Exercise 4.1
1. Check whether the following are quadratic equations: (i) (x + 1)2 = 2(x – 3) (ii) x2 – 2x = (–2) (3 – x) (iii) (x – 2)(x + 1) = (x – 1)(x + 3) (iv) (x – 3)(2x +1) = x(x + 5) (v) (2x – 1)(x – 3) = (x + 5)(x – 1) (vi) x2 + 3x + 1 = (x – 2)2 (vii) (x + 2)3 = 2x (x2 – 1) (viii) x3 – 4x2 – x + 1 = (x – 2)3
Solutions: (i) Given, (x + 1)2 = 2(x – 3) By using the formula for (a+b)2 = a2+2ab+b2 ⇒ x2 + 2x + 1 = 2x – 6 ⇒ x2 + 7 = 0 Since the above equation is in the form of ax2 + bx + c = 0. Therefore, the given equation is quadratic equation.
(ii) Given, x2 – 2x = (–2) (3 – x) ⇒ x2 – 2x = -6 + 2x ⇒ x2 – 4x + 6 = 0 Since the above equation is in the form of ax2 + bx + c = 0. Therefore, the given equation is quadratic equation.
(iii) Given, (x – 2)(x + 1) = (x – 1)(x + 3) By multiplication ⇒ x2 – x – 2 = x2 + 2x – 3 ⇒ 3x – 1 = 0 Since the above equation is not in the form of ax2 + bx + c = 0. Therefore, the given equation is not a quadratic equation.
(iv) Given, (x – 3)(2x +1) = x(x + 5) By multiplication ⇒ 2x2 – 5x – 3 = x2 + 5x ⇒ x2 – 10x – 3 = 0 Since the above equation is in the form of ax2 + bx + c = 0. Therefore, the given equation is quadratic equation.
(v) Given, (2x – 1)(x – 3) = (x + 5)(x – 1) By multiplication ⇒ 2x2 – 7x + 3 = x2 + 4x – 5 ⇒ x2 – 11x + 8 = 0 Since the above equation is in the form of ax2 + bx + c = 0. Therefore, the given equation is quadratic equation.
(vi) Given, x2 + 3x + 1 = (x – 2)2 By using the formula for (a-b)2=a2-2ab+b2 ⇒ x2 + 3x + 1 = x2 + 4 – 4x ⇒ 7x – 3 = 0 Since the above equation is not in the form of ax2 + bx + c = 0. Therefore, the given equation is not a quadratic equation.
(vii) Given, (x + 2)3 = 2x(x2 – 1) By using the formula for (a+b)3 = a3+b3+3ab(a+b) ⇒ x3 + 8 + x2 + 12x = 2x3 – 2x ⇒ x3 + 14x – 6x2 – 8 = 0 Since the above equation is not in the form of ax2 + bx + c = 0. Therefore, the given equation is not a quadratic equation.
(viii) Given, x3 – 4x2 – x + 1 = (x – 2)3 By using the formula for (a-b)3 = a3-b3-3ab(a-b) ⇒ x3 – 4x2 – x + 1 = x3 – 8 – 6x2 + 12x ⇒ 2x2 – 13x + 9 = 0 Since the above equation is in the form of ax2 + bx + c = 0. Therefore, the given equation is quadratic equation.
2. Represent the following situations in the form of quadratic equations: (i) The area of a rectangular plot is 528 m2. The length of the plot (in metres) is one more than twice its breadth. We need to find the length and breadth of the plot. (ii) The product of two consecutive positive integers is 306. We need to find the integers. (iii) Rohan’s mother is 26 years older than him. The product of their ages (in years) 3 years from now will be 360. We would like to find Rohan’s present age. (iv) A train travels a distance of 480 km at a uniform speed. If the speed had been 8 km/h less, then it would have taken
Solutions: (i) Let us consider, Breadth of the rectangular plot = x m Thus, the length of the plot = (2x + 1) m. As we know, Area of rectangle = length × breadth = 528 m2 Putting the value of length and breadth of the plot in the formula, we get, (2x + 1) × x = 528 ⇒ 2x2 + x =528 ⇒ 2x2 + x – 528 = 0 Therefore, the length and breadth of plot, satisfies the quadratic equation, 2x2 + x – 528 = 0, which is the required representation of the problem mathematically.
