II. b2 - 33b + 272 = 01). a b3). a = b or the relationship cannot be determined4). a ≥ b

II. b2 - 33b + 272 = 01). a < b2). a > b3). a = b or the relationship

1.	II. b2 - 33b + 272 = 01). a < b2). a > b3). a = b or the relationship cannot be determined4). a ≥ b
Answer» I. a² - 36A + 323 = 0 ⇒ a² - 19a - 17a + 323 = 0 ⇒ a(a – 19) - 17(a - 19) = 0 ⇒ (a - 19)(a - 17) = 0 Then, a = 19 or a = 17 II. b² - 33b + 272 = 0 ⇒ b² - 17b – 16b + 272 = 0 ⇒ b(b - 17) - 16(b - 17) = 0 ⇒ (b - 17)(b - 16) = 0 Then, b = 17 or b = 16 So, when a = 19, a > b for b = 17 and a > b for b = 16 And when a = 17, a = b for b = 17 and a > b for b = 16 ∴ We can observe that a ≥ b.

Answer»

I. a² - 36A + 323 = 0

⇒ a² - 19a - 17a + 323 = 0

⇒ a(a – 19) - 17(a - 19) = 0

⇒ (a - 19)(a - 17) = 0

Then, a = 19 or a = 17

II. b² - 33b + 272 = 0

⇒ b² - 17b – 16b + 272 = 0

⇒ b(b - 17) - 16(b - 17) = 0

⇒ (b - 17)(b - 16) = 0

Then, b = 17 or b = 16

So, when a = 19, a > b for b = 17 and a > b for b = 16

And when a = 17, a = b for b = 17 and a > b for b = 16

∴ We can observe that a ≥ b.