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Quadratic Equations Test - 5

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Quadratic Equations Test - 5
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  • Question 1
    1 / -0
    The equation ax2 + bx + c = 0 has equal roots. Based on this information, which of the following is correct?
    Solution
    General rules:
    If b2 – 4ac = 0, then there are two equal real roots.
    If b2 – 4ac > 0, then there are two distinct real roots.
    If b2 – 4ac < 0, then there are no real roots.
    Hence, option 1 is the correct answer.
  • Question 2
    1 / -0
    If n = , then what is the value of n?
    Solution
    n =
    n =
    Squaring both sides:
    n2 = 6 + n
    n2 – n – 6 = 0
    n2 - 3n + 2n - 6 = 0
    n(n – 3) + 2(n – 3) = 0
    (n + 2) (n – 3) = 0
    This means that n = 3, as n ≠ -2.
  • Question 3
    1 / -0
    Which among the following is a root of the quadratic equation 17x2 – 153x + 340 = 0?
    Solution
    We have to find the roots of the quadratic equation 17(x2 – 9x + 20) = 0, which can also be written as x2 - 9x + 20 = 0.
    Comparing it with ax2 + bx + c = 0, we get
    a = 1, b = –9 and c = 20

    x =

    =

    =

    =

    x = 5, 4
    So, x = 5 and x = 4 are the roots of the given equation.
  • Question 4
    1 / -0
    How many positive integers will 'k' take, such that the roots of the quadratic equation 2x2 - 13x + k = 0 are real?
    Solution
    Since the roots are real, therefore b2 - 4ac ≥ 0
    169 - 8k ≥ 0.
    8k ≤ 169
    k ≤ 21.1
    So, 'k' will take 21 positive integers (from 1 to 21).
  • Question 5
    1 / -0
    If 3 is a real root of the equation x2 + bx – 15 = 0 and equation x2 + bx + q = 0 has equal roots, then what is the value of q?
    Solution
    x = 3 is a real root of the equation, so:
    x2 + bx – 15 = 0
    32 + b × 3 – 15 = 0
    9 + 3b – 15 = 0
    3b – 6 = 0
    3b = 6
    b = 2
    Now, the equation x2 + bx + q = 0 becomes x2 + 2x + q = 0 --- (I)
    Since eq (I) has equal roots:
    b2 – 4ac = 0
    22 – 4 × 1 × q = 0
    4 – 4q = 0
    -4q = -4
    q = 1
  • Question 6
    1 / -0
    What is the value of p - q, if the roots of the quadratic equation px2 - (p + q)x + q = 0 are reciprocals?
    Solution
    Suppose, the roots are α and .
    Product = = 1
    ⇒ q = p
    p - q = 0
  • Question 7
    1 / -0
    If the roots of the quadratic equation ax2 + bx + c = 0 are negative to each other, then
    Solution
    In this case, the sum of the roots is zero.
    ⇒ -
    c ≠ 0, and if c = 0, then one root must be zero and the other root may or may not be zero.
    For quadratic equation a ≠ 0
    So, option 3 is the correct answer.
  • Question 8
    1 / -0
    If one of the roots of 3x2 + ax + 36 = 0 is thrice the other, then the value of a is ___.
    Solution
    Given equation is:
    3x2 + ax + 36 = 0

    Let one root be x, then the other root would be 3x.
    Now, x × 3x = 12 (Product of roots = )
    3x2 = 12
    x2 =
    x2 = 4 ... eq (1)
    and x + 3x = (Sum of the roots = )
    4x =
    12x = -a
    Squaring both sides, we get
    144x2 = a2 … eq (2)
    Putting the value of eq. (1) in eq. (2):
    144 × 4 = a2
    a2 = 576
    a = + 24
    Hence, option 2 is the answer.
  • Question 9
    1 / -0
    The roots of the equation can be found by solving
    Solution


    It can also be written as:



    Squaring both sides above, we get:
    25x2 + 49 + 70x = 3x
    25x2 + 49 + 67x = 0
    25x2 + 67x + 49 = 0
  • Question 10
    1 / -0
    If the equation tx2 + sx + u = 0 has equal roots, then what is the value of u?
    Solution
    tx2 + sx + u = 0 has equal roots, so:
    Discriminant, D = b2 - 4ac = 0
    So, s2 - 4tu = 0
    s2 = 4 × t × u

    u =
  • Question 11
    1 / -0
    If α and β are the roots of the equation 3x2 + 4x + 7 = 0, then the value of is
    Solution
    Sum of roots,

