# 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
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
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
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
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
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
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
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
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
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