4

Q1. Solve the below equation by the perfect square method: 5x² - 6x - 2 = 0

• 1)
• 2)
• 3)
• 4)

Solution

 

Q2.

Solution

Q3. To make 6x² + 5x = 0 a perfect square, we need to add and subtract 

• 1)
• 2)
• 3)
• 4)

Solution

 

Q4. Solve the below equation by the factorisation method: x² - 9x + 20 = 0

• 1) x = 4 or x = -5
• 2) x = -4 or x = 5
• 3) x = -4 or x = -5
• 4) x = 4 or x = 5

Solution

x² - 9x + 20 = 0 x² - 5x - 4x + 20 = 0 x(x - 5) - 4(x - 5) = 0 (x - 4)(x - 5) = 0 x = 4 or x = 5

Q5. Solve the following equation

Solution

     

Q6. Solve the below equation by the perfect square method: 2x² - 5x + 3 = 0

• 1)
• 2)
• 3)
• 4)

Solution

 

Q7. Find the solution of the quadratic equation by the formula method. x² + 4x - 2 = 0

• 1)
• 2)
• 3)
• 4)

Solution

 

Q8. Form a quadratic equation whose roots are and.

Solution

Sum of roots(S) = Product of roots (P) = Required equation is x²-Sx+P=0 6x² - 13x - 5=0

Q9. Solve the quadratic equation by the completing the square method. x² + 4x + 1 = 0 

• 1)
• 2)
• 3)
• 4)

Solution

 

Q10. The sum of the reciprocals Rehman"s ages 3 years ago and five years from now is . Find his present age.

Solution

Let Rehman"s present age =  years Rehman's age 3 years ago  =- 3 years Five years from now  = + 5 years According to question                                     Rejecting negative value Hence Rehmans age =7 Yrs  

Q11. Solve the quadratic equation by the formula method. x² + 2x - 3 = 0

• 1) x = 3 or 1
• 2) x = −3 or −1
• 3) x = −3 or 1
• 4) x = 2 or −1

Solution

 

Q12. Find the solution of the quadratic equation by the formula method. x² + 7x + 12 = 0 

• 1) x = −4 or −3
• 2) x = 4 or 3
• 3) x = −4 or 3
• 4) x = 4 or −3

Solution

 

Q13. If 8 is a root of the equation x² - 10x + k = 0, then the value of k is:

• 1) 8
• 2) -8
• 3) 2
• 4) 16

Solution

Given, 8 is a root of the equation x² - 10x + k = 0. (8)² - 10(8) + k = 0 64 - 80 + k = 0 k = 16

Q14. One of the roots of the quadratic equation 6x² - x - 2 = 0 is:

• 1)
• 2) -1
• 3)
• 4)

Solution

6x² - x - 2 = 0 6x² - 4x + 3x - 2 = 0 2x(3x - 2) + 1(3x - 2) = 0 (3x - 2) (2x + 1) = 0 Thus, the required one root of the given quadratic equation is .

Q15. Solve the following quadratic equations by factorization :

Solution

Q16. Solve the quadratic equation by the completing the square method. x² + 8x + 1 = 0

• 1)
• 2)
• 3)
• 4)

Solution

 

Q17. Solve for x:

Solution

  Consider equation    

Q18. Solve the quadratic equation by the formula method. 3x² + 9x - 2 = 0

• 1)
• 2)
• 3)
• 4)

Solution

 

Q19.  

Solution

 

Q20. Find the solution of the quadratic equation by the formula method. x² + 2x - 5 = 0

• 1)
• 2)
• 3)
• 4)

Solution

 

Q21. Solve the following quadratic equations by factorization :

Solution

Q22.

Solution

    

Q23. Solve the following quadratic equation by the factorisation method: x² + 2√2x - 6 = 0 

• 1) x = 3√2, −√2
• 2) x = −3√2, √2
• 3) x = 3√2, √2
• 4) x = −3√2, −√2

Solution

x² + 2√2x - 6 = 0 x² + 3√2x - √2x - 6 = 0 x(x + 3√2) - √2(x + 3√2) = 0 (x + 3√2)(x - √2) = 0

Q24. A train covers a distance of 90 km at a uniform speed. Had the speed been 15 km/hr more it would have taken 30 minutes less for the journey. Find the original speed of the train.