    Product of roots,



    Hence, the desired answer is .
  • Question 12
    1 / -0
    What is the sum of the solutions of the equation 1 = ?
    Solution
    1 =
    1 =
    x(x + 3) = 6 (x + 3) - 8
    x2 + 3x = 6x + 18 - 8
    x2 - 3x - 10 = 0
    x2 - 5x + 2x - 10 = 0
    x(x - 5) + 2(x - 5) = 0
    (x + 2)(x - 5) = 0
    x = -2 or x = 5
    Thus, required sum = 3
    Hence, answer option 2 is correct.
  • Question 13
    1 / -0
    The product of two consecutive positive odd integers is 63, the smaller integer being x. Which of the following quadratic equations represents this situation?
    Solution
    The smaller of the two consecutive positive odd integers is x.
    Then, the second odd integer is x + 2.
    According to the question:
    x (x + 2) = 63
    x2 + 2x = 63 or x2 + 2x – 63 = 0 is the required quadratic equation.
  • Question 14
    1 / -0
    Which of the following is the solution set of the quadratic equation ?
    Solution

    (2y – 5)(y + 1) = (y – 1)(3y – 7)
    2y2 – 3y – 5 = 3y2 – 10y + 7
    y2 – 7y + 12 = 0
    y2 – 3y – 4y + 12 = 0
    (y – 3)(y – 4) = 0
    y = 3, 4
  • Question 15
    1 / -0
    A positive number is increased by 16 and then squared. The resulting number is 368 more than twice the square of the original number. Which among the following is possibly the original number?
    Solution
    Let the original number be x.
    Now, as per given information:
    (x + 16)2 = 2x2 + 368
    x2 + 256 + 32x = 2x2 + 368
    x2 - 32x + 112 = 0
    (x - 28)(x - 4) = 0
    x = 28 or 4
  • Question 16
    1 / -0
    In a right-angled triangle ABC, the length of the longest side is 10 cm. The perpendicular of this triangle is 2 cm more in length than its base. If x represents the length of the perpendicular, then which of the following quadratic equations represents this case?
    Solution
    According to the question: perpendicular (p) = x cm, Base (b) = (x – 2) cm, Hypotenuse (h) = 10 cm.
    Also, according to Pythagoras theorem:
    h2 = b2 + p2


    So, (10)2 = (x – 2)2 + x2
    100 = x2 – 2(2)x + 22 + x2
    2x2 – 4x + 4 – 100 = 0
    2x2 – 4x – 96 = 0
    x2 - 2x - 48 = 0
    Thus, x2 - 2x - 48 = 0 is the required equation.
  • Question 17
    1 / -0
    Rachel can cover a distance of 56 km at some constant speed in a certain time. If she increases her speed by 1 km/hr, she will be able to cover the same distance in an hour less. What would be the time taken by her to cover a distance of 104 km at the increased speed?
    Solution
    Let Rachel's original speed be x km/hr.
    Time taken to cover 56 km at x km/hr = t hours
    Condition 1:
    56 = x × t

     ... (1)

    Condition 2:
    56 = (x + 1) × (t – 1)
    56 = xt – x + t – 1 ... (2)
    Put the value of (1) in (2):
    56 = 56 - x + - 1
    -x2 + 56 - x = 0
    x2 + x - 56 = 0
    x² + 8x - 7x - 56 = 0
    x(x + 8) - 7(x + 8) = 0
    (x - 7) (x + 8) = 0
    x = 7, -8
    Since speed cannot be negative, the required speed is 7 km/hr.
    When the speed is increased by 1 km/hr, it becomes = 7 + 1 = 8 km/hr
    Time taken to cover 104 km at 8 km/hr = = 13 hours
  • Question 18
    1 / -0
    A boat covers 50 km upstream and 40 km downstream in total 7.5 hours. The speed of boat in still water is 15 km/hr. If the speed of stream is x km/hr, then which of the following quadratic equations correctly represents this case?
    Solution
    Let x km/hr be the speed of the stream.

