II. y2 + 31y – 140 = 01). x y3). x = y OR the relationship cannot be determined4). x ≥ y

II. y2 + 31y – 140 = 01). x < y2). x > y3). x = y OR the relationshi

1.	II. y2 + 31y – 140 = 01). x < y2). x > y3). x = y OR the relationship cannot be determined4). x ≥ y
Answer» I. x² + 22x – 104 = 0 ⇒ x² + 26x – 4X – 104 = 0 ⇒ x(x + 26) – 4(x + 26) = 0 ⇒ (x + 26)(x – 4) = 0 Then, x = - 26 or x = + 4 II. y² + 31y – 140 = 0 ⇒ y² + 35Y – 4y – 140 = 0 ⇒ y(y + 35) – 4(y + 35) = 0 ⇒ (y + 35)(y – 4) = 0 Then, y = - 35 or y = + 4 So, when x = - 26, x > y for y = - 35 and x < y for y = + 4 And when x = + 4, x > y for y = - 35 and x = y for y = + 4 ∴ So, we can observe that no clear RELATIONSHIP cannot be determined between x and y.

Answer»

I. x² + 22x – 104 = 0

⇒ x² + 26x – 4X – 104 = 0

⇒ x(x + 26) – 4(x + 26) = 0

⇒ (x + 26)(x – 4) = 0

Then, x = - 26 or x = + 4

II. y² + 31y – 140 = 0

⇒ y² + 35Y – 4y – 140 = 0

⇒ y(y + 35) – 4(y + 35) = 0

⇒ (y + 35)(y – 4) = 0

Then, y = - 35 or y = + 4

So, when x = - 26, x > y for y = - 35 and x < y for y = + 4

And when x = + 4, x > y for y = - 35 and x = y for y = + 4

∴ So, we can observe that no clear RELATIONSHIP cannot be determined between x and y.