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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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