    Time taken moving upstream = hours

    Time taken moving downstream = hours
    According to question:



    50(15 + x) + 40(15 - x) = 7.5[(15)2 - x2]
    750 + 50x + 600 - 40x = 7.5[225 - x2]
    1350 + 10x = 1687.5 - 7.5x2
    7.5x2 + 10x + 1350 - 1687.5 = 0
    7.5x2 + 10x - 337.5 = 0
    75x2 + 100x - 3375 = 0
    3x2 + 4x - 135 = 0
  • Question 19
    1 / -0
    There are two squares. Side of one square measures 4 cm more in length than that of the other. The sum of their areas is 400 cm2. Which among the following represents the area of the larger square?
    Solution
    Let the side of the smaller square be x cm.
    Then, length of each side of the larger square = (x + 4) cm
    According to problem:
    (x + 4)2 + x2 = 400 { Area of square = Side × Side}
    x2 + 8x + 16 + x2 = 400
    2x2 + 8x - 384 = 0
    x2 + 4x - 192 = 0
    (x + 16)(x - 12) = 0
    x = 12, x ≠ -16
    So, length of side of the larger square = (12 + 4) cm = 16 cm
    Area = (16)2 cm2 = 256 cm2
  • Question 20
    1 / -0
    A bought some notebooks. These notebooks cost him Rs. 300. After a few days, the cost of each notebook dropped by Rs. 5. At this price, he could buy 5 more notebooks for the same amount. What was the original cost of a notebook?
    Solution
    Let the cost of each notebook be Rs. x.
    Also, let the number of notebooks bought for Rs. 300 be y.

    So, y = … (1)

    When the cost of each notebook decreases by Rs. 5, 5 more notebooks could be bought.

    So, y + 5 =

    y = - 5 ... (2)

    Comparing both the equations:



    300(x - 5) = x(325 - 5x)
    300x - 1500 = 325x - 5x2
    5x2 - 25x - 1500 = 0
    x2 - 5x - 300 = 0
    x2 - 20x + 15x - 300 = 0
    x(x - 20) + 15(x - 20) = 0
    (x + 15) (x - 20) = 0
    x = -15, 20
    Negative value is rejected, so x = 20.
  • Question 21
    1 / -0
    Which of the following equations does not have real roots?
    Solution
    (1) x2 + x - 5 = 0
    D = 1 - 4(1)(-5) = 1 + 20 = 21 > 0
    This equation has real roots.

    (2) 5x2 - 3x + 1 = 0
    D = (-3)2 - 4(5)(1) = 9 - 20 = -11 < 0
    This equation does not have real roots

    (3) -x2 + 2x + 3 = 0
    D = (2)2 - 4(-1)(3) = 4 + 12 = 16 > 0
    This equation has real roots.

    (4) x2 + 3x - 12 = 0
    D = (3)2 - 4(-12)(1) = 9 + 48 = 57 > 0
    This equation has real roots.

    Option 2 is the answer.
  • Question 22
    1 / -0
    Directions: Read the given statements carefully.

    Statement I: The quadratic equation 108x2 – 288x + 192 = 0 has real and equal roots.
    Statement II: If the equation 18x2 + 12kx + 8 = 0, where k > 0, has equal roots, then both the roots are equal to .
    Solution
    Statement I: Discriminant = b2 – 4ac or D = b2 – 4ac
    Case I: If D > 0, then roots are real and unequal.
    Case II: If D = 0, then roots are real and equal.
    Case III: If D < 0, then roots are non-real and unequal.
    Consider the quadratic equation: 12(9x2 – 24x + 16) = 0
    9x2 - 24x + 16 = 0
    Comparing it with ax2 + bx + c = 0, we get a = 9, b = –24 and c = 16.
    Now, D = b2 – 4ac
    D = (–24)2 – 4 × 9 × 16
    D = 576 – 576 = 0
    Since D = 0, the roots of this quadratic equation are real and equal.
    Thus, 9x2 – 24x + 16 = 0 is the required quadratic equation.

