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- WORKSHEET 1 - Quadratic Equations

1.  Check whether the following are quadratic equations:
  • Solution:(i) (x + 1)2 = 2(x – 3)
    x2 + 2x + 1 = 2x – 6
    x2 + 7 = 0
    This is of the form ax2 + bx + c = 0
    \ It is a quadratic equation.
    (ii) x2 – 2x = -2(3 – x)
    x2 – 2x = -6 + 2x
    x2 – 2x -2x + 6 = 0
    x2 – 4x + 6. This is of the form ax2 + bx + c = 0
    It is a quadratic equation.
    (iii) (x - 2)(x + 1) = (x - 1)(x + 3)
    x2 – 2x + x - 2 = x2 + 3x – x - 3
    – x – 2x +1 = 0
    - 3x + 1 = 0
    It is not of the form ax2 + bx + c
    It is not a quadratic equation.
 2.  Check whether the following are quadratic equations:
(i) (x – 3) (2x + 1) = x (x + 5)
(ii) (2x – 1) (x – 3) = (x + 5) (x – 1) (iii) x2 + 3x + 1 = (x – 2)2
  • Solution:(i)(x – 3) (2x + 1) = x (x + 5)
    2x2 + x – 6x – 3 = x2 + 5x
    x2 - 3 = 0
    This is of the form ax2 + bx + c = 0
    Hence it is a quadratic equation.
    (ii)(2x – 1) (x – 3) = (x + 5) (x – 1)
    2x2 – 6x – x + 3 = x2 – x + 5x – 5
    2x2 – 7x + 3 - x2 –4x + 5 = 0
    x2 – 11x + 8 = 0
    This is of the form ax2 + bx + x = 0
    Hence it is a quadratic equation.
    (iii) x2 + 3x + 1 = (x – 2)2
    x2 + 3x + 1 = x2 – 4x +4
    7x – 3 = 0
    This is not of the form ax2 + bx + c = 0
    \ It is not a quadratic equation.
    (iv) (x+2)3 = 2x(x2 – 1)
    x2 + 23 + 3x2 ´ 2 + 3x ´ 4 = 2x3 – 2x
    x3 + 8 + 6x2 + 12x = 2x3 – 2x
    x3 – 6x2 – 14x – 8 = 0
    Its not of the form ax2 + bx + c = 0
    It is not a quadratic equation.
    (v) x3 – 4x2 – x + 1 = (x – 2)3
    x3 – 4x2 – x + 1 = x3 – 8 - 3x2 ´ 2 + 3x ´ 22
    x3 – 4x2 – x + 1 = x3 – 8 – 6x2 + 12x
    2x2 –13x + 1 = 0
    This of the form ax2 + bx + c = 0
    Hence it is a quadratic equation.
 3.  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.
  • Solution:(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.
    Let l be length (in metres) and b be the breadth (in metres) of the rectangle.
    Given
    l = 1 + 2b …………………(1)
    lb = 528 m2 ----- (2)
    (Area of rectangle = length ´ breadth)
    Substituting (1) in (2)
    (1+2b)b = 528
    b +2b2l = 528
    2b2 + b – 528 = 0
    This is the quadratic equation where the breadth is in metres.
    (ii) The product of two consecutive positive integers is 306. We need need to find the integers
    Let the 2 consecutive positive integer be x, x + 1
    x(x + 1) = 306
    x2 + x - 306 = 0 is the quadratic equation where x is the smallest
    positive integer
 4.  Represent the following situations in the form of quadratic equations:
(i) 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.
  • Solution:(i) 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.
    Let the Rohans present age be x
    Rohans mothers present age = x + 26
    After 3 years,
    Rohans age = x + 3
    Rohans mothers age = x + 26 + 3
    = x + 29
    (x + 3)(x + 29) = 360 (given)
    x2 + 3x + 29x + 87 = 360
    x2 + 32x – 273 = 0 where x is Rohans present age (in years)
    (ii) 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.
    Let the speed of the train be x km / hr
    \ Time taken = 
    If speed = (x – 8) km/ h
    Time taken for the same distance =  hr
    Given 9;  -  = 3
    480x – 480 (x – 8) = 3(x – 8)x
    480x – 480x + 480 x 8 = 3x2 – 24x
    3x2 – 24x – 3840 = 0
    x2 – 8x - 1280 = 0 is the required quadratic equation where x is the speed of the train in km / hr.
 5.  Find the roots of the following quadratic equations by factorization:
(i) x2 – 3x – 10 = 0 (ii) 2x2 + x – 6 = 0
(iii) + 7x + 5 = 0 (iv) 2x2 – x +  = 0
  • Solution:(i)x2 – 3x – 10 = 0
    x2 – 5x + 2x - 10 = 0
    x(x – 5) + 2(x – 5) = 0
    (x – 5)(x + 2) = 0
    \ x = -2, x = 5
    Thus –2, 5 are the roots of the equation.