Solution

  Let the original speed of train be km/h. then increased speed  km/hr                                                            usual time hours                                                                                 New time hours According to the question                                                                                                                                      Rejecting the negative value, we have x = 45 Hence usual speed of train = 45 km/hr.

Q25. Find the value of m so that the quadratic equation mx(x - 7) + 49 = 0 has two equal roots.

Solution

The given quadratic equation is mx(x-7)+49=0 mx²-7mx+49=0 a = m, b = -7m and c = 49 It is given that the roots of this equation are equal. D = b² - 4ac=0 (-7m)² - 4(m)(49)=0 49m² - 4(m)(49)=0 49m(m-4)=0 49m=0 or m - 4 = 0 m = 0 or m=4 For m = 0 equation will not be quadratic. Hence the value of m = 4.

Q26. The speed of a boat in still water is 11 km/hr. It can go 12 km upstream and return downstream to the original point in 2hrs 45 min. Find the speed of the stream.

Solution

Let the speed of the stream = x km/hr Speed of the boat in still water = 11 km/hr Speed of the boat downstream = (11 + x) km/hr Speed of the boat upstream = (11 - x) km/hr According to the given information, 121 - x² = 96 x² = 25 x = 5 Rejecting negative value, as speed cannot be negative. So, x = 5. Hence, the speed of the stream is 5 km/hr.

Q27. A person on tour has Rs. 306 for his daily expenses. If he extends his tour for four days, he has to cut down his daily expenses by Rs. 3. Find the original duration of the tour.

Solution

Let original duration of tour be x days Daily expenses initially= Daily expenses later = Accordingly,    360 4 = 3(x(x + 4)) x² + 4x - 480 = 0 (x + 24) (x - 20) = 0 x = -24 or x = 20 X = -24 not possible x = 20 Original duration of tour = 20 days

Q28. Solve by method of completing square.

Solution

The coefficient of  is not perfect square. So we divide the equation through out by 3 divide and multiply coefficient of  by 2 Add and subtract  to the above equation, we have,              Required solution =1,

Q29. (2x - y)² =

• 1) 4x² - 4xy + y²
• 2) 4x² + 4xy + y²
• 3) 4x² - 2xy + y²
• 4) 2x² - 4xy + y²

Solution

(2x - y)² = 4x² - 4xy + y² 

Q30. Using quadratic formula solve for

Solution

Here  A = 9                B = -3(a + b)              C = a b                       Using quadratic formula, we have                                                                                                               are the required solution.

Q31. The positive root of is:

• 1) 4
• 2) 5
• 3) 3
• 4) 7

Solution

_{Thus, the positive root of the given equation is 5.}

Q32. A plane left 30 minutes late then its scheduled time and in order to reach the destination 1500 km away in time it had to increase the speed by 250 km/h from the usual speed. Find its usual speed.

Solution

Let usual speed of plane=x km/hr Usual time taken= Increased speed  New time= According to question,   By factorisation method, x = -1000 , because speed can't be negative. Hence, usual speed of plane is 750 km/hr.

Q33. Solve the quadratic equation by the completing the square method. x² + x - 6 = 0 

• 1) x = 3 or −2
• 2) x = −3 or 2
• 3) x = −3 or −2
• 4) x = 3 or 2

Solution

 

Q34. Using Quadratic formula ,solve the following quadratic equation

Solution

    Comparing this equation with  we have                  =       =      =      =              So the given roots of the given equation are real and given by                                                    

Q35. Find the roots of the equation , x 0,

Solution

Q36. Find two positive numbers whose squares have the difference 48 and the sum of the numbers is 12.

Solution

Since sum of the numbers is 12, the two numbers are x and 12 - x. By question, x² - (12 - x)²=48 x² - 144 - x² + 24x = 48 24x = 144 + 48 x = = 8 The two numbers are 8 and 4.

