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Directions: In the following questions, two equations numbered I and II are given. You have to solve both the equations and give an answer. I. a2 + 16a + 55 = 0 II. b2 + 7b + 12 = 0

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Directions: In the following questions, two equations numbered I and II are given. You have to solve both the equations and give an answer.

I. a2 + 16a + 55 = 0

II. b2 + 7b + 12 = 0
1). a < b
2). a > b
3). a ≤ b
4). a ≥ b


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Answered by on | Votes 2 |

I. a2 + 16a + 55 = 0

⇒ a2 + 11a + 5a + 55 = 0

⇒ a(a + 11) + 5(a + 11) = 0

⇒ (a + 11)(a + 5) = 0

Then, a = -11 or a = -5

II. b2 + 7b + 12 = 0

⇒ b2 + 4b + 3b + 12 = 0

⇒ b(b + 4) + 3(b + 4) = 0

⇒ (b + 4)(b + 3) = 0

Then, b = -4 or b = -3

So, when a = -11, a < b for b = -4 and a < b for b = -3

And when a = -5, a < b for b = -4 and a < b for b = -3

∴ We can observe that a < b.

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