NCERT Solutions For Class 10 Maths Chapter 4 Quadratic Equations Ex 4.3


Access complimentary NCERT Solutions for Class 10 Mathematics Chapter 4, Exercise 4.3 on Quadratic Equations. These solutions are a valuable aid when completing homework assignments. Our team of knowledgeable educators at School Notes Wallah has diligently prepared the Class 10 Mathematics NCERT Solutions for Exercise 4.3. You will find comprehensive and insightful explanations to all the questions in Chapter 4 of the NCERT textbook, focusing on Quadratic Equations.

Topics and Sub Topics in Class 10 Maths Chapter 4 Quadratic Equations:

Section NameTopic Name
4Quadratic Equations
4.1Introduction
4.2Quadratic Equations
4.3Solution of a Quadratic Equation by Factorisation
4.4Solution of a Quadratic Equation by Completing the Square
4.5Nature of Roots
4.6Summary

You can also download the free PDF of Chapter 4 Ex 4.3 Quadratic Equations NCERT Solutions or save the solution images and take the print out to keep it handy for your exam preparation.

BoardCBSE
TextbookNCERT
ClassClass 10
SubjectMaths
ChapterChapter 4
Chapter NameQuadratic Equations
ExerciseEx 4.3
Number of Questions Solved11
CategoryNCERT Solutions

NCERT Solutions For Class 10 Maths Chapter 4 Quadratic Equations Ex 4.3

NCERT Solutions for Class 10 Maths Chapter 4 Quadratic Equations Ex 4.3 are part of NCERT Solutions for Class 10 Maths. Here we have given NCERT Solutions for Class 10 Maths Chapter 4 Quadratic Equations Ex 4.3.

             Exercise 4.3
Q.1.    Find the rootsof 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+ 3 = 0
(iv) 2x2 + x + 4 = 0
Sol.     (i) 2x2 – 7x + 3 = 0
                 
                 
         (ii) 2x2 + x – 4 = 0
                   We have:
                   2x2 + x – 4 = 0
                   Dividing throughout by 2,
                   
                   
         
         
Q.2.    Find the roots of the following quadratic equations, using the quadratic formula:
              (i) 2x2 – 7x + 3 = 0
              (ii) 2x2 + x – 4 = 0
              (iii) 
              (iv) 2x2 + x + 4 = 0
Sol.       (i) 2x2 – 7x + 3 = 0
                   Comparing the given equation with ax2 + bx + c = 0, we have
                   a = 2
                   b = –7
                   c = 3
                   ∴ b2 – 4ac = (–7)2 – 4 (2)(3)
                   = 49 – 24 = 25 0
                   Since b2 – 4ac > 0
                   ∴ The given equation has real roots. The roots are given by
                   
                   Taking +ve sign,
                   
                   Taking –ve sign,
                   
                   Thus, the roots of the given equation are
                   
         (ii) 2x2 + x – 4 = 0
               Comparing the given equation with ax2 + bx + c = 0 we have:
               a = 2
               b = 1
               c = –4
               ∴ b2 – 4ac = (1)2 – 4(2)(–4)
               = 1 + 32
               = 33 > 0
               Since b2 – 4ac > 0
               ∴ The given equation has equal roots. The roots are given by
                   
         
                   Since b2 – 4ac = 0
                   ∴ The given equation has real roots, which are given by:
                   
         (iv) 2x2 + x + 4 = 0
                   Comparing the given equation with ax2 + bx + c = 0 we have:
                   a = 2
                   b = 1
                   c = 4
                   ∴ b2 – 4ac = (1)2 – 4(2)(4)
                   = 1 – 32
                   = –31 > 0
                   Since b2 – 4ac is less than 0, therefore the given equation does not have real roots.
Q.3.    Find the roots of the following equations:
         
                           
                    
Q.4.    The sum of the reciprocals of Rehman’s ages, (in years) 3 years ago and 5 years from now is 
              Find his present age.
Sol.      Let the present age of Rehman = x
             ∴ 3 years ago Rehman’s age = (x – 3) years
             5 years later Rehman’s age = (x + 5) years
             Now, according to the condition,
                     
             ⇒ 3 [x + 5 + x – 3] = (x – 3) (x + 5)
             ⇒ 3[2x + 2] = x2 + 2x – 15
             ⇒ 6x + 6 = x2 + 2x – 15
             ⇒ x2 + 2x – 6x – 15 – 6 = 0
             ⇒ x2 – 4x – 21 = 0
             Now, comparing (1) with ax2 + bx + c = 0, we have:
             a = 1
             b = – 4
             c = – 21
             b2 – 4ac = ( – 4)2 – 4 (1) ( – 21)
             = 16 + 84
             = 100
             Since,
             