Q37. Solve  by factorization method.

Solution

   

Q38.

Solution

Q39. Solve for x: 6x² + 7x - 10 = 0

Solution

6x² + 7x - 10 = 0 6x² + 12x - 5x - 10 = 0 6x(x + 2) - 5(x + 2) = 0 (6x - 5)(x + 2) = 0 x = -2,

Q40. Find two consecutive positive integers, sum of whose squares is 365.

Solution

Let two consecutive positive integers be and . According to the question, By factorisation method, we get Since the integers are positive. x = -14 is not acceptable. Hence, the required positive integers 3 and 14.

Q41.

Solution

Q42. Find the roots of the quadratic equation: a²b²x² + b²x - a²x - 1 = 0.

Solution

a²b²x² + b²x - a²x - 1 = 0 b²x(a²x + 1) - 1(a²x + 1) = 0 (b²x - 1) (a²x + 1) = 0 x =

Q43. The sum of the reciprocals of Rehman's ages (in years) 3 years ago and 5 years from now is 1/3. Find his present age.

Solution

Let the present age of Rehman be x years. So, 3 years ago Rehman's age = (x - 3) years And 5 years hence, Rehman's age = (x + 5) years According to question, we have: But x, being the age cannot be negative. So, x = 7. Thus, the present age of Rehman is 7 years.

Q44. Which of the following equations has two distinct real roots?

• 1) 2x² - 3 x + = 0
• 2) 5x² - 3x + 1 = 0
• 3) x² + 3x + 2 = 0
• 4) x² + x - 5 = 0

Solution

A quadratic equation has two distinct roots if D = b² - 4ac > 0. Consider the quadratic equation x² + x - 5 = 0. Here, D = (1)² - 4(1)(-5) = 1 + 20 = 21 > 0 Thus, this equation has two distinct real roots.

Q45. Find two numbers whose sum is 27 and product is 182.

Solution

Let first number be x then second number be 27 - x. According to question, x(27 - x) = 182 27x  - x² = 182 x²  + 27x +182 = 0 x²  -14x - 13x + 182 = 0 x(x - 14) - 13(x - 14) = 0 (x - 14)(x - 13) = 0 x = 14 or x = 13 Hence the required numbers are 13 and 14.

Q46. Find the solution of the quadratic equation by the formula method. x² + 5x - 10 = 0

• 1)
• 2)
• 3)
• 4)

Solution

 

Q47.

Solution

 

Q48. The product of Tanay's age (in years) five years ago and his age ten years later is 16. Determine Tanay's present age.

Solution

Let the present age of Tanay be x years. Then, from the given information, we have: (x - 5) (x + 10) = 16 x² + 5x - 50 = 16 x² + 5x - 66 = 0 x² - 6x +11x - 66 = 0 x(x - 6) +11(x - 6) =0 (x - 6)(x +11) = 0 x = -11 or 6 Rejecting x = -11, as age cannot be negative. Present age of Tanay is 6 years

Q49. Two water taps together can fill a tank in hours. The tap of larger diameter takes 10 hours less than the smaller one to fill the tank separately. Find the time in which each tap can separately fill the tank.

Solution

Let the larger tap can fill the tank in x hours. Then, the smaller tap will fill the tank in (x + 10) hours. = 15 (rejecting negative value) Larger tap can fill the tank in 15 hours Smaller tap can fill the tank in 25 hours

Q50. A motor boat whose speed is 36 km/h in still water takes 1 hour more to go 48 km upstream than to return downstream to the same spot. Find the speed of the stream.

Solution

Let the speed of the stream be km/h Therefore speed of boat upstream = And speed of boat in downstream Time taken to go upstream “             “           “  downstream According to question Using the quadratic formula. Since is the speed of stream, it cannot be negative so we ignore the root = - 108. Therefore= 12 gives the speed of stream as 12 Km/Hr  

Q51. The product of the digits of a two digit positive number is 24. If 18 is added to the number then the digits of the number are interchanged. Find the number.

