## NCERT Class 10 Maths## Chapter 4: Quadratic Equations## Exercise 4.1
- (x + 1)
^{2}= 2(x - 3) - x
^{2}- 2x = (-2) (3 - x) - (x - 2)(x + 1) = (x - 1)(x + 3)
- (x - 3)(2x +1) = x(x + 5)
- (2x - 1)(x - 3) = (x + 5)(x - 1)
- x
^{2}+ 3x + 1 = (x - 2)^{2} - (x + 2)
^{3}= 2x (x^{2}- 1) - x
^{3}- 4x^{2}- x + 1 = (x - 2)^{3}
(x + 1) Since, the highest power of x is 2. Therefore, the given equation is quadratic.
x Since, the highest power of x is 2. Therefore, the given equation is quadratic.
(x - 2)(x + 1) = (x - 1)(x + 3) Since, the highest power of x is 1. Therefore, the given equation is not quadratic.
(x - 3)(2x + 1) = x(x + 5) Since, the highest power of x is 2. Therefore, the given equation is quadratic.
(2x - 1)(x - 3) = (x - 1)(x + 5) Since, the highest power of x is 2. Therefore, the given equation is quadratic.
x Since, the highest power of x is 1. Therefore, the given equation is not quadratic.
(x + 2)^3 = 2x(x Since, the highest power of x is 3. Therefore, the given equation is not quadratic.
x^3 - 4x^(2 )- x + 1 = (x - 2)^3 Since, the highest power of x is 2. Therefore, the given equation is quadratic.
- The area of a rectangular plot is 528 m
^{2}. 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. - The product of two consecutive positive integers is 306. We need to find the integers.
- 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.
- 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 3 hours more to cover the same distance. We need to find the speed of the train.
I. Let the breadth of the rectangular plot be x (in metres). It is known that the length of the plot is one more than two times the breadth, Therefore the length of the rectangular plot = 2x + 1 We know that, Area of Rectangle = Length × Breadth Hence, the required equation will be: 528 = x(2x + 1) II. Let the smaller integer be x. It is known that both the integers are consecutive, Therefore the other integer will be = x + 1. It is given that the product of these two integers is 306. Hence, the required equation will be: 306 = x(x + 1) III. Let the present age of Rohan (in years) be x. Then, the present age of his mother (in years) = x + 26. It is given that the product of their ages 3 years into the future would be 360. Hence, the required equation will be: IV. Let the speed of the train (in km/h) be x. We know that, Using this knowledge, we can conclude that the time taken by the train = 480/x It is given that reducing the speed of the train by 8 km/h would have resulted in 3 hours of extra time required for the journey. Hence, the required equation will be: ## Exercise 4.2
- x
^{2}- 3x - 10 = 0 - 2x
^{2}+ x - 6 = 0 - √2x
^{2}+ 7x + 5√2 = 0 - 2x
^{2}- x + 1/8 = 0 - 100x
^{2}- 20x + 1 = 0
x OR x + 2 = 0 Hence, the roots of the given equation are 5 and -2.
2x OR x + 2 = 0 Hence, the roots of the given equation are 3/2 and -2.
Hence, the roots of the given equation are -5/√2 and -√2.
Hence, the roots of the given equations are ¼ and ¼.
Hence, the roots of the given equations are 1/10 and 1/10.
**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.****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 Rs 750. We would like to find out the number of toys produced on that day.**
I. Let the number of marbles John had at the start be x. Then the number of marbles Jivanti had at the start will be 45 - x. It is given that both of them lost 4 marbles and the product of their marbles becomes 124. Therefore (x - 5)(45 - x - 5) = 124 Hence, John and Jivanti had 9 and 36 marbles respectively or 36 and 9 marbles respectively. II. Let the number of toys produced in a day by the cottage industry be x. Then, the cost of production of each toy (in rupees) = 55 - x It is given that on a particular day, the total cost of production was Rs 750. Therefore Hence, the number of toys produced in the cottage industry on the given day was either 25 or 30.
Let one of the two numbers be x. Then, the other number will be = 27 - x It is given that the product of these numbers is 182. Therefore, x(27 - x) = 182 Hence, the two numbers are 13 and 14 respectively or 14 and 13 respectively.
Let the smaller positive integer be x. Then the other positive integer = x + 1 It is given that the sum of the squares of the given two numbers is 365. Therefore, (x But, it is given to us that the integers are positive. Therefore, x = -14 is rejected. Hence, the required positive integers are 13 and 13 + 1 = 14.
Let the base of the right triangle (in cm) be x. Then, the altitude of the right triangle (in cm) = x - 7 Hypotenuse = 13 cm (Given) According to the Pythagoras Theorem, Base^(2 )+ Altitude Therefore, x But, x is length and can never be negative. Therefore, x = -15 is rejected. Hence, the base of the given right triangle is 12 cm and the altitude is 12 - 7 = 5 cm.
