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NCERT Solutions for Class 10th Maths: Chapter 4 - Quadratic Equations




In this page we have NCERT Solutions for Class 10th Maths: Chapter 4 - Quadratic Equations for EXERCISE 4.1 on page number 73 and 74. Hope you like them and do not forget to like , social_share and comment at the end of the page.
Question 1
Check whether the following are quadratic equations
  1. (x+1)2=2(x-3)
  2. x2 -2x=(-2)(3-x)
  3. (x-2)(x+1)=(x-1)(x+3)
  4. (x – 3)(2x +1) = x(x + 5)
  5. (2x – 1)(x – 3) = (x + 5)(x – 1)
  6.  x2 + 3x + 1 = (x – 2)2
  7. (x + 2)3 = 2x (x2 – 1)
  8. x3 – 4x2x + 1 = (x – 2)3
Solution 1
we know that
Quadratic equation
ax2 +bx+c   =0     where a≠0
  1. (x+1)2=2(x-3) We know that
      (a+b)2=a2+b2+2ab
    ⇒x2 + 2x+1=2x-6
    Simplifying it
    ⇒;x2 +7=0
    Since it is of a quadratic form : ax2 +bx+c   =0     where a≠0
    with b=0
    So it is a quadratic equation
  2.  x2 -2x=(-2)(3-x)
    Simplifying it
    x2 -2x=-6+2x
    ⇒x2 -4x+6=0
    Since it is a quadratic form
    ax2 +bx+c   =0     where a≠0
    So it is a quadratic equation
  3.  x2 -2x=(-2)(3-x)
    (x-2)(x+1)=(x-1)(x+3)
    Multiplying both the factors
    ⇒x2 -2x+2+x= x2 +3x-x-3
    Simplifying
    -3x+1=0
    It is not of the quadratic form
    ax2 +bx+c   =0     where a≠0
    So it is not a quadratic equation
  4. (x – 3)(2x +1) = x(x + 5)
    Multiplying both the factors
    2x2+x-6x-3=x2+5x
    Simplifying
    x2 -10x-3=0
    Since it is a quadratic form
    ax2 +bx+c   =0     where a≠0
    ⇒So it is a quadratic equation
  5. (2x – 1)(x – 3) = (x + 5)(x – 1)
    Multiplying both the factors on both sides
    2x2 -6x-x+3=x2 –x+5x-5
    ⇒x2 -11x +8=0
    Since it is a quadratic form
    ax2 +bx+c   =0     where a≠0
    So it is a quadratic equation
  6. x2 + 3x + 1 = (x – 2)2
    We know that
    (a+b)2=a2+b2+2ab
    x2 + 3x + 1 =x2-4x+4
    7x-3=0
    Since it is not of quadratic form
    ax2 +bx+c   =0     where a≠0
    So it is a not quadratic equation
  7. (x + 2)3 = 2x (x2 – 1)
    Important formula you must have remembered in old classes
    (a+b)3= a3 +b3+3ab2+3a2b
    ⇒x3 +8+6x2+12x=2x3 -2x
    Simplifying
    x3-6x2 -14x-8=0
    Since it is not of quadratic form
    ax2 +bx+c   =0     where a≠0
    So it is a not quadratic equation
  8. x3 – 4x2x + 1 = (x – 2)3
    Important formula  you must have remembered in  old classes
    (a-b)3= a3 -b3+3ab2-3a2b
    x3 – 4x2x + 1 =x3-8-6x2+12x
    Simplifying
    2x2 -13x+9=0
    Since it is a quadratic form
    ax2 +bx+c   =0     where a≠0
    So it is a quadratic equation


Question 2
Represent the following situations in the form of quadratic equations :
  1. The area of a rectangular plot is 528 m2. The length of the plot (in meters) is one more than twice its breadth. We need to find the length and breadth of the plot.
  2. The product of two consecutive positive integers is 306. We need to find the Integers.
  3. 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.
  4. 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
Solution
  1. Let the breath of the plot= x  m
    As per given condition in the question
    Length  =2x+1
    Now we know that Area is given by
    A=LB
    A=528 m2
    So
    528=(2x+1)x
    ⇒ x2+2x-528=0
    Which is a quadratic equation
  2. let the two consecutive positive integers are x and x+1
    The product of these would be
    x(x+1)
    It is given that product is 306
    So
    ⇒ x(x+1)=306
    ⇒ x2+x-306=0
    ⇒ Which is a quadratic equation
  3. Let Rohan present age=x year
    Then Rohan Mother present age would =x+26
    After 3 year,
    Rohan age would be =x+3
    Rohan mother’s age would be =x+26+3=x+29
    According to question, The product of their ages (in years) =360
    Then
    (x+3)(x+29)= 360
    Simplifying
    ⇒ x2 +29x +3x+87=360
    ⇒ x2 +32x -273=0
    Which is a quadratic equation
  4. Let the speed of the train is x km/hr
    Now distance travelled by the train=480 km
    Few important formula here
    Speed=Distance/time
    Or Time= Distance /Speed
    Case I
    Time taken to travel 480 km by train will be =480/x
    Case II
    Now the speed of the train is reduced by 8 km/hr,
    So speed would (x-8)
    Now Time taken to travel 480 km will be =480/x-8
    Now as per the question
    480/(x-8) -  480/x  =3
    ⇒ [480x-480(x-8)]/x(x-8)  =3
    ⇒ 480x-480x+3840=3x(x-8)
    ⇒ 3x2 -24x-3840=0
    ⇒ Which is a quadratic equation

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Reference Books for class 10

Given below are the links of some of the reference books for class 10 math.

  1. Oswaal CBSE Question Bank Class 10 Hindi B, English Communication Science, Social Science & Maths (Set of 5 Books)
  2. Mathematics for Class 10 by R D Sharma
  3. Pearson IIT Foundation Maths Class 10
  4. Secondary School Mathematics for Class 10
  5. Xam Idea Complete Course Mathematics Class 10

You can use above books for extra knowledge and practicing different questions.


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Class 10 Maths Class 10 Science

Practice Question

Question 1 What is $1 - \sqrt {3}$ ?
A) Non terminating repeating
B) Non terminating non repeating
C) Terminating
D) None of the above
Question 2 The volume of the largest right circular cone that can be cut out from a cube of edge 4.2 cm is?
A) 19.4 cm3
B) 12 cm3
C) 78.6 cm3
D) 58.2 cm3
Question 3 The sum of the first three terms of an AP is 33. If the product of the first and the third term exceeds the second term by 29, the AP is ?
A) 2 ,21,11
B) 1,10,19
C) -1 ,8,17
D) 2 ,11,20