Quantitative Aptitude: Cubic Equations Set 1

With the changing pattern in exam, One should be ready for each type of questions. So we are providing you with questions on Cubic Equations.

Directions: In the following questions, two equations numbered are given in variables x and y. You have to solve both the equations and find out the relationship between x and y. Then give answer accordingly –

  1. I. x3 + 7x2 + 16x + 12 = 0,
    II. y3 – 2y2 – 5y + 6 = 0
    A) If x > y
    B) If x < y
    C) If x ≥ y
    D) If x ≤ y
    E) If x = y or relation cannot be established
    View Answer
    Option D
    Solution:

    In these equations, you will have to first satisfy the equation with some root.
    Like start with 1: Put x = 1 in x3 + 7x2 + 16x + 12 = 0 and see if it equals to 0. It does not. Next move on to -1, again not, next put 2 and then -2
    -2 satisfies the equation: -8 + 28 – 32 + 12 = 0. Since -2 satisfies the equation so (x+2) is a factor of this equation.
    So x = -2, -3
    Next divide x3 + 7x2 + 16x + 12 = 0 by (x+2) . It gives x2 + 5x + 6 which is a quadratic equation. Its roots are -2 and -3
    Now same with y3 – 2y2 – 5y + 6 = 0
    Putting y = 1, satisfies the equation, so (y-1) is a factor of this equation> Now divide y3 – 2y2 – 5y + 6 = 0 by (y-1) gives y2 – y – 6 = 0 which is a quadratic equation with roots -2 and 3
    So y = 1, -2, 3
    Now put on number line
    -3……-2…..1……3
    so x ≤ y
  2. I. x3 – x2 – 2x = 0,
    II. y3 – 7y – 6 = 0
    A) If x > y
    B) If x < y
    C) If x ≥ y
    D) If x ≤ y
    E) If x = y or relation cannot be established
    View Answer
    Option E
    Solution:

    x3 – x2 – 2x = 0 becomes
    x (x2 – x – 2) = 0
    So x = 0 and (x2 – x – 2) = 0
    Second is a quadratic equation with roots -1 and 2
    So x = 0, -1, 2
    Now y3 – 7y – 6 = 0
    Put y = 1 – Does not satisfy.
    Put y = -1 – Satisfies the equation so (y+1) is a factor
    Divide y3 – 7y – 6 = 0 by (y+1) gives y2 – y – 6 = 0
    So y = -1, -2, 3
    Put on number line
    -2……-1..…0……2……3 so no relation
  3. I. x3 – 6x2 + 11x – 6 = 0,
    II. y3 + 9y2 + 26y + 24 = 0
    A) If x > y
    B) If x < y
    C) If x ≥ y
    D) If x ≤ y
    E) If x = y or relation cannot be established
    View Answer
    Option A
    Solution:

    For x3 – 6x2 + 11x – 6 = 0, x = 1 satisfies the equation so (x-1) is a factor
    Divide x3 – 6x2 + 11x – 6 = 0 by (x-1) => gives x2 – 5x + 6 = 0
    Gives x = 1, 2, 3
    Similarly for second equation, y = -2, -3, -4
    put on number line
    So x > y
  4. I. x3 + x2 – 4x – 4 = 0,
    II. y3 – 7y2 + 14y – 8 = 0
    A) If x > y
    B) If x < y
    C) If x ≥ y
    D) If x ≤ y
    E) If x = y or relation cannot be established
    View Answer
    Option E
    Solution:

    x = -1 satisfies x3 + x2 – 4x – 4 = 0
    So (x+1) is a factor. On dividing gives x2 – 4 = 0
    So x = -1, 2, -2
    Similarly for second equation
    y = 1, 2, 4
  5. I. 3x3 – 13x2 + 18x – 8 = 0,
    II. 3y3 + 14y2 – 32 = 0
    A) If x > y
    B) If x < y
    C) If x ≥ y
    D) If x ≤ y
    E) If x = y or relation cannot be established
    View Answer
    Option E
    Solution:

    x = 1 satisfies 3x3 – 13x2 + 18x – 8 = 0 so (x-1) is a factor. On dividing gives 3x2 – 10x + 8 = 0 which is quadratic equation with roots x = 2, 4/3
    So x = 1, 2, 4/3
    Similarly with second equation, y = -2, -4, 4/3
  6. I. 2x3 + 15x2 + 13x – 30 = 0,
    II. 2y3 + 22y2 + 5y – 18 = 0
    A) If x > y
    B) If x < y
    C) If x ≥ y
    D) If x ≤ y
    E) If x = y or relation cannot be established
    View Answer
    Option E
    Solution:

    x = -6, -5/2, 1
    y = -9/2, -2, 1
    Put on number line
    -6 … -9/2 …. -5/2 …. -2…. 1
  7. I. 3x3 + 25x2 + 68x + 60 = 0,
    II. 3y3 + 5y2 – 4y – 4 = 0
    A) x > y
    B) x < y
    C) x ≥ y
    D) x ≤ y
    E) x = y or relationship cannot be determined
    View Answer
    Option D
    Solution:

    x = -10/3, -3, -2
    y = -2, -2/3, 1
    Put on number line
    -10/3….-3…. -2…. -2/3….1
  8. I. x3 + 2x2 – 19x – 20 = 0,
    II. 2y3 + 15y2 + 28y + 15 = 0
    A) If x > y
    B) If x < y
    C) If x ≥ y
    D) If x ≤ y
    E) If x = y or relation cannot be established
    View Answer
    Option E
    Solution:

    x = -5, -1, 4
    y = -5, -3/2, -1
    Put on number line
    -5…. -3/2…-1….. 4
  9. I. 5x3 – 12x2 + x + 6 = 0,
    II. 5y3 + 28y2 + 35y + 12 = 0
    A) If x > y
    B) If x < y
    C) If x ≥ y
    D) If x ≤ y
    E) If x = y or relation cannot be established
    View Answer
    Option C
    Solution:

    x = -3/5, 1, 2
    y = -4, -1, -3/5
    Put on number line
    -4…..-1…. -3/5…1. 2
  10. I. 2x3 – 13x2 + 22x – 8 = 0,
    II. 2y3 + 9y2 + 3y – 4 = 0
    A) x > y
    B) x < y
    C) x ≥ y
    D) x ≤ y
    E) x = y or relationship cannot be determined
    View Answer
    Option C
    Solution:

    x = 2 satisfies 2x3 – 13x2 + 22x – 8 = 0. On dividing gives 2x2 – 9x + 4 = 0
    x = 4 , 1/2, 2
    y = -1 satisfies 2y3 + 9y2 + 3y – 4 = 0. On dividing gives 2y2 + 7y – 4 = 0
    y = -4, 1/2, -1
    Put on number line
    -4……-1….. 1/2……2…. 4

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