Solution

Let the digit at ten's place = x Then, digit at one's place = Original number = 10x + On interchanging the digits, new number = + x According to the question, 10x + + 18 = + x 10x² + 24 + 18x = 240 + x² 9x² + 18x - 216 = 0 x² + 2x - 24 = 0 (x + 6) (x - 4) = 0 x = -6 or 4 But x can't be negative, so, x = 4 Digit at ten's place = x = 4 and digit at one's place == 6 Thus, the original number is 46.

Q52. Which of the following in not a quadratic equation?

• 1) (x + 2)² = 5x² - 4
• 2) x(2x + 3) = x² + 1
• 3) (x - 2)² + 1 = 2x - 3
• 4) x(x + 1) + 8 = (x + 2) (x - 2)

Solution

Simplify each option. Consider,  x(x + 1) + 8 = (x + 2) (x - 2) Clearly, this is a linear equation and not a quadratic equation.

Q53. By increasing the speed of a bus by 10 km/hr, it takes one and half hours less to cover a journey of 450 km. Find the original speed of the bus.

Solution

Let speed of the bus be x km/hr Time t = 450/x If speed is x + 10, then time T = 450/(x + 10) By question, 450/x - 450/(x + 10) =3/2 450 (x + 10) -450 x = (3/2) (x² + 10x) 4500 2 = 3 x² + 30x x² + 10x - 3000 = 0 x (x + 60) - 50x - 3000 = 0 x (x + 60) - 50 (x + 60) = 0 x = 50 (asx = -60 not possible )

Q54. Solve the quadratic equation by the completing the square method. x² + 16x + 9 = 0 

• 1)
• 2)
• 3)
• 4)

Solution

 

Q55. Solve the below equation by the perfect square method: x² - 4x + 2 = 0

• 1)
• 2)
• 3)
• 4)

Solution

 

Q56. Solve the quadratic equation by the formula method. x² + x - 2 = 0

• 1) x = 2 or 1
• 2) x = 2 or −1
• 3) x = −2 or 1
• 4) x = −2 or −1

Solution

x = −2 or x = 1

Q57. A person can buy 15 books less for Rs. 900 when the price of book goes up by Rs. 3. Find the original price and the no. of copies he would buy at initial price.

Solution

Therefore the number of copies that he can buy with the initial price

Q58. Solve the following quadratic equations by factorization :

Solution

 

Q59. Two positive numbers differ by 3 and their product is 504. Find the numbers. OR Solve for x: 10ax² - 6x + 15ax - 9 = 0 ; a 0

Solution

Let the numbers be x, x + 3. x (x + 3) = 504 x² + 3x - 504 = 0 (x + 24)(x - 21) = 0 x = 21, - 24 The numbers are positive so we reject x = -24. Hence, the numbers are 21, 24. OR 10ax² - 6x + 15ax - 9 = 0 2x(5ax - 3) + 3(5ax - 3) = 0 (5ax-3) (2x + 3) = 0 5ax - 3 = 0 or 2x + 3 = 0 5ax = 3 or 2x = -3 or

Q60.

Solution

 

Q61. Solve the below equation by the factorisation method: x² - 7x + 12 = 0 

• 1) x = -4 or x = -3
• 2) x = 4 or x = -3
• 3) x = 4 or x = 3
• 4) x = -4 or x = 3

Solution

x² - 7x + 12 = 0 x² - 4x - 3x + 12 = 0 x(x - 4) - 3(x - 4) = 0 (x - 4)(x - 3) = 0 x = 4 or x = 3

Q62. If x² + 2 kx + 4 = 0 has a root x = 2, then the value of k is?

• 1) -2
• 2) -4
• 3) -1
• 4) 2

Solution

It is given that x² + 2 kx + 4 = 0 has a root x = 2. (2)² + 2k(2) + 4 = 0 4 + 4k + 4 = 0 4k + 8 = 0 4k = -8 k = -2

Q63. Solve the quadratic equation by the formula method. x² + 3x - 7 = 0

• 1)
• 2)
• 3)
• 4)

Solution

 

Q64. Solve the following quadratic equations by factorization : 4x² + 5x = 0

Solution

Consider the given quadratic equation, 4x² + 5x = 0 x (4x + 5) = 0 ⇒x = 0 or 4x + 5 = 0 ⇒x = 0 or 4x = -5

Q65.