(ii) Let us consider, The first integer number = x Thus, the next consecutive positive integer will be = x + 1 Product of two consecutive integers = x × (x +1) = 306 ⇒ x2 + x = 306 ⇒ x2 + x – 306 = 0 Therefore, the two integers x and x+1, satisfies the quadratic equation, x2 + x – 306 = 0, which is the required representation of the problem mathematically.
(iii) Let us consider, Age of Rohan’s = x years Therefore, as per the given question, Rohan’s mother’s age = x + 26 After 3 years, Age of Rohan’s = x + 3 Age of Rohan’s mother will be = x + 26 + 3 = x + 29 The product of their ages after 3 years will be equal to 360, such that (x + 3)(x + 29) = 360 ⇒ x2 + 29x + 3x + 87 = 360 ⇒ x2 + 32x + 87 – 360 = 0 ⇒ x2 + 32x – 273 = 0 Therefore, the age of Rohan and his mother, satisfies the quadratic equation, x2 + 32x – 273 = 0, which is the required representation of the problem mathematically.
(iv) Let us consider, The speed of train = x km/h And Time taken to travel 480 km = 480/x km/hr As per second condition, the speed of train = (x – 8) km/h Also given, the train will take 3 hours to cover the same distance. Therefore, time taken to travel 480 km = (480/x)+3 km/h As we know, Speed × Time = Distance Therefore, (x – 8)(480/x )+ 3 = 480 ⇒ 480 + 3x – 3840/x – 24 = 480 ⇒ 3x – 3840/x = 24 ⇒ x2 – 8x – 1280 = 0 Therefore, the speed of the train, satisfies the quadratic equation, x2 – 8x – 1280 = 0, which is the required representation of the problem mathematically.
Exercise 4.2
1. Find the roots of the following quadratic equations by factorisation: (i) x2 – 3x – 10 = 0 (ii) 2x2 + x – 6 = 0 (iii) √2 x2 + 7x + 5√2 = 0 (iv) 2x2 – x +1/8 = 0 (v) 100x2 – 20x + 1 = 0
Solutions: (i) Given, x2 – 3x – 10 =0 Taking LHS, =>x2 – 5x + 2x – 10 =>x(x – 5) + 2(x – 5) =>(x – 5)(x + 2) The roots of this equation, x2 – 3x – 10 = 0 are the values of x for which (x – 5)(x + 2) = 0 Therefore, x – 5 = 0 or x + 2 = 0 => x = 5 or x = -2
(ii) Given, 2x2 + x – 6 = 0 Taking LHS, => 2x2 + 4x – 3x – 6 => 2x(x + 2) – 3(x + 2) => (x + 2)(2x – 3) The roots of this equation, 2x2 + x – 6=0 are the values of x for which (x x + 2)(2x – 3) = 0 Therefore, x + 2 = 0 or 2x – 3 = 0 => x = -2 or x = 3/2
(iii) √2 x2 + 7x + 5√2=0 Taking LHS, => √2 x2 + 5x + 2x + 5√2 => x (√2x + 5) + √2(√2x + 5)= (√2x + 5)(x + √2) The roots of this equation, √2 x2 + 7x + 5√2=0 are the values of x for which (√2x + 5)(x + √2) = 0 Therefore, √2x + 5 = 0 or x + √2 = 0 => x = -5/√2 or x = -√2
(iv) 2x2 – x +1/8 = 0 Taking LHS, =1/8 (16x2 – 8x + 1) = 1/8 (16x2 – 4x -4x + 1) = 1/8 (4x(4x – 1) -1(4x – 1)) = 1/8 (4x – 1)2 The roots of this equation, 2x2 – x + 1/8 = 0, are the values of x for which (4x – 1)2= 0 Therefore, (4x – 1) = 0 or (4x – 1) = 0 ⇒ x = 1/4 or x = 1/4
(v) Given, 100x2 – 20x + 1=0 Taking LHS, = 100x2 – 10x – 10x + 1 = 10x(10x – 1) -1(10x – 1) = (10x – 1)2 The roots of this equation, 100x2 – 20x + 1=0, are the values of x for which (10x – 1)2= 0 ∴ (10x – 1) = 0 or (10x – 1) = 0 ⇒x = 1/10 or x = 1/10
2. Solve the problems given in Example 1. Represent the following situations mathematically: (i) John and Jivanti together have 45 marbles. Both of them lost 5 marbles each, and the product of the number of marbles they now have is 124. We would like to find out how many marbles they had to start with. (ii) A cottage industry produces a certain number of toys in a day. The cost of production of each toy (in rupees) was found to be 55 minus the number of toys produced in a day. On a particular day, the total cost of production was ` 750. We would like to find out the number of toys produced on that day.