    Statement II: 2(9x2 + 6kx + 4) = 0 has equal roots.
    Determinant, D = 0 or b2 - 4ac = 0 or b2 = 4ac
    36k2 = 4 × 4 × 9
    36k2 = 36 × 4
    k2 = 4
    k = 2
    But k > 0
    k = 2
    The equation becomes:
    9x2 + 12x + 4 = 0
    9x2 + 6x + 6x + 4 = 0
    3x(3x + 2) + 2(3x + 2) = 0
    (3x + 2)(3x + 2) = 0

    x =
  • Question 23
    1 / -0
    Which among the following is a factor of the polynomial 9x2 + 63x + 108?
    Solution
    9(x2 + 7x + 12) = 0
    x2 + 4x + 3x + 12 = 0
    x(x + 4) + 3(x + 4) = 0
    (x + 3) (x + 4) = 0
    So, factors of x2 + 7x + 12 = 0 are (x + 3) and (x + 4).

    Alternative solution: Comparing 9(x2 + 7x + 12) = 0 with ax2 + bx + c = 0, we get a = 1, b = 7 and c = 12.

    x =

    =

    =

    =

    x = –3, –4

    Thus, (x + 3) and (x + 4) are the factors of the given polynomial.
  • Question 24
    1 / -0
    Directions: Determine whether the given statements are true (T) or false (F) and choose the correct option accordingly.

    (i) is the solution set for the quadratic equation = (a + b)2.


    (ii) Two friends, A and B, are discussing their ages. The age of B is square of the age of A. B will be 3 times as old as A after 5 years. The difference between their current ages is 20 years.


    (iii) Two taps, A and B, together can fill a tank in 10 hours. Tap A, which is bigger, can fill the tank in 16 hours less than tap B, which is smaller. If x is the number of hours required to fill the tank by the smaller tap, then this case can be represented by equation x2 + 36x + 160 = 0.

    (iv) The quadratic equation has the discriminant equal to 0.
    Solution
    (i) False

    = (a + b)2

    abx2 - (a + b)2x + (a + b)2 = 0
    abx2 - (a2 + b2 + 2ab)x + (a + b)2 = 0
    abx2 - a2x - b2x - 2abx + (a + b)2 = 0
    abx2 - a2x - abx - b2x - abx + (a + b)2 = 0
    abx2 - a(a + b)x - b(b + a)x + (a + b)2 = 0
    ax(bx - ( a+ b)) - (a + b) (bx - (a + b)) = 0
    (ax - (a + b)) (bx - ( a + b)) = 0

    x =

    (ii) True

    Let the age of A be x.
    So, B's age = x2
    After 5 years: x2 + 5 = 3(x + 5)
    x2 + 5 – 3x – 15 = 0
    x2 – 3x – 10 = 0
    x = -2, 5
    x is +ve, so x = 5
    So, A's age = 5 years
    B's age = 25 years
    Difference = (25 – 5) years = 20 years

    (iii) False
    Let x be the number of hours taken to fill the tank by the smaller tap alone.
    Then, x – 16 is the time (in hours) to fill the tank by the larger tap alone.

    The smaller tap fills of the tank in 1 hour.

    The larger tap fills of the tank in 1 hour.

    In 10 hours of running together, the two will fill the complete tank.

    As per given information:



    10x + 10x – 160 = x2 – 16x
    x2 – 16x – 20x + 160 = 0
    x2 – 36x + 160 = 0

    (iv) True



    (x2 + 4x + 4) = 0

    x2 + 4x + 4 = 0

    If a quadratic equation is ax2 + bx + c = 0, then discriminant (D) = b2 - 4ac.
    Consider the equation x2 + 4x + 4 = 0, where a = 1, b = 4, and c = 4.
    D = b2 - 4ac = (4)2 - 4 × 1 × 4 = 0
  • Question 25
    1 / -0
    There are two rectangles, ABCD and PQRS. ABCD lies inside PQRS such that there is a uniform width all around ABCD. The difference between the areas of these rectangles is 150 cm2. If the length of rectangle ABCD is 27 cm and width is 20 cm, then what is the perimeter of rectangle PQRS?
    Solution

    Area of rectangle ABCD = (27 × 20) cm2 = 540 cm2
    Area of rectangle PQRS = (540 + 150) cm2 = 690 cm2
    Let the width of the shaded region be x cm.
    So, (27 + 2x) (20 + 2x) = 690
    4x2 + 94x – 150 = 0
    2x2 + 47x – 75 = 0
    2x2 + 50x – 3x – 75 = 0
    (2x – 3) (x + 25) = 0
    x = 1.5, x = –25
    On neglecting the –ve value of x, we get
    x = 1.5 cm
    Dimensions of rectangle PQRS are 30 cm × 23 cm.
    Perimeter = 106 cm
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