NCERT Solutions For Class 10 Maths Chapter                Quadratic Equations Ex 4.1

Class 10 Maths Chapter 4 Quadratic Equations - Comprehensive NCERT Solutions

Welcome to the comprehensive NCERT Solutions for Class 10 Maths Chapter 4 Quadratic Equations. This collection of solutions encompasses all the questions, answers, images, and step-by-step explanations found in Chapter 4 of the Class 10 Mathematics textbook.

If you are a Class 10 student utilizing the NCERT Textbook to study Mathematics, then you will undoubtedly encounter Chapter 4 Quadratic Equations. Upon completing this lesson, it is natural to seek answers to the associated textbook questions. Rest assured, as here you can find the complete set of NCERT Solutions for Class 10 Maths Chapter 4 Quadratic Equations conveniently compiled in one place.

NCERT Solutions For Class 10 Maths Chapter 4 Quadratic Equations Ex 4.1

The NCERT Solutions for Exercise 4.1 of Chapter 4, Quadratic Equations in Class 10 Maths, are included as part of the NCERT Solutions for Class 10 Maths. Below you will find the NCERT Solutions specifically dedicated to Class 10 Maths Chapter 4, Quadratic Equations Exercise 4.1.

BoardCBSE
TextbookNCERT
ClassClass 10
SubjectMaths
ChapterChapter 4
Chapter NameQuadratic Equations
ExerciseEx 4.1
Number of Questions Solved2
CategoryNCERT Solutions