    (ii) 2x2 + x – 6 = 0
    2x2 + 4x – 3x – 6 = 0
    2x(x + 2) – 3(x + 2) = 0
    (x + 2) (2x - 3) = 0
    Þ x = -2 (or) x =  are the roots of the equation
    (iii) + 7x + 5 = 0
     x2 + 5x + 2x + 5 = 0
    x() + (x + 5) = 0
    (x + )(x + 5) = 0
    \ x = -, x = -
    \ The roots are -, -.
    (iv) 2x2 – x +  = 0
    16x2 – 8x + 1 = 0
    16x2 – 4x – 4x + 1 = 0
    4x(4x – 1) – (4x – 1) = 0
    (4x –1)2 = 0
    x = , 
    \ The root of the equation are , .
    (v) 100x2 – 20x + 1 = 0
    100x2 – 10x – 10x + 1 = 0
    10x(10x – 1) –1 (10x – 1) = 0
    (10x – 1)2 = 0
    x = , 
    \ The roots of the equation are, .
 6.  Solve the following situation mathematically:
(i) 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.
  • Solution:(i) Let the number of marbles John had be x
    Then the number of marbles Jivanti had = 45 – x.
    The number of marbles left with John, when he lost 5 marbles = x – 5
    The number of marbles left with Jivanti, when she lost 5 marbles = 45 – x – 5
                                                                                               = 40 – x
    Therefore, their product = (x – 5)(40 – x)
                                     =40x – x2 – 200 + 5x
                                     = -x2 + 45x – 200
    So, -x2 + 45x – 200 = 124
    i.e., -x2 + 45x – 324 = 0
    i.e., x2 - 45x + 324 = 0
    Therefore, the number of marbles John had, satisfies the quadratic equation
    x2 – 45x + 324 = 0
    Which is the required representation of the problem mathematically.
    Now by solving we get, x2 – 9x – 36 x + 324 = 0
    x(x – 9) - 36(x – 9) = 0
    (x - 9)(x - 36) = 0
    \ x = 9 (or) 36.
    \ The number of marbles John had is 9 (or) 36.
    (ii) Let the number of toys produced on that day be x.
    Therefore, the cost of production (in rupees) of each toy that day = 55 – x
    So, the total cost of production (in rupees) that day = x(55 – x)
    Therefore, x(55 – x) = 750
    i.e., 55x – x2 = 750
    i.e., -x2 + 55x – 750 = 0
    i.e., x2 - 55x + 750 = 0
    Therefore, the number of toys produced that day satisfies the quadratic equation
    x2 – 55x + 750 = 0
    which is the required representation of the problem mathematically.
    Now by solving we get, x2 – 25x – 30x + 750 = 0
    x(x – 25)- 30(x – 25) = 0
    (x - 25)(x - 30) = 0
    x = 25, x = 30.
    The number of toys produced in a day = 25 (or) 30.
 7.  Find two numbers whose sum is 27 and product is 182.
  • Solution:Let the two numbers be x and y
    x + y = 27 …………………….. (1)
    xy = 182 …………………….. (2)
    from (1) x = 27 – y
    Substituting in (2) we get,
    (27 – y)y = 182
    27y – y2 = 182
    y2 – 27y + 182 = 0
    y2 – 13y - 14y + 182 = 0
    y(y – 13) – 14(y – 13) = 0
    (y – 14)(y - 13) = 0
    y = 14 (or) y = 13
    If y = 14, x = 27 – 14
    = 13
    If y = 13, x = 27 – 13
    = 14
    Hence the two numbers are 13 and 14.
 8.  Find two consecutive positive integers, sum of whose squares is 365.
  • Solution:Let the 2 consecutive positive nos be x, x+1
    x2 + (x + 1)2 = 365
    x2 + x2 + 2x + 1 = 365
    2x2 + 2x – 364 = 0
    x 2 + x – 182 = 0 (Dividing by 2)
    Þ x2 + 14x – 13x – 182 = 0
    x(x + 14) – 13(x + 14) = 0
    (x – 13)(x + 14) = 0
    Þ x = 13 (or) x = -14.
    Since the positive integers cannot be -14
    Hence the positive integers are 13 and 14.
 9.  The altitude of a right triangle is 7 cm less that its base. If the hypotenuse is 13 cm, find the other two sides.
  • Solution:In the right angled D
    Let (BC) the base be x cm and altitude (AB) be (x – 7) cm
    In the right angled D ,
    (AB)2 + (BC)2 = (AC)2 (Pythagoras theorem)
    (x - 7)2 + x2 = 132
    x2 – 14x + 49 + x2 = 169
    2x2 – 14x – 120 = 0
    x2 – 7x – 60 = 0 (Dividing by 2 )
    x2 – 12x +5x - 60 = 0
    x(x – 12) + 5(x – 12) = 0
    (x – 12)(x + 5) = 0
    => x = -5 (or) x = 12
    Since the side (base) cannot be negative, x ¹ -5
    We assume x = 12
    => Base (BC) = 12 cm
    Altitude = x – 7 = 5 cm
    Thus the two sides of the triangle are 5 cm and 12 cm.
 10.  A cottage industry produces a certain number of pottery articles in a day. It was observed on a particular day that the cost of production of each article(in rupees) was 3 more than twice the number of articles produced on that day. If the total cost of production on that day was Rs. 90, find the number of articles produced and the cost of each article.

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