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Direction: In the given question, two equations numbered l and II are given. You have to solve both the equations and mark the appropriate answer I. a2 + 41a + 400 = 0 II. b2 - 20b + 99 = 0

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Direction: In the given question, two equations numbered l and II are given. You have to solve both the equations and mark the appropriate answer

I. a2 + 41a + 400 = 0

II. b2 - 20b + 99 = 0
1). a < b
2). a > b
3). a ≤ b
4). a ≥ b


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Answered by on | Votes 3 |

I. a2 + 41a + 400 = 0

⇒ a2 + 25a + 16a + 400 = 0

⇒ a(a + 25) + 16(a + 25) = 0

⇒ (a + 25) (a + 16) = 0

Then, a = (-25) or a = (-16)

II. b2 - 20b + 99 = 0

⇒ b2 - 11b - 9b + 99 = 0

⇒ b(b - 11) - 9(b - 11) = 0

⇒ (b - 9) (b - 