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 – × + 1 = (x – 2)3
Sol. (i) (x + 1)2 = 2(x – 3)
              We have:
              (x + 1)2 = 2 (x – 3) x2 + 2x + 1 = 2x – 6
              ⇒ x2 + 2x + 1 – 2x + 6 = 0
              ⇒ x2 + 70
              Since x2 + 7 is a quadratic polynomial
              ∴ (x + 1)2 = 2(x – 3) is a quadratic equation.
         (ii) x2– 2x = (–2) (3 – x)
               We have:
               x2 – 2x = (– 2) (3 – x)
               ⇒ x2 – 2x = –6 + 2x
               ⇒ x2 – 2x – 2x + 6 = 0
               ⇒ x2 – 4x + 6 = 0
               Since x2 – 4x + 6 is a quadratic polynomial
               ∴ x2 – 2x = (–2) (3 – x) is a quadratic equation.
         (iii) (x – 2) (x + 1) = (x – 1) (x + 3)
                 We have:
                 (x – 2) (x + 1) = (x – 1) (x + 3)
                 ⇒ x2 – x – 2 = x2 + 2x – 3
                 ⇒ x2 – x – 2 – x2 – 2x + 3 = 0
                 ⇒ –3x + 1 = 0
                 Since –3x + 1 is a linear polynomial
                 ∴ (x – 2) (x + 1) = (x – 1) (x + 3) is not quadratic equation.
         (iv) (x – 3) (2x + 1) = x(x + 5)
                 We have:
                 (x – 3) (2x + 1) = x(x + 5)
                 ⇒ 2x2 + x – 6x – 3 = x2 + 5x
                 ⇒ 2x2 – 5x – 3 – x2 – 5x – 0
                 ⇒ x2 + 10x – 3 = 0
                 Since x2 + 10x – 3 is a quadratic polynomial
                 ∴ (x – 3) (2x + 1) = x(x + 5) is a quadratic equation.
         (v) (2x – 1) (x – 3) = (x + 5) (x – 1)
                 We have:
                 (2x – 1) (x – 3) = (x + 5) (x – 1)
                 ⇒ 2x2 – 6x – x + 3 = x2 – x + 5x – 5
                 ⇒ 2x2 – x2 – 6x – x + x – 5x + 3 + 5 = 0
                 ⇒ x2 – 11x + 8 = 0
                 Since x2 – 11x + 8 is a quadratic polynomial
                 ∴ (2x – 1) (x – 3) = (x + 5) (x – 1) is a quadratic equation.
         (vi) x2 + 3x + 1 = (x – 2)2
                 We have:
                 x2 + 3x + 1 = (x – 2)2
                 ⇒ x2 + 3x + 1 = x2 – 4x + 4
                 ⇒ x2 + 3x + 1 – x2 + 4x – 4 =0
                 ⇒ 7x – 3 = 0
                 Since 7x – 3 is a linear polynomial.
                 ∴ x2 + 3x + 1 = (x – 2)2 is not a quadratic equation.
         (vii) (x + 2)3 = 2x(x2 – 1)
                  We have:
                  (x + 2)3 = 2x(x2 – 1)
                  x3 + 3x2(2) + 3x(2)2 + (2)3 = 2x3 – 2x
                  ⇒ x3 + 6x2 + 12x + 8 = 2x3 – 2x
                  ⇒ x3 + 6x2 + 12x + 8 – 2x3 + 2x = 0
                  ⇒ –x3 + 6x2 + 14x + 8 = 0
                  Since –x3 + 6x2 + 14x + 8 is a polynomial of degree 3
                  ∴ (x + 2)3 = 2x(x2 – 1) is not a quadratic equation.
         (viii) x3 – 4x2 – x + 1 = (x – 2)3
                   We have:
                   x3 – 4x2 – x + 1 = (x – 2)3
                   ⇒ x3 – 4x2 – x + 1 = x3 + 3x2(– 2) + 3x(– 2)2 + (– 2)3
                   ⇒ x3 – 4x2 – x + 1 = x3 – 6x2 + 12x – 8
                   ⇒ x3 – 4x2 – x – 1 – x3 + 6x2 – 12x + 8 = 0
                   2x2 – 13x + 9 = 0
                   Since 2x2 – 13x + 9 is a quadratic polynomial
                   ∴ x3 – 4x2 – x + 1 = (x – 2)3 is a quadratic equation.
Q.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 3 hours more to cover the same distance. We need to find the speed of the train.
Sol. (i) Let the breadth = x metres
              Length = 2 (Breadth) + 1
              Length = (2x + 1) metres
              Since Length × Breadth = Area
              ∴ (2x + 1) × x = 528
              2x2 + x = 528
              2x2 + x – 528 = 0
              Thus, the required quadratic equation is
              2x2 + x – 528 = 0
         (ii) Let the two consecutive numbers be x and (x + 1).
               ∵ Product of the numbers = 306
               ∴ x (x + 1) = 306
               ⇒ x2 + x = 306
               ⇒ x2 + x – 306 = 0
               Thus, the required equdratic equation is
               x2 + x – 306 = 0
         (iii) Let the present age = x
                 ∴ Mother’s age = (x + 26) years
                 After 3 years
                 His age = (x + 3) years
                 Mother’s age = [(x + 26) + 3] years
                 = (x + 29) years
                 According to the condition,
                 
                 ⇒ (x + 3) × (x + 29) = 360
                 ⇒ x2 + 29x + 3x + 87 = 360
                 ⇒ x2 + 29x + 3x + 87 – 360 = 0
                 ⇒ x2 + 32x – 273 = 0
                 Thus, the required quadratic equation is
                 x2 + 32x – 273 = 0
         (iv) Let the speed of the tram = u km/hr
                 Distance covered = 480 km
                 Time taken = Distance + Speed
                 = (480 ÷ u) hours
                 
                 In second case,
                 Speed = (u – 8) km/ hour
                 
                 According to the condition,
                 
                 ⇒ 480u – 480(u – 8) = 3u(u – 8)
                 ⇒ 480u – 480u + 3840 = 3u2 – 24u
                 ⇒ 3840 – 3u2 + 24u = 0
                 ⇒ 1280 – u2 + 8u = 0
                 ⇒ –1280 + u2 – 8u = 0
                 ⇒ u2 – 8u – 1280 = 0
                 Thus, the required quadratic equation is
                 u2 – 8u – 1280 = 0