Solutions: (i) Let us say, the number of marbles John have = x. Therefore, number of marble Jivanti have = 45 – x After losing 5 marbles each, Number of marbles John have = x – 5 Number of marble Jivanti have = 45 – x – 5 = 40 – x Given that the product of their marbles is 124. ∴ (x – 5)(40 – x) = 124 ⇒ x2 – 45x + 324 = 0 ⇒ x2 – 36x – 9x + 324 = 0 ⇒ x(x – 36) -9(x – 36) = 0 ⇒ (x – 36)(x – 9) = 0 Thus, we can say, x – 36 = 0 or x – 9 = 0 ⇒ x = 36 or x = 9 Therefore, If, John’s marbles = 36, Then, Jivanti’s marbles = 45 – 36 = 9 And if John’s marbles = 9, Then, Jivanti’s marbles = 45 – 9 = 36
(ii) Let us say, number of toys produced in a day be x. Therefore, cost of production of each toy = Rs(55 – x) Given, total cost of production of the toys = Rs 750 ∴ x(55 – x) = 750 ⇒ x2 – 55x + 750 = 0 ⇒ x2 – 25x – 30x + 750 = 0 ⇒ x(x – 25) -30(x – 25) = 0 ⇒ (x – 25)(x – 30) = 0 Thus, either x -25 = 0 or x – 30 = 0 ⇒ x = 25 or x = 30 Hence, the number of toys produced in a day, will be either 25 or 30.
3. Find two numbers whose sum is 27 and product is 182.
Solution: Let us say, first number be x and the second number is 27 – x. Therefore, the product of two numbers x(27 – x) = 182 ⇒ x2 – 27x – 182 = 0 ⇒ x2 – 13x – 14x + 182 = 0 ⇒ x(x – 13) -14(x – 13) = 0 ⇒ (x – 13)(x -14) = 0 Thus, either, x = -13 = 0 or x – 14 = 0 ⇒ x = 13 or x = 14 Therefore, if first number = 13, then second number = 27 – 13 = 14 And if first number = 14, then second number = 27 – 14 = 13 Hence, the numbers are 13 and 14.
4. Find two consecutive positive integers, sum of whose squares is 365.
Solution: Let us say, the two consecutive positive integers be x and x + 1. Therefore, as per the given questions, x2 + (x + 1)2 = 365 ⇒ x2 + x2 + 1 + 2x = 365 ⇒ 2x2 + 2x – 364 = 0 ⇒ x2 + x – 182 = 0 ⇒ x2 + 14x – 13x – 182 = 0 ⇒ x(x + 14) -13(x + 14) = 0 ⇒ (x + 14)(x – 13) = 0 Thus, either, x + 14 = 0 or x – 13 = 0, ⇒ x = – 14 or x = 13 since, the integers are positive, so x can be 13, only. ∴ x + 1 = 13 + 1 = 14 Therefore, two consecutive positive integers will be 13 and 14.
