** 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. 8/√x + 9/(√x +1) = 7,

II. 9/√y – 3/√y = 2

A) x > y

B) x < y

C) x ≥ y

D) x ≤ y

E) x = y or relation cannot be established

View Answer

** Option B**

Solution:

8/√x + 9/(√x +1) = 7

[8(√x +1) + 9√x]/[√x * (√x +1)] = 7

17√x + 8 = 7 (x + √x)

7x – 10√x – 8 = 0

7x – 14√x + 4√x – 8 = 0

7√x (√x – 2) + 4 (√x – 2) = 0

√x cannot be -4/7

So √x = 2, so x = 4

9/√y – 3/√y = 2

(9 – 3)/√y = 2

Gives √y = 3, so y = 9

- I. 9/√x + 3/√x = √x + 1,

II. 4y^{2} + 5y – 6 = 0

A) x > y

B) x < y

C) x ≥ y

D) x ≤ y

E) x = y or relation cannot be established

View Answer

** Option A**

Solution:

9/√x + 3/√x = √x + 1

12/√x = √x + 1

x + √x – 12 = 0

x + 4√x – 3√x – 12 = 0

√x(√x + 4) – 3 (√x + 4) = 0

√x cannot be -4, So √x = 3 => x = 9

4y^{2} + 5y – 6 = 0

4y^{2} + 5y – 6 = 0

Gives y = -2, 3/4

Put all values on number line and analyze the relationship

-2… 3/4… 9

- I. 6x
^{2} + 13x + 6 = 0,

II. 6y^{2} – y – 2 = 0

A) x > y

B) x < y

C) x ≥ y

D) x ≤ y

E) x = y or relation cannot be established

View Answer

** Option B**

Solution:

6x^{2} + 13x + 6 = 0

6x^{2} + 9x + 4x + 6 = 0

Gives x = -2/3, -3/2

6y^{2} – y – 2 = 0

6y^{2} + 3y – 4y – 2 = 0

Gives y = -1/2, 2/3

Put all values on number line and analyze the relationship

-3/2… -2/3… -1/2… 2/3

- I. 3x
^{2} + 14x – 5 = 0,

II. 3y^{2} – 11y + 6 = 0

A) x > y

B) x < y

C) x ≥ y

D) x ≤ y

E) x = y or relation cannot be established

View Answer

** Option B**

Solution:

3x^{2} + 14x – 5 = 0

3x^{2} + 15x – x – 5 = 0

Gives x = -5, 1/3

3y^{2} – 11y + 6 = 0

3y^{2} – 9y – 2y + 6 = 0

Gives y = 2/3, 3

Put all values on number line and analyze the relationship

-5… 1/3… 2/3… 3

- I. 6x
^{2} + 5x – 1 = 0,

II. 3y^{2} – 10y + 3 = 0

A) x > y

B) x < y

C) x ≥ y

D) x ≤ y

E) x = y or relation cannot be established

View Answer

** Option B**

Solution:

6x^{2} + 5x – 1 = 0

6x^{2} + 6x – x – 1 = 0

Gives x = -1, 1/6

3y^{2} – 10y + 3 = 0

3y^{2} – 9y – y + 3 = 0

Gives y = 1/3, 3

Put all values on number line and analyze the relationship

-1… 1/6… 1/3… 3

- I. 12x
^{2} – 5x – 3 = 0,

II. 3y^{2} – 11y + 6 = 0

A) x > y

B) x < y

C) x ≥ y

D) x ≤ y

E) x = y or relation cannot be established

View Answer

** Option E**

Solution:

12x^{2} – 5x – 3 = 0

12x^{2} + 4x – 9x – 3 = 0

Gives x = -1/3, 3/4

3y^{2} – 11y + 6 = 0

3y^{2} – 9y – 2y + 6 = 0

Gives y = 2/3, 3

Put all values on number line and analyze the relationship

-1/3… 2/3… 3/4… 3

- I. 6x
^{2} + 7x + 2 = 0,

II. 15y^{2} – 38y – 40 = 0

A) x > y

B) x < y

C) x ≥ y

D) x ≤ y

E) x = y or relation cannot be established

View Answer

** Option E**

Solution:

6x^{2} + 7x + 2 = 0

6x^{2} + 4x + 3x + 2 = 0

Gives x = -2/3, -1/2

15y^{2} – 38y – 40 = 0

15y^{2} + 12y – 50y – 40 = 0

Gives y = -4/5, 10/3

Put all values on number line and analyze the relationship

-4/5… -2/3… -1/2… 10/3

- I. 3x
^{2} – 25x + 52 = 0,

II. 2y^{2} – 7y + 3 = 0

A) x > y

B) x < y

C) x ≥ y

D) x ≤ y

E) x = y or relation cannot be established

View Answer

** Option A**

Solution:

3x^{2} – 25x + 52 = 0

3x^{2} – 12x – 13x + 52 = 0

Gives x = 4, 13/3

2y^{2} – 7y + 3 = 0

2y^{2} – 6y – y + 3 = 0

So y = 1/2, 3

Put all values on number line and analyze the relationship

1/2… 3… 4… 13/3

- I. x
^{2} = 1156,

II. y = √1156

A) x > y

B) x < y

C) x ≥ y

D) x ≤ y

E) x = y or relation cannot be established

View Answer

** Option D**

Solution:

x^{2} = 1156,

So x = -34, 34

y = √1156

So y = 34

Put all values on number line and analyze the relationship

-34… 34

- I. x
^{2} – √3969 = √6561,

II. y^{2} – √1296 = √4096

A) x > y

B) x < y

C) x ≥ y

D) x ≤ y

E) x = y or relation cannot be established

View Answer

** Option E**

Solution:

x^{2} – √3969 = √6561

x^{2} – 63 = 81

x^{2} = 144

So x = -12, 12

y^{2} – √1296 = √4096

y^{2} – 36 = 64

y^{2} = 100

So y = -10, 10

Put all values on number line and analyze the relationship

-12… -10….10…. 12

tq

9√x + 3/√x = √x + 1, check 2nd one it has mistake, but correct in solu

Yes divide symbol