Solution

Let the first number be x then the second number be 15 - x. According to the given condition,

Q66. A train travels at a uniform speed for a distance of 63 km and then travels a distance of 72 km at an average speed of 6 km/h more than its original speed. If it takes 3 hours to complete the total journey, what is the original speed of the train?

Solution

Let original speed of the train be x km/h. Then, time taken to travel 63 km = 63/x hours New speed = (x + 6) km/hr Time taken to travel 72 km = 72/(x + 6) hours According to the question, x² - 39x - 126 = 0 (x - 42)(x + 3) = 0 x = -3 or x = 42 As the speed cannot be negative, x = 42 Thus, the average speed of the train is 42 km/hr.

Q67. Sum of the area of two squares is 468 m². If the difference of their perimeters is 24 m, find the sides of the two squares.

Solution

Let the side of the squares be x and y meters. Areas of first square and second square are x² and y² respectively. Perimeters of first and second squares are 4x and 4y respectively. From the given information, we have: x² + y² = 468… (1) 4x - 4y = 24 x - y = 6 y = x - 6 Substituting the value of y in (1), we get x² + (x - 6)² = 468 x² + x² +36 - 12x = 468 2x² - 12x - 432 = 0 x² - 6x - 216 = 0 (x - 18)(x + 12) = 0 x = 18 or x = -12 As the side cannot be negative, x = 18 Hence, side of the first square is 18 m. Side of the second square is y = (18 - 6) m = 12 m

Q68. A two digit number is such that the product of the digits is 18. When 63 is subtracted from the number, the digits interchange their places. Find the number.

Solution

Let unit's digit is x and ten's digit = Number = On interchanging the digits the number becomes 10x + x² + 7x - 18 = 0 x² - 2x + 9x - 18 = 0 x(x - 2) + 9(x - 2) = 0 _{(x - 2) (x + 9) = 0} x = -9 or x = 2 Neglecting the negative value, x = 2. Therefore, the number is 92.

Q69. For what values of k, the roots of the quadratic equation (k + 4) x² + (k + 1)x + 1 = 0 are equal?

Solution

(k + 4)x² + (k + 1)x + 1 = 0 a = k + 4, b = k + 1, c = 1 For equal roots, D = 0 b² - 4ac = 0 (k + 1)² - 4(k + 4) ×1 = 0 k² + 2k + 1 - 4k - 16 = 0 k² - 2k - 15 = 0 k² - 5k + 3k - 15 = 0 k(k - 5) + 3(k - 5) = 0 (k - 5) (k + 3) = 0 k = 5 or k = -3 Thus, for k = 5 or k = - 3, the given quadratic equation has equal roots.

Q70.

Solution

Q71. The difference of squares of two natural numbers is 45. The square of the smaller number is four times the larger number. Find the numbers.

Solution

Let the two numbers be x and y such that y < x. From the given conditions, we have: x² - y² = 45 … (1) y² = 4x … (2) Using (2) in (1), we have: x² - 4x - 45 = 0 (x - 9)(x + 5) = 0 x = 9 (Rejecting the negative value of x) From (2), we get, y = 6 or -6 Rejecting the negative value, we have, y = 6. Thus, the two numbers are 9 and 6.

Q72. Find the roots of the following quadratic equation by the factorization method: 4

Solution

Q73. Some students arranged a picnic, the budget for food was Rs. 240. Because four students of the group failed to go, the cost of food to each student got increased by Rs. 5. How many students went for picnic?

Solution

Let number of student planned for picnic Budget for food = Rs. 240 Students left = 4 Hence, No. of students who went for picnic = x - 4 Earlier cost for each student =   New cost for each student = According to given information,   Number of students went for the picnic = 16 - 4 = 12 Hence,12 students went for picnic.

Q74. Find the roots of the quadratic equation 3x² - 14x + 8 = 0.