5. 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 us say, the base of the right triangle be x cm. Given, the altitude of right triangle = (x – 7) cm From Pythagoras theorem, we know, Base2 + Altitude2 = Hypotenuse2 ∴ x2 + (x – 7)2 = 132 ⇒ x2 + x2 + 49 – 14x = 169 ⇒ 2x2 – 14x – 120 = 0 ⇒ x2 – 7x – 60 = 0 ⇒ x2 – 12x + 5x – 60 = 0 ⇒ x(x – 12) + 5(x – 12) = 0 ⇒ (x – 12)(x + 5) = 0 Thus, either x – 12 = 0 or x + 5 = 0, ⇒ x = 12 or x = – 5 Since sides cannot be negative, x can only be 12. Therefore, the base of the given triangle is 12 cm and the altitude of this triangle will be (12 – 7) cm = 5 cm.
6. A cottage industry produces a certain number of pottery articles in a day. It was observed on a particular day that the cost of production of each article (in rupees) was 3 more than twice the number of articles produced on that day. If the total cost of production on that day was Rs.90, find the number of articles produced and the cost of each article.
Solution: Let us say, the number of articles produced be x. Therefore, cost of production of each article = Rs (2x + 3) Given, total cost of production is Rs.90 ∴ x(2x + 3) = 90 ⇒ 2x2 + 3x – 90 = 0 ⇒ 2x2 + 15x -12x – 90 = 0 ⇒ x(2x + 15) -6(2x + 15) = 0 ⇒ (2x + 15)(x – 6) = 0 Thus, either 2x + 15 = 0 or x – 6 = 0 ⇒ x = -15/2 or x = 6 As the number of articles produced can only be a positive integer, therefore, x can only be 6. Hence, number of articles produced = 6 Cost of each article = 2 × 6 + 3 = Rs 15.
Exercise 4.3
1. Find the roots of the following quadratic equations, if they exist, by the method of completing the square: (i) 2x2 – 7x +3 = 0 (ii) 2x2 + x – 4 = 0 (iii) 4x2 + 4√3x + 3 = 0 (iv) 2x2 + x + 4 = 0
Solutions: (i) 2x2 – 7x + 3 = 0 ⇒ 2x2 – 7x = – 3 Dividing by 2 on both sides, we get ⇒ x2 -7x/2 = -3/2 ⇒ x2 -2 × x ×7/4 = -3/2 On adding (7/4)2 to both sides of equation, we get ⇒ (x)2-2×x×7/4 +(7/4)2 = (7/4)2-3/2 ⇒ (x-7/4)2 = (49/16) – (3/2) ⇒(x-7/4)2 = 25/16 ⇒(x-7/4)2 = ±5/4 ⇒ x = 7/4 ± 5/4 ⇒ x = 7/4 + 5/4 or x = 7/4 – 5/4 ⇒ x = 12/4 or x = 2/4 ⇒ x = 3 or x = 1/2
(ii) 2x2 + x – 4 = 0 ⇒ 2x2 + x = 4 Dividing both sides of the equation by 2, we get ⇒ x2 +x/2 = 2 Now on adding (1/4)2 to both sides of the equation, we get, ⇒ (x)2 + 2 × x × 1/4 + (1/4)2 = 2 + (1/4)2 ⇒ (x + 1/4)2 = 33/16 ⇒ x + 1/4 = ± √33/4 ⇒ x = ± √33/4 – 1/4 ⇒ x = (± √33-1)/4 Therefore, either x = (√33-1)/4 or x = (-√33-1)/4
(iii) 4x2 + 4√3x + 3 = 0 Converting the equation into a2+2ab+b2 form, we get, ⇒ (2x)2 + 2 × 2x × √3 + (√3)2 = 0 ⇒ (2x + √3)2 = 0 ⇒ (2x + √3) = 0 and (2x + √3) = 0 Therefore, either x = -√3/2 or x = -√3/2.
(iv) 2x2 + x + 4 = 0 ⇒ 2x2 + x = -4 Dividing both sides of the equation by 2, we get ⇒ x2 + 1/2x = 2 ⇒ x2 + 2 × x × 1/4 = -2 By adding (1/4)2 to both sides of the equation, we get ⇒ (x)2 + 2 × x × 1/4 + (1/4)2 = (1/4)2 – 2 ⇒ (x + 1/4)2 = 1/16 – 2 ⇒ (x + 1/4)2 = -31/16 As we know, the square of numbers cannot be negative. Therefore, there is no real root for the given equation, 2x2 + x + 4 = 0.
