**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. 4x
^{2} + 5x â€“ 6 = 0,

II. 2y^{2} + 11y + 12 = 0

A) If x > y

B) If x < y

C) If x â‰¥ y

D) If x â‰¤ y

E) If x = y or relation cannot be established

View Answer

** Option E**

Solution:

4x^{2} + 5x â€“ 6 = 0

4x^{2} + 8x â€“ 3x â€“ 6 = 0

Gives x = -2, 3/4

2y^{2} + 11y + 12 = 0

2y^{2} + 8y + 3y + 12 = 0

Gives y = -4, -3/2

- I. 12x
^{2} – 49x + 30 = 0,

II. 6y^{2} – 35y + 50 = 0

A) If x > y

B) If x < y

C) If x â‰¥ y

D) If x â‰¤ y

E) If x = y or relation cannot be established

View Answer

** Option E**

Solution:

12x^{2} – 49x + 30 = 0

12x^{2} – 9x â€“ 40x + 30 = 0

Gives x = 3/4, 10/3

6y^{2} – 35y + 50 = 0

6y^{2} – 15y â€“ 20y + 50 = 0

Gives y = 5/2, 10/3

- I. 4x
^{2} + 13x + 10 = 0,

II. 4y^{2} â€“ 7y – 15 = 0

A) If x > y

B) If x < y

C) If x â‰¥ y

D) If x â‰¤ y

E) If x = y or relation cannot be established

View Answer

** Option D**

Solution:

4x^{2} + 13x + 10 = 0

4x^{2} + 8x + 5x + 10 = 0

Gives x = -2, -5/4

4y^{2} â€“ 7y – 15 = 0

4y^{2} – 12y + 5y – 15 = 0

Gives y = -5/4, 3

- I. 12x
^{2} â€“ 5x â€“ 3 = 0,

II. 6y^{2} + y – 2 = 0

A) If x > y

B) If x < y

C) If x â‰¥ y

D) If x â‰¤ y

E) If x = y or relation cannot be established

View Answer

** Option E**

Solution:

12x^{2} â€“ 5x â€“ 3 = 0

12x^{2} + 4x â€“ 9x â€“ 3 = 0

Gives x = -1/3, 3/4

6y^{2} + y – 2 = 0

6y^{2} – 3y + 4y – 2 = 0

Gives y= -2/3, 1/2

- I. 3x
^{2} + 7x â€“ 6 = 0,

II. 3y^{2} â€“ 11y + 6 = 0

A) If x > y

B) If x < y

C) If x â‰¥ y

D) If x â‰¤ y

E) If x = y or relation cannot be established

View Answer

** Option D**

Solution:

Explanation:

3x^{2} + 7x â€“ 6 = 0

3x^{2} + 9x â€“ 2x â€“ 6 = 0

Gives x = -3, 2/3

3y^{2} â€“ 11y + 6 = 0

3y^{2} â€“ 9y â€“ 2y + 6 = 0

Gives y = 2/3, 3

- I. 5x
^{2} â€“ 36x â€“ 32 = 0,

II. 3y^{2} + 16y + 20 = 0

A) If x > y

B) If x < y

C) If x â‰¥ y

D) If x â‰¤ y

E) If x = y or relation cannot be established

View Answer

** Option A**

Solution:

5x^{2} – 36x â€“ 32 = 0

5x^{2} + 4x – 40x â€“ 32 = 0

Gives x = -4/5, 8

3y^{2} + 16y + 20 = 0

3y^{2} + 6y + 10y + 20 = 0

Gives y = -10/3, -2

- I. 3x
^{2} – (6 + 2âˆš3)x + 4âˆš3 = 0,

II. 3y^{2} – (2 + 3âˆš3)y + 2âˆš3 = 0

A) x > y

B) x < y

C) x â‰¥ y

D) x â‰¤ y

E) x = y or relationship cannot be determined

View Answer

** Option E**

Solution:

3x^{2} – (6 + 2âˆš3)x + 4âˆš3 = 0

(3x^{2} – 6x) â€“ (2âˆš3x – 4âˆš3) = 0

3x (x- 2) – 2âˆš3 (x â€“ 2) = 0,

So x = 2, 2âˆš3/3 (1.15)

3y^{2} – (2 + 3âˆš3)y + 2âˆš3 = 0

(3y^{2} – 2y) – (3âˆš3y – 2âˆš3) = 0

y (3y â€“ 2) – âˆš3 (3y â€“ 2) = 0

So x = 2/3, âˆš3 (1.73)

- I. 2x
^{2} + (4 + âˆš2)x + 2âˆš2 = 0

II. y^{2} – (1 + 3âˆš3)y + 3âˆš3 = 0

A) If x > y

B) If x < y

C) If x â‰¥ y

D) If x â‰¤ y

E) If x = y or relation cannot be established

View Answer

** Option B**

Solution:

2x^{2} + (4 + âˆš2)x + 2âˆš2 = 0

(2x^{2} + 4x) + (âˆš2x + 2âˆš2) = 0

2x (x + 2) + âˆš2 (x + 2) = 0

So x = -2, -âˆš2/2 (-0.7)

y^{2} – (1 + 3âˆš3)y + 3âˆš3 = 0

(y^{2} – y) – (3âˆš3y – 3âˆš3) = 0

y (y â€“ 1) – 3âˆš3 (y â€“ 1) = 0

So, y = 1, 3âˆš3 (5.2)

- I. x
^{2} + (3 + 2âˆš2)x + 6âˆš2 = 0

II. 5y^{2} – (1 + 5âˆš2)y + âˆš2 = 0

A) If x > y

B) If x < y

C) If x â‰¥ y

D) If x â‰¤ y

E) If x = y or relation cannot be established

View Answer

** Option B**

Solution:

x^{2} + (3 + 2âˆš2)x + 6âˆš2 = 0

(x^{2} + 3x) + (2âˆš2x + 6âˆš2) = 0

x (x + 3) + 2âˆš2 (x + 3) = 0

So x = -3, -2âˆš2

5y^{2} – (1 + 5âˆš2)y + âˆš2 = 0

(5y^{2} – y) – (5âˆš2y – âˆš2) = 0

y (5y â€“ 1) – 3âˆš2 (5y â€“ 1) = 0

So, y = 1/5, 3âˆš2

- I. 2x
^{2} + (4 + 2âˆš6)x + 4âˆš6 = 0

II. 5y^{2} + (10 + âˆš6)y + 2âˆš6 = 0

A) If x > y

B) If x < y

C) If x â‰¥ y

D) If x â‰¤ y

E) If x = y or relation cannot be established

View Answer

** Option D**

Solution:

2x^{2} + (4 + 2âˆš6)x + 4âˆš6 = 0

(2x^{2} + 4x) + (2âˆš6x + 4âˆš6) = 0

2x (x + 2) + 2âˆš6 (x + 2) = 0

So x = -2, -âˆš6

5y^{2} + (10 + âˆš6)y + 2âˆš6 = 0

(5y^{2} + 10y) + (âˆš6y + 2âˆš6) = 0

5y (y + 2) + âˆš6 (y + 2) = 0

So, y = -2, -âˆš6/5

q 1 ????????// kya answer hi hoga?????????// -2 is < -1.5 so e nhi hoga kya

Yes its E

Alternate roots

okk////////// thank you

some questions are confusing but solvable thx

question 10 . answer should be D part. x<=y

ty)))

Mam 10nth question me E Option hoga plz chech kariye

D hi hai

Root 6 is -2.4

-2 will come in middle

thnk u mam :)))

thank u mam

thanku

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