Solution

3x² - 14x + 8 = 0 3x² - 12x - 2x + 8 = 0 3x(x - 4) -2 (x - 4) = 0 (3x - 2) (x - 4) = 0 x = , x = 4

Q75. Solve for x:

Solution

Q76. Find the values of k for which the following quadratic equation has two equal roots: 2x² + kx + 3 = 0

Solution

Here, a = 2, b = k, c = 3 Since, the roots are equal, D = 0 Or, k² - 24 = 0

Q77. Find the solution of the quadratic equation by the formula method. x² + 9x - 3 = 0

• 1)
• 2)
• 3)
• 4)

Solution

 

Q78. The angry Arjun carried some arrows for fighting with Bheeshm. With half the arrows, he cut down the arrows thrown by Bheeshm on him and with six other arrows he killed the charioter of Bheesham. With one arrow each he knocked down respectively the rath, flag and bow of Bheeshm. Finally with one more than four times the square root of arrows he laid Bheeshm unconscious on an arrow bed. Find total number of arrows Arjun had.

Solution

Q79. Two pipes can together fill a tank in minutes. If one pipe takes 3 minutes more than the other to fill it, find the time in which each pipe can fill the tank.

Solution

Let the faster pipe take x minutes to fill the tank. Hence, the slower pipe takes x + 3 mins In 1 min the faster pipe fill 1/x of the tank and the slower pipe fills 1/(x + 3) of the tank. Part of the tank filled in 40/13 mins is (40/13) (1/(x+3)) Accoedingly,40/13x +40/13 (x + 3) = 1 40 [13(x + 3) + 13x] = 13 x 13 (x + 3) 13x² - 41x - 120 = 0 13x² - 65x + 24x - 120 = 0 13x (x - 5) + 24 (x - 5) = 0 (13x + 24) (x - 5) = 0 x = 5 [x = -13/24 is not possible]. Therefore, Faster pipe can fill in x = 5 minutes and Slower pipe in x + 3 = 5 + 3 = 8 mins

Q80. The sum of the squares of two positive integers is 208. If the square of the larger number is 18 times the smaller, find the numbers.

Solution

Let x and y be two positive integers. Since the sum of the squares of two positive integers is 208, we have x² + y² = 208…(1) Also given that the square of the larger number is 18 times the smaller. ⇒ x² = 18y…(2) Substituting the value of x² from equation (2) in equation (1), we get 18y + y² = 208 ⇒ y² + 18y = 208 ⇒ y² + 18y - 208 = 0 ⇒ y² + 26y - 8y - 208 = 0 ⇒ y(y + 26) - 8(y + 26)= 0 ⇒ (y - 8) (y + 26) = 0 ⇒ y - 8 = 0 or y + 26 = 0 ⇒ y = 8 or y = -26 Since the numbers are positive integers, y = 8 And hence, x² = 18y = 18 × 8 = 144 ⇒ x = 12 Thus, the positive integers are 12 and 8.  

Q81.  Solve the following equation            

Solution

Put  Then,        

Q82. A plane left 30 min late due to bad weather, but in order to reach its destination 1500 km away on time it has to increase its speed by 250 km/hr. Find the original speed.

Solution

Let the original speed of the plane be x. Distance = 1500 km Time = hrs Given that in order to reach its destination, the plane had to increase its speed by 250 km/hr. New speed = (x + 250) km/hr New Time = According to question, we have: x² + 250x - 750000 = 0 x² + 1000x - 750x - 750000 = 0 x(x + 1000) - 750(x + 1000) = 0 (x + 1000) (x - 750) = 0 x = -1000 or 750 But x, being the speed, cannot be negative. So, x = 750 Thus, the original speed of the plane was 750 km/hr.