2. Find the roots of the quadratic equations given in Q.1 above by applying the quadratic formula. (i) 2x2 – 7x + 3 = 0 On comparing the given equation with ax2 + bx + c = 0, we get, a = 2, b = -7 and c = 3 By using quadratic formula, we get,
⇒ x = (7±√(49 – 24))/4
⇒ x = (7±√25)/4
⇒ x = (7±5)/4
⇒ x = (7+5)/4 or x = (7-5)/4
⇒ x = 12/4 or 2/4
∴ x = 3 or 1/2
(ii) 2x2 + x – 4 = 0
On comparing the given equation with ax2 + bx + c = 0, we get, a = 2, b = 1 and c = -4 By using quadratic formula, we get,
⇒x = (-1±√1+32)/4
⇒x = (-1±√33)/4
∴ x = (-1+√33)/4 or x = (-1-√33)/4
(iii) 4x2 + 4√3x + 3 = 0
On comparing the given equation with ax2 + bx + c = 0, we get a = 4, b = 4√3 and c = 3 By using quadratic formula, we get,
⇒ x = (-4√3±√48-48)/8
⇒ x = (-4√3±0)/8
∴ x = -√3/2 or x = -√3/2
(iv) 2x2 + x + 4 = 0
On comparing the given equation with ax2 + bx + c = 0, we get, a = 2, b = 1 and c = 4 By using quadratic formula, we get
⇒ x = (-1±√1-32)/4
⇒ x = (-1±√-31)/4
As we know, the square of a number can never be negative. Therefore, there is no real solution for the given equation.
3. Find the roots of the following equations: (i) x-1/x = 3, x ≠ 0 (ii) 1/x+4 – 1/x-7 = 11/30, x = -4, 7
Solution: (i) x-1/x = 3 ⇒ x2 – 3x -1 = 0 On comparing the given equation with ax2 + bx + c = 0, we get a = 1, b = -3 and c = -1 By using quadratic formula, we get,
⇒ x = (3±√9+4)/2
⇒ x = (3±√13)/2
∴ x = (3+√13)/2 or x = (3-√13)/2
(ii) 1/x+4 – 1/x-7 = 11/30
⇒ x-7-x-4/(x+4)(x-7) = 11/30 ⇒ -11/(x+4)(x-7) = 11/30 ⇒ (x+4)(x-7) = -30 ⇒ x2 – 3x – 28 = 30 ⇒ x2 – 3x + 2 = 0 We can solve this equation by factorization method now, ⇒ x2 – 2x – x + 2 = 0 ⇒ x(x – 2) – 1(x – 2) = 0 ⇒ (x – 2)(x – 1) = 0 ⇒ x = 1 or 2
4. 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 us say, present age of Rahman is x years. Three years ago, Rehman’s age was (x – 3) years. Five years after, his age will be (x + 5) years. Given, the sum of the reciprocals of Rehman’s ages 3 years ago and after 5 years is equal to 1/3. ∴ 1/x-3 + 1/x-5 = 1/3 (x+5+x-3)/(x-3)(x+5) = 1/3 (2x+2)/(x-3)(x+5) = 1/3 ⇒ 3(2x + 2) = (x-3)(x+5) ⇒ 6x + 6 = x2 + 2x – 15 ⇒ x2 – 4x – 21 = 0 ⇒ x2 – 7x + 3x – 21 = 0 ⇒ x(x – 7) + 3(x – 7) = 0 ⇒ (x – 7)(x + 3) = 0 ⇒ x = 7, -3 As we know, age cannot be negative. Therefore, Rahman’s present age is 7 years.
5. In a class test, the sum of Shefali’s marks in Mathematics and English is 30. Had she got 2 marks more in Mathematics and 3 marks less in English, the product of their marks would have been 210. Find her marks in the two subjects.
