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 –**

- I. x
^{3} + 7x^{2} + 16x + 12 = 0,

II. y^{3} – 2y^{2} – 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 x^{3} + 7x^{2} + 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 x^{3} + 7x^{2} + 16x + 12 = 0 by (x+2) . It gives x^{2} + 5x + 6 which is a quadratic equation. Its roots are -2 and -3

Now same with y^{3} – 2y^{2} – 5y + 6 = 0

Putting y = 1, satisfies the equation, so (y-1) is a factor of this equation> Now divide y^{3} – 2y^{2} – 5y + 6 = 0 by (y-1) gives y^{2} – 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

- I. x
^{3} – x^{2} – 2x = 0,

II. y^{3} – 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:

x^{3} – x^{2} – 2x = 0 becomes

x (x^{2} – x – 2) = 0

So x = 0 and (x^{2} – x – 2) = 0

Second is a quadratic equation with roots -1 and 2

So x = 0, -1, 2

Now y^{3} – 7y – 6 = 0

Put y = 1 – Does not satisfy.

Put y = -1 – Satisfies the equation so (y+1) is a factor

Divide y^{3} – 7y – 6 = 0 by (y+1) gives y^{2} – y – 6 = 0

So y = -1, -2, 3

Put on number line

-2……-1..…0……2……3 so no relation

- I. x
^{3} – 6x^{2} + 11x – 6 = 0,

II. y^{3} + 9y^{2} + 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 x^{3} – 6x^{2} + 11x – 6 = 0, x = 1 satisfies the equation so (x-1) is a factor

Divide x^{3} – 6x^{2} + 11x – 6 = 0 by (x-1) => gives x^{2} – 5x + 6 = 0

Gives x = 1, 2, 3

Similarly for second equation, y = -2, -3, -4

put on number line

So x > y

- I. x
^{3} + x^{2} – 4x – 4 = 0,

II. y^{3} – 7y^{2} + 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 x^{3} + x^{2} – 4x – 4 = 0

So (x+1) is a factor. On dividing gives x^{2} – 4 = 0

So x = -1, 2, -2

Similarly for second equation

y = 1, 2, 4

- I. 3x
^{3} – 13x^{2} + 18x – 8 = 0,

II. 3y^{3} + 14y^{2} – 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 3x^{3} – 13x^{2} + 18x – 8 = 0 so (x-1) is a factor. On dividing gives 3x^{2} – 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

- I. 2x
^{3} + 15x^{2} + 13x – 30 = 0,

II. 2y^{3} + 22y^{2} + 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

- I. 3x
^{3} + 25x^{2} + 68x + 60 = 0,

II. 3y^{3} + 5y^{2} – 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

- I. x
^{3} + 2x^{2} – 19x – 20 = 0,

II. 2y^{3} + 15y^{2} + 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

- I. 5x
^{3} – 12x^{2} + x + 6 = 0,

II. 5y^{3} + 28y^{2} + 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

- I. 2x
^{3} – 13x^{2} + 22x – 8 = 0,

II. 2y^{3} + 9y^{2} + 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 2x^{3} – 13x^{2} + 22x – 8 = 0. On dividing gives 2x^{2} – 9x + 4 = 0

x = 4 , 1/2, 2

y = -1 satisfies 2y^{3} + 9y^{2} + 3y – 4 = 0. On dividing gives 2y^{2} + 7y – 4 = 0

y = -4, 1/2, -1

Put on number line

-4……-1….. 1/2……2…. 4

in question 10 equation 2nd it is 14 or 4 as a constant as i am solving a question i am getting here 14 instead of 4 please check

4 only

Put y = -1

-2 + 9 – 3 – 4 = 0

ok

mam ek doubt hai mera…………..jab cubic equations ke root three hain to why we compare only 2 ???

Nhi 2 nhi.

All 3 compare kiye h

ok mam and jab 1 root given ho to bhi 3 ko hi compare karna hai na ???

1 root given means ?

I dint get

mam ek type ke ques hote hain jinme one root given hota hai and humein remaining 2 find karne hote hain

jese mam x^3 – 5 x^2 – 2x + 24/ (x + 2) = 0

Isme 2 lo. x+2 to cancel ho jaega

And also -2 put karke denominator 0 a ra h to not defined ho jaega

ok mam