Let the number of pottery articles produced in a day by the cottage industry be x. Then, the cost of production of each pottery article (in rupees) = 2x + 3 It is given that on a particular day, the total cost of production was Rs 90. Therefore x(2x + 3) = 90 (Number of articles produced in the day × Cost of each article = Total cost) 2x But, the number of articles produced cannot be in fractions. Therefore, x = 15/2 is rejected. Hence, the number of articles produced in the cottage industry on the given day was 6 and the cost of each article = 90/6 = Rs 15. ## Exercise 4.3
- 2x
^{2}- 7x + 3 = 0 - 2x
^{2}+ x - 4 = 0 - 4x
^{2}+ 4√3x + 3 = 0 - 2x
^{2}+ x + 4 = 0
I. 2x We need the coefficient of x2 to be 1. So, we will divide both sides by 2. Add and subtract (7/4) Hence, the roots are x = ½ and x = 3. II. 2x We need the coefficient of x2 to be 1. So, we will divide both sides by 2. Add and subtract (1/4) Hence, the roots are x = -√3/2 and x = -√3/2. IV. 2x We need the coefficient of x2 to be 1. So, we will divide both sides by 2. Add and subtract (1/4)2 on the LHS. √(-31) is an imaginary number. Hence, roots are not real for the given quadratic equation.
**2x**^{2}- 7x + 3 = 0**2x**^{2}+ x - 4 = 0**4x**^{2}+ 4 √3x + 3 = 0**2x**^{2}+ x + 4 = 0
I. On comparing the given quadratic equation with ax a = 2, b = -7 and c = 3 Find the discriminant d = b Since, d > 0. Therefore, two real roots α and β exist for the given equation. Hence, the roots are 3 and ½. I. On comparing the given quadratic equation with ax a = 2, b = 1 and c = -4 Find the discriminant d = b Since, d > 0. Therefore, two real roots α and β exist for the given equation.
a = 4, b = 4√3 and c = 3 Find the discriminant Since, d = 0. Therefore, two equal and real roots α and β exist for the given equation. IV. On comparing the given quadratic equation with ax2 + bx + c = 0, we get a = 2, b = 1 and c = 4 Find the discriminant d = b Since, d < 0. Therefore, real roots do not exist for this equation.
- x - 1/x = 3, x ≠ 0
- 1/(x + 4) - 1/(x - 7) = 11/30, x ≠ -4, 7
I. Multiply both sides by x. x x On comparing the given quadratic equation with ax2 + bx + c = 0, we get a = 1, b = -3and c = -1 Find the discriminant d = b Since, d > 0. Therefore, two real roots α and β exist for the given equation. On comparing the given quadratic equation with ax2 + bx + c = 0, we get a = 1, b = -3and c = 2 Find the discriminant d = b Since, d > 0. Therefore, two real roots α and β exist for the given equation. Hence, the roots are 2 and 1.
Let the present age of Rehman (in years) be x. Reciprocal of his age 3 years ago = 1/(x - 3) Reciprocal of his age 5 years later = 1/(x + 5) On comparing the given quadratic equation with ax a = 1, b = -4 and c = 21 Find the discriminant Since, d > 0. Therefore, two real roots α and β exist for the given equation. Since, age cannot be negative, therefore x = -3 is rejected. Hence, Rehman's present age (in years) is x = 7.
Let Shefali's marks in Mathematics be x. Then, Shefali's marks in English = 30 - x. It is given that if she had gotten 2 marks more in Mathematics and 3 less marks in English, the product of the marks would be 210. Therefore (x + 2)(30 - x - 3) = 210 On comparing the given quadratic equation with ax2 + bx + c = 0, we get a = 1, b = -25 and c = 156 Find the discriminant d = b Since, d > 0. Therefore, two real roots α and β exist for the given equation. When x = 12, 30 - x = 30 - 12 = 18 When x = 13, 30 - x = 30 - 13 = 17 Hence, Shefali got either 13 marks in Mathematics and 17 marks in English OR 12 marks in Mathematics and 18 marks in English.
Let the longer side of the rectangular field (in metres) be x. Then the shorter side of the rectangular field (in metres) = x - 30 The diagonal of the rectangle = 60 + (x - 30) The diagonal of the rectangular field and two sides form a right triangle. According to Pythagoras Theorem, Hypotenuse Shorter side = x - 30 = 120 - 30 = 90 m Hence, the longer side of the rectangular field is 120 m while the shorter side is 90 m.
Let the larger number = x. Then the square of smaller number = 8x Since, the difference of the squares of the two numbers is 180. Therefore x On comparing the given quadratic equation with ax2 + bx + c = 0, we get a = 1, b = -8 and c = -180 Find the discriminant d = b Since, d > 0. Therefore, two real roots α and β exist for the given equation. But, x cannot be negative as that will result in the square of smaller number to be negative which is not possible. Therefore, x = 18.
