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