Solution: Let us say, the marks of Shefali in Maths be x. Then, the marks in English will be 30 – x. As per the given question, (x + 2)(30 – x – 3) = 210 (x + 2)(27 – x) = 210 ⇒ -x2 + 25x + 54 = 210 ⇒ x2 – 25x + 156 = 0 ⇒ x2 – 12x – 13x + 156 = 0 ⇒ x(x – 12) -13(x – 12) = 0 ⇒ (x – 12)(x – 13) = 0 ⇒ x = 12, 13 Therefore, if the marks in Maths are 12, then marks in English will be 30 – 12 = 18 and the marks in Maths are 13, then marks in English will be 30 – 13 = 17.
6. The diagonal of a rectangular field is 60 metres more than the shorter side. If the longer side is 30 metres more than the shorter side, find the sides of the field.
Solution: Let us say, the shorter side of the rectangle be x m. Then, larger side of the rectangle = (x + 30) m
As given, the length of the diagonal is = x + 30 m
Therefore,
⇒ x2 + (x + 30)2 = (x + 60)2
⇒ x2 + x2 + 900 + 60x = x2 + 3600 + 120x
⇒ x2 – 60x – 2700 = 0
⇒ x2 – 90x + 30x – 2700 = 0
⇒ x(x – 90) + 30(x -90) = 0
⇒ (x – 90)(x + 30) = 0
⇒ x = 90, -30
However, side of the field cannot be negative. Therefore, the length of the shorter side will be 90 m.
and the length of the larger side will be (90 + 30) m = 120 m.
7. The difference of squares of two numbers is 180. The square of the smaller number is 8 times the larger number. Find the two numbers.
Solution: Let us say, the larger and smaller number be x and y respectively. As per the question given, x2 – y2 = 180 and y2 = 8x ⇒ x2 – 8x = 180 ⇒ x2 – 8x – 180 = 0 ⇒ x2 – 18x + 10x – 180 = 0 ⇒ x(x – 18) +10(x – 18) = 0 ⇒ (x – 18)(x + 10) = 0 ⇒ x = 18, -10 However, the larger number cannot considered as negative number, as 8 times of the larger number will be negative and hence, the square of the smaller number will be negative which is not possible. Therefore, the larger number will be 18 only. x = 18 ∴ y2 = 8x = 8 × 18 = 144 ⇒ y = ±√144 = ±12 ∴ Smaller number = ±12 Therefore, the numbers are 18 and 12 or 18 and -12.
8. A train travels 360 km at a uniform speed. If the speed had been 5 km/h more, it would have taken 1 hour less for the same journey. Find the speed of the train.
Solution: Let us say, the speed of the train be x km/hr. Time taken to cover 360 km = 360/x hr. As per the question given, ⇒ (x + 5)(360/x – 1) = 360 ⇒ 360 – x + 1800/x-5 = 360 ⇒ x2 + 5x – 1800 = 0 ⇒ x2 + 45x – 40x – 1800 = 0 ⇒ x(x + 45) -40(x + 45) = 0 ⇒ (x + 45)(x – 40) = 0 ⇒ x = 40, -45 As we know, the value of speed cannot be negative. Therefore, the speed of train is 40 km/h.
9. Two water taps together can fill a tank in
= 75/8 hours by both the pipes together.
Therefore,
1/x + 1/x-10 = 8/75
x-10+x/x(x-10) = 8/75
⇒ 2x-10/x(x-10) = 8/75
⇒ 75(2x – 10) = 8x2 – 80x
⇒ 150x – 750 = 8x2 – 80x
⇒ 8x2 – 230x +750 = 0
⇒ 8x2 – 200x – 30x + 750 = 0
⇒ 8x(x – 25) -30(x – 25) = 0
⇒ (x – 25)(8x -30) = 0
⇒ x = 25, 30/8
Time taken by the smaller pipe cannot be 30/8 = 3.75 hours, as the time taken by the larger pipe will become negative, which is logically not possible.
Therefore, time taken individually by the smaller pipe and the larger pipe will be 25 and 25 – 10 =15 hours respectively.
= (-(-6) ± √(-62-4(2)(3)) )/ 2(2)
= (6±2√3 )/4
= (3±√3)/2
Therefore the roots for the given equation are (3+√3)/2 and (3-√3)/2
