) (9, 12) c) (6, 9) d) (12, 15) Q6. Robert can do a piece of work in x days. Mary can do the same work in (x + 16) days. When both of them work together they can do the work in 15 days. What is the value of x? a) 22 b) 23 c) 24 d) 25 Page 4 of 11

5 Solutions Page 5 of 11

6 S1. We have to factorize 2x² -7x -39. In order to do so we have to split the second term. The product of the coefficient of x² and the constant term is as follows. Coefficient of x² = 2 Constant = -39 The product = 2 (-39) The coefficient of x is -7 = -78 We have to find two numbers such that their product is -78 and their sum is -7. The numbers are -13 and 6. {-13 6 = -78, = -7} The expression can be factored as follows: 2x² - 7x 39 = 2x² - 13x + 6x - 39 = (2x² - 13x) + (6x 39) Because x is the common factor in (2x 2-13x) and 3 is the common factor in (6x- 39), we can rewrite the above as: (2x. x - 13x) + (2. 3. x 3. 13) = x (2x 13) + 3 (2x 13) Taking out (2x - 13), which is the common factor, the above expression becomes: (2x 13) (x + 3) The factors are (2x - 13) and (x + 3). Therefore the answer is a). S2. Given: {- 2, 3} are roots of the quadratic equation We have x = - 2 and x = 3 x + 2 = 0 and x 3 = 0. Since 0 0 = 0, (x + 2) (x - 3) = 0 Page 6 of 11

7 (x + 2).(x - 3) = 0 x.x + 2.x - (x ) = 0 x² + 2x - (3x + 6) = 0 x² + 2x- 3x - 6 = 0 x²- x- 6=0 Therefore the answer is c). S3. Given: 5(3y + 1)² + 6(3y + 1) 8 = 0 Let us expand the equation 5(9y² +6y+1)+18y+6-8 = 0 45y² +30y+5+18y-2 = 0 45y² +48y+3 = 0 From this equation, a = coefficient of y 2 = 45 b = coefficient of y = 48 c = constant = 3 We need to find two numbers whose product is (a.c) and sum is b. i.e. Product (a.c) is 45 3 = 135 Sum = 48 The numbers are 45 and 3 Rewriting the equation, we get 45y²+ 48y +3 = 0 45y² + 45y + 3y + 3 = 0 45y(y + 1) + 3(y + 1) =0 Page 7 of 11

8 (y + 1)(45y + 3) =0 y + 1=0 or 45y + 3 =0 y = 1 or y = = 15 1 y = { 1, } i.e. the answer is a). 15 S4. To find the nature of the root, we examine the discriminant b² 4ac Given: 2x 2 + 6x + 3= 0 a = coefficient of x 2 = 2 b = coefficient of x = 6 c = constant = 3 b 2 4ac = (6) = = 12 > 0 (i.e. positive) If the value of the discriminant is positive and not a perfect square, the roots will be irrational and unequal. Therefore the answer is d). S5. Let one of the numbers be x. Given: The difference between the numbers is 3 If one of the numbers is x, the other will be x 3. Also given: Sum of their squares =117 x 2 + (x 3) 2 = 117 x 2 + x 2 6x + 9 = Page 8 of 11

9 2x 2 6x = 0 2x 2 6x 108 = 0 a = 2, b = 6, c = 108 We need to find two numbers whose product is (a.c) and sum is b. i.e. Product (a.c) is = -216 Sum = -6 The two numbers whose product is -216 and sum is -6 are: -18 and x 2-6x 108 =0 2x 2 18x + 12x 108 =0 (2.x.x 2.9x) + (12.x 12.9) = 0 2x(x 9) 12(x 9) =0 (x 9) (2x 12) =0 x 9 = 0 or 2x 12 =0 x = 9 or 2x =12 x = 9 or x = 12 2 = 6 The numbers are 6 and 9. (Verify the sum of their squares 36 and 81 is 117) The answer is c). Page 9 of 11

10 S6. We have been given that Robert can do the piece of work in x days. In one day he can do 1 part of the work. x Let us take A = 1 x Mary can do the same piece of work in (x+16) days. Let us take B = 1 x +16 If they work together in one day they can complete: A + B = { 1 x + 1 x + 16 } part of the work = x+ 16+ x xx ( + 16) = 2x + 16 xx ( + 16) Given: They can complete the same work together in 15 days. Therefore in one day they can finish 1 th of the work x + 16 = 1 xx ( + 16) 15 15(2x + 16) = 1.x(x + 16) 30x = x 2 +16x x 2 30x + 16x 240 = 0 (by shifting all terms to left hand side) x 2 14x 240 = 0 Product is = -240 Sum = -14 The two numbers whose product is -240 and sum is -14 are: -24 and 10 Rewriting the equation, we get: x 2 14x 240 =0 (x 2 24x) + (10x 240) = 0 Page 10 of 11

11 x(x 24) + 10(x 24) = 0 (x + 10) (x 24) =0 (x + 10) = 0 or (x 24) = 0 x = 10 or x = 24 Since x cannot be negative, x? 10 x = 24. Robert can complete the piece of work in 24 days. The answer is c). Page 11 of 11