**Directions(1-10):** Find the value of x and y and then compare their values.

- I. x² + 3x – 4 = 0

II. y² + 10y + 24 = 0

y > xx > yNo relationx ≥ yy ≥ xOption D

𝐈. x² + 3x – 4 = 0

x² + 4x – x – 4 = 0

x(x + 4) – 1 (x + 4) = 0

(x − 1) (x + 4) = 0

x = 1, −4

II. y² + 10y + 24 = 0

y² + 4y + 6y + 24 = 0

y(y + 4) + 6(y + 4) = 0

(y + 6) (y + 4) = 0

y = −4, −6

x ≥ y - I. 48/x² − 14/x + 1 = 0

𝐈𝐈. 45/y² + 1/y = 2

y ≥ xx ≥ yy > xx > yNo relationOption D

𝐈. 48/x² − 14/x + 1 = 0

⇒ x^2 − 14x + 48 = 0

⇒ x^2 − 8x − 6x + 48 = 0

⇒ x(x − 8) − 6(x − 8) = 0

⇒ (x − 8)(x − 6) = 0

x = 8, 6

𝐈𝐈. 45/y² + 1/y = 2

⇒ 2y^2 − y − 45 = 0

⇒ 2y^2 − 10y + 9y − 45 = 0

⇒ 2y(y − 5) + 9(y − 5) = 0

⇒ (2y + 9)(y − 5) = 0

y = 5, − 9/2

x > y - I. x² + 2x – 35 = 0

II. y² + 15y + 56 = 0

No relationx > yy ≥ xx ≥ yy > xOption D

I.x² + 2x – 35 = 0

=> x² + 7x – 5x – 35 = 0

=> x (x + 7) – 5 (x + 7) = 0

=> (x – 5) (x + 7) = 0

x = 5, –7

II. y² + 15y + 56 = 0

=> y² + 7y + 8y + 56 = 0

=>y (y + 7) + 6 (y + 7) = 0

=>(y + 8) (y + 7) = 0

y = – 8, – 7

x ≥ y - I.2x² + 7x + 5 = 0

II. 3y² + 12y + 9 = 0

x > yy ≥ xNo relationx ≥ yy > xOption C

I.2x² + 7x + 5 = 0

=> 2x² + 2x + 5x + 5 = 0

=> 2x (x + 1) + 5 (x + 1) = 0

=> (2x + 5) (x + 1) = 0

𝑥 = – 5/2 , – 1

II. 3y² + 12y + 9 = 0

=>3y² + 9y + 3y + 9 = 0

=> 3y (y + 3) +3 (y + 3) = 0

=>(3y + 3) (y + 3) = 0

y = –1, – 3

No relation can be established. - I.(x – 12)² = 0

II. y² – 21y + 108 = 0

y > xx > yy ≥ xNo relationx ≥ yOption E

I.(x – 12)² = 0

=> x – 12 = 0

x = 12

II. y² – 21y + 108 = 0

=> y² – 12y – 9y + 108 = 0

=>y (y – 12) – 9 (y – 12) = 0

=>(y – 9) (y – 12) = 0

y = 9, 12

x ≥ y - I.x² – 17x + 72 = 0

II. y² – 27y + 180 = 0

x ≥ yy ≥ xy > xx > yNo relationOption C

I.x² – 17x + 72 = 0

x² – 9x – 8x + 72 = 0

x(x – 9) – 8 (x – 9) = 0

(x – 8) (x – 9) = 0

x = 8, 9

II. y² – 27y + 180 = 0

y² – 12y – 15y + 180 = 0

y(y – 12) – 15 (y – 12) = 0

(y – 15) (y – 12) = 0

y = 15, 12

y > x - I.2x² – 5x + 3 = 0

II. 3y² – 4y + 1 = 0

y > xNo relationx > yx ≥ yy ≥ xOption D

I.2x² – 5x + 3 = 0

=> 2x² – 2x – 3x + 3 = 0

=>2x (x – 1) – 3(x – 1) = 0

=> (x – 1) (2x – 3) = 0

x = 1, 3/2

II. 3y² – 4y + 1 = 0

=>3y² – 3y – y + 1 = 0

=>3y(y – 1) –1 (y – 1) = 0

=> (3y – 1) (y – 1) = 0

=>𝑦 = 1/3 , 1

x≥y - I. (x − 2)² = 9

𝐈𝐈. (2y + 8)² = 16

y ≥ xy > xNo relationx > yx ≥ yOption C

𝐈. x² − 16x + 64 = 0

=> x² − 8x – 8x + 64 = 0

=>x(x − 8) − 8(x − 8) = 0

=> (x − 8) (x − 8) = 0

x = 8, 8

𝐈𝐈. y² − 16y + 63 = 0

=>y² − 7y – 9y + 63 = 0

=> y(y − 7) − 9(y − 7) = 0

=> (y – 9) (y – 7) = 0

y = 9, 7

No relation can be established between x & y - I. x² − 16x + 64 = 0

𝐈𝐈. y² − 16y + 63 = 0

y > xx ≥ yNo relationx > yy ≥ xOption D

𝐈. (x − 2)² = 9

⇒ (x − 2) = ± 3

⇒ x = 5, −1

𝐈𝐈. (2y + 8)² = 16

=>(2y + 8) = ± 4

⇒ y = −2, −6

x > y - I. 25/x² − 15/x + 2 = 0

𝐈𝐈. 40/y² + 1 = 13/y

x > yNo relationy > xy ≥ xx ≥ yOption D

𝐈. 25/x² − 15 x + 2 = 0

⇒ 2x^2 − 15x + 25 = 0

⇒ 2x^2 − 10x − 5x + 25 = 0

=>2x (x − 5) − 5(x − 5) = 0

=>(2x − 5)(x − 5) = 0

x = 5/2 , 5

𝐈𝐈. 40/y² + 1 = 13/y

⇒ y^2 − 13y + 40 = 0

⇒ y^2 − 8y − 5y + 40 = 0

⇒ y(y − 8) − 5(y − 8) = 0

=>(y − 5)(y − 8) = 0

y = 5, 8

y ≥ x