Q.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.
Sol.      Let, Shefali’s marks in Mathematics = x
             ∴ Marks in English = (30 – x)                                                                 [∵Sum of their marks in Eng. and Maths = 30]
             Now, according to the condition,
             (x + 2) × [(30 – x) – 3] = 210
             ⇒ (x + 2) × (30 – × – 3) = 210
             ⇒(x + 2) ( – x + 27) 210
             ⇒ – x2 + 25x + 54 = 210
             ⇒ – x2 + 25x + 54 – 210 = 0
             ⇒ – x2 + 25x – 156 = 0
             ⇒ x2 – 25x + 156 = 0 ...(1)
             Now, comparing (1) with ax2 + bx + c = 0
                   a = 1
                   b = – 25
                   c = 156
                   b2 – 4ac = ( – 25)2 – 4(1) (156)
                   = 625 – 624 = 1
             Since,
                   
                   When x = 13, then 30 – 13 = 17
                   When x = 12, then 30 – 12 = 18
                   Thus, marks in Maths = 13, marks in English = 17
                   marks in Maths = 12, marks in English = 18
Q.6.    The diagonal of a rectangular field is 60 metres more than the shorter 30 metres more than the shorter side, find the sides of the field.
Sol.      Let the shorter side (i.e., breadth) = x metres.
             ∴ The longer side (length) = (x + 30) metres.
             In a rectangle,
                   
             ⇒ (x + 60)2 = 2x2 + 60x + 900
             ⇒ x2 + 120x + 3600 = 2x2 + 60x + 900
             ⇒ 2x2 – x2 + 60x – 120x + 900 – 3600 = 0
             ⇒ x2 – 60x – 2700 = 0
             Comparing (1) with ax2 + bx + c = 0
             a = 1
             b= –60
             c = –2700
             ∴ b2 – 4ac = (–60)2 – 4(1) (–2700)
             ⇒ b2 – 4ac = 3600 + 10800
             ⇒ b2 – 4ac = 14400
             Since,
             
                   Since breadth cannot be negative,
                   ∴ x ≠ – 30 ⇒ x = 90
                   ∴ x + 30 = 90 + 30=120
                   Thus, the shorter side = 90 m
                   The longer side = 120 m.
Q.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.
Sol.      Let the larger number be x. Since,
             (smaller number)2 = 8 (larger number)
             ⇒ (smaller number)2 = 8x
             
             
Q.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.
Sol.      Let the uniform speed of the train be x km/hr.
             
             When speed is 5 km/hr more then time is 1 hour less.
             
             
Q.9.    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.
Sol.      Let the smaller tap fills the tank in × hours.
             ∴ The larger tap fills the tank (x – 10) hours.
             
             
Q.10.    An express train takes 1 hour less than a passenger train to travel 132 km between Mysore a Bangalore (without taking into consideration the time they stop at intermediate stations). If the average speed of the express train is 11 km/h more than that of the passenger train, find average speed of the two trains.
Sol.      Let the average speed of the passenger train be x km/ hr
             ∴ Average speed of the express train = (x + 11) km/hr
             Total distance covered = 132 km
             
             According to the condition, we get
             
             But average speed cannot be negative
             ∴ x – 44
             ∴ x = 33
             ⇒ Average speed of the passenger train = 33 kin/hr
             ∴ Average speed of the express train = (x + 11) km/hr
             = (33 + 11) km/hr
             = 44 km/hr

Q.11.   Sum of the areas of two squares is 468 m2. If the difference of their perimeters is 24 m, find sides of the two squares.
Sol.        Let the side of the smaller square be × m.
             ⇒ Perimeter of the smaller square = 4x m.
             ∴ Perimeter of the larger square = (4x + 24) m
             
             ∴ Area of the smaller square = (side)2 = (x)2 m2
             Area of the larger square = (x + 6)2 m2
             According to the condition,
             x2 + (x + 6)2 = 468
             ⇒ x2 + x2 + 12x + 36 = 468
             ⇒ 2x2 + 12x – 432 = 0
             ⇒ x2 + 6x – 216 = 0
             Comparing (1) with ax2 + bx + c = 0
             ∴ a = 1
             b = 6
             c = – 216
             ∴ b2 – 4ac = (6)2 – 4 (1) (–216)
             = 36 + 864 = 900
             Since,
             
             But the length of a square cannot be negative
             ∴ x = 12
             ⇒ Length of the smaller square = 12 m
             Thus, the length of the larger square = x + 6
             = 12 + 6
             = 18m