Q83. Solve for x:

Solution

8x + 12 - 6x² - 9x = 5x 6x² + 6x - 12 = 0 x² + x - 2 = 0 (x + 2) (x - 1) = 0 x = -2, 1

Q84. Solve for x:

Solution

Q85. The quadratic equation whose roots are real and equal is:

• 1) 3x² - 5x + 2 = 0
• 2) x² - 4x + 4 = 0
• 3) 2x² - 4x + 3 = 0
• 4) x² - 2 - 6 = 0

Solution

Consider the equation x ² - 4x + 4 = 0 x ² - 4x + 4 = 0 (x-2)²=0 x=2 Thus, the roots of the equation x ² - 4x + 4 = 0 are equal and real.

Q86. Solve for x: 9x² - 3(a + b)x + ab = 0

Solution

9x² - 3ax - 3bx + ab = 0 3x(3x - a) - b(3x - a) = 0 (3x-a)(3x-b)=0 3x-a=0 or 3x-b=0 x =  

Q87. For what value(s) of p does the equation px² + (p - 1)x + (p - 1) = 0 have a repeated root?

Solution

Q88. Solve for x:

Solution

(x + 2 + 2x + 2) (x + 4) = 4(x² + 3x + 2) (3x + 4) (x + 4) = 4x² + 12x + 8 x² - 4x - 8 = 0 x = = = 2 + 2, 2 - 2

Q89. For what value of k the equation 4x² - 2 (k + 1) x + (k + 1) = 0 has real and equal roots?

Solution

Comparing the given equation with the general form of a quadratic equation ax² + bx + c = 0, we get a = 4, b = -2 (k + 1), c = k + 1 D = 0, for real and equal roots b² - 4ac = 0 4 (k + 1)² - 16 (k + 1) = 0 4 (k + 1) (k + 1 - 4) = 0 (k + 1)(k - 3) = 0 k = -1, 3 Therefore, for k = -1, 3 the given quadratic equation has real and equal roots.

Q90. Using quadratic formula solve the following quadratic equation

Solution

Comparing the equation with  we have         Now compute the discriminant.              So the given equation has real roots                                         

Q91. The altitude of a right triangle is 7 cm less than its base. If the hypotenuse is 13 cm, find the other two sides.

Solution

Let the base of the triangle  then altitude of the triangle Hypotenuse = 13 cm By Pythagoras theorem, (Hypotenuse) ² = (Base)² + (altitude)² Length can not be negative. Hence, x = 12 cm Altitude = x - 7 = 12 - 7 = 5 cm Therefore the length of the other two sides are 12 cm and 5 cm.

Q92. A takes 6 days less than the time taken by B finish a piece of work. If both A and B together can finish it in 4 days, find the time taken by B to finish the work.

Solution

Let the number of days taken by A to finish the work be x. Then, number of days taken by B to finish the work = x + 6 Now, work done by A in one day = and work done by B in one day = Now, we have: x = 6 (neglecting x = -4) Thus, time taken by B to finish the work = x + 6 = 12 days

Q93. The length of a rectangular plot is greater than thrice its breadth by 2 m. The area of the plot is 120 sq. m. Find the length and breadth of the plot.

Solution

Let the breadth of the rectangular plot be x m, then length = (3x + 2) m Area = 120 length × breadth = 120 Rejecting the negative value, as breadth can't be negative. Breadth of the rectangular plot = 6 m Length of the rectangular plot = (3 6 + 2) m = 20 m

Q94.   A factory produces certain pieces in a day. It was observed on a particular day that the cost of production of each piece (in rupees) was 3 more than twice the number of pieces produced in the day. If the total cost of production on that day was Rs. 90, find the number of pieces produced and cost of each piece.

Solution

  Let the number of pieces per day = n Cost of one pieces is C = 2n + 3 Total cost = Rs. 90  ... Given

Q95. Solve the given equation for x:

Solution

Q96. State whether the following equation is a quadratic equation or not_:

Solution

  Yes, it is a quadratic equation. 

Q97. Three consecutive positive integers are taken such that the sum of the square of the first and the product of the other two is 154. Find the integers.

Solution

Let the three consecutive positive integers be x, x + 1, x + 2. x² + (x + 1) (x + 2) = 154 x² + (x² + 3x + 2) = 154 2x² + 3x - 152 = 0 2x² - 16x + 19x - 152 = 0 2x (x - 8) + 19 (x - 8) = 0 (x - 8)(2x + 19) = 0 x = 8 or -19/2 But, x is a positive integer. So, x = 8. Thus, the numbers are 8, 9, 10.