Let the speed of the train (in km/h) = x. Distance travelled = 360 km We know that, On comparing the given quadratic equation with ax2 + bx + c = 0, we get a = 1, b = 5 and c = -1800 Find the discriminant d = b Since, d > 0. Therefore, two real roots α and β exist for the given equation.
Let the time taken by the tap of larger diameter to fill the tank (in hours) = x Then the time taken by the tap of smaller diameter to fill the tank (in hours) = x + 10 Amount of water filled in the tank by the tap of smaller diameter in an hour = 1/(x + 10) Amount of water filled in the tank by the tap of larger diameter in an hour = 1/x It is given that if the two taps can fill the tank in hours = 75/8 together. Therefore On comparing the given quadratic equation with ax a = 4, b = -35 and c = -375 Find the discriminant d = b Since, d > 0. Therefore, two real roots α and β exist for the given equation. But time taken cannot be negative, so x = -25/4 is rejected. Therefore x = 15 Hence, time taken by tap of larger diameter = 15 hrs and time taken by tap of smaller diameter = 15 + 10 = 25 hrs.
Let the average speed of the passenger train (in km/h) = x. Then the average speed of the express train (in km/h) = x + 11 Time taken by passenger train to cover 132 km = 132/x Time taken by express train to cover 132 km = 132/(x + 11) It is given that express train takes 1 hour less to complete the journey. Therefore On comparing the given quadratic equation with ax2 + bx + c = 0, we get a = 1, b = 11 and c = -1452 Find the discriminant d = b Since, d > 0. Therefore, two real roots α and β exist for the given equation. But the speed of the trains cannot be negative, so x = -44 is rejected. Hence, average speed of passenger train = 33 km/h and average speed of express train = 33 + 11 = 44 km/h
Let the length of side of larger square (in m) = x Perimeter of smaller square = 4x - 24 Then the length of side of smaller square = (4x - 24)/4 = x - 6 The sum of the area of both of the squares is given to be 468 m x On comparing the given quadratic equation with ax2 + bx + c = 0, we get a = 1, b = -6 and c = -198 Find the discriminant Since, d > 0. Therefore, two real roots α and β exist for the given equation. But, length cannot be negative, so x = -4 is rejected. Hence, side of the larger square = 18 m, and side of the smaller square = x - 6 = 18 - 6 = 12 m. ## Exercise 4.4
**2x**^{2}- 3x + 5 = 0**3x**^{2}- 4 √ 3 x + 4 = 0**2x**^{2}- 6x + 3 = 0
I. On comparing the given quadratic equation with ax a = 2, b = -3 and c = 5 Find the discriminant d =? b? Since, d < 0. The roots for the given equation will be imaginary. II. On comparing the given quadratic equation with ax2 + bx + c = 0, we get a = 3, b = -4√3 and c = 4 Find the discriminant d = b Since, d = 0. The roots for the given equation will be real and equal.
a = 2, b = -6 and c = 3 Find the discriminant Since, d > 0. The roots for the given equation will be real.
- 2x
^{2}+ kx + 3 = 0 - kx (x - 2) + 6 = 0
I. Since, it is given that the given equation has equal roots. Therefore, discriminant will be 0. On comparing the given quadratic equation with ax a = 2, b = k and c = 3 II. kx(x - 2) + 6 = 0 kx Since, it is given that the given equation has equal roots. Therefore, discriminant will be 0. On comparing the given quadratic equation with ax a = k, b = -2k and c = 6 d = b
Let the required rectangular mango grove have length of x m. Then its breadth = x/2 m. Area of rectangle = length × breadth But length cannot be negative, so x = -40 is rejected. Hence, the length of rectangular mango grove = 40 m, and breadth of rectangular mango grove = 40/2 = 20 m.
Let the present age of one friend (in years) = x. Then the present age of the other friend = 20 - x. The ages of the two friends four years ago would be (x - 4) and (20 - x - 4) respectively. The product of their ages 4 years ago is given to be 48. Therefore (x - 4)(16 - x) = 48 On comparing the given quadratic equation with ax2 + bx + c = 0, we get a = 1, b = -20 and c = 112 Find the discriminant d = b Since, d < 0. The roots for the given equation do not exist. Hence, the given situation is not possible.
Let the length of the rectangular park = x m. Perimeter of rectangular park = 80 80 = 2(length) + 2(breadth) 80 - 2x = 2(breadth) Then, breadth of rectangular field = (80 - 2x)/2 = 40 - x Area = Length × Breadth 400 = x(40 - x) On comparing the given quadratic equation with ax2 + bx + c = 0, we get a = 1, b = -40 and c = 400 Find the discriminant d = b Since, d = 0. Therefore, roots are real and equal for the given equation. Hence, length of the rectangular park = 20 m. Breadth of the rectangular park = 40 - x = 40 - 20 = 20 m. Next TopicClass 10 Maths Chapter 5 |