Q98. The sum of the squares of two consecutive positive odd numbers is 130. Find the numbers.

Solution

Let the consecutive numbers be x, x + 2. x² + (x + 2)² = 130 x² + x² + 4x + 4 = 130 2x² + 4x - 126 = 0 x² + 2x - 63 = 0 x² + 9x - 7x - 63 = 0 x(x + 9) - 7(x + 9) = 0 (x + 9) (x - 7) = 0 x = -9 or x = 7 Neglecting the negative value, we get, x = 7. The numbers are 7 and 9.

Q99. Find the roots of quadratic equation .

Solution

Let us split middle term Therefore, the roots are .

Q100. Solve for x:

Solution

Q101. Solve the following equation for x. 9x² - 9 (a + b) x + (2a² + 5ab + 2b²) = 0

Solution

9x² - 9(a + b) x + (2a²+ 5ba + 2b²) = 0 9x² - 9(a + b)x + (2a + b) (a + 2b) = 0 9x² - 3(2a + b)x - 3 (a + 2b)x + (2a + b) (a + 2b) = 0 3x[3x - (2a + b)] - (a + 2b) [3x - (2a + b)] = 0 [3x - (2a + b)] [3x - (a + 2b)] = 0 x =

Q102.

Solution

Q103. Sum of the areas of two squares is 468 m². If the difference of their perimeters is 24m, find the sides of the two squares.

Solution

Sum of the areas of two squares = 468 m² Let a and b be the sides of the two squares. ⇒a² + b² = 468…(1) Also given that, the difference of their perimeters = 24m ⇒4a - 4b = 24 ⇒a - b = 6 ⇒a = b + 6…(2) We need to find the sides of the two squares. Substituting the value of a from equation (2) in equation (1), we get (b + 6)² + b² = 468 ⇒b² + 6² + 2 × b × 6 + b² = 468 ⇒2b² + 36 + 12b = 468 ⇒2b² + 36 + 12b - 468 = 0 ⇒2b² + 12b - 432 = 0 ⇒b² + 6b - 216 = 0 ⇒b² + 18b - 12b - 216 = 0 ⇒b(b + 18) - 12(b + 18) = 0 ⇒(b + 18)(b - 12) = 0 ⇒b + 18 = 0 or b - 12 = 0 ⇒b = -18 = 0 or b = 12 Side cannot be negative and hence b = 12 m. Therefore, a = b + 6 = 12 + 6 = 18 m. Therefore the sides of the two squares are 12m and 18m.

Q104. A person has a rectangular garden whose area is 100 sq m. He fences three sides of the garden with 30 m barbed wire. On the fourth side, the wall of his house is constructed; find the dimensions of the garden.

Solution

Let the length and breadth of garden be x m and y m respectively. _{Area of the garden = 100 sq m} xy = 100 sq m or y =  Suppose the person builds his house along the breadth of the garden. Then, we have: 2x + y = 30   2x² - 30x + 100 = 0 x² - 15x + 50 = 0 x² - 10x - 5x + 50 = 0 x(x - 10) - 5(x - 10) = 0 (x - 10) (x - 5) = 0 x = 10, x = 5 When x = 10 m, we have: y = 10 m When x = 5 m, we have: y = 20 m Thus, the dimensions of the garden are 10 m 10 m or 5 m 20 m.

Q105. Find value of p such that the quadratic equation (p - 12)x² - 2(p - 12)x + 2 = 0 has equal roots.

Solution

For equal roots, D = 0, i.e., b² - 4ac = 0 [-2(p - 12)]² - 4(p - 12) 2 = 0 4(p - 12)² - 8(p - 12) = 0 4(p - 12) (p - 12 - 2) = 0 (p - 12) (p - 14) = 0 p = 12 or p = 14 Rejecting p = 12, as then the given equation will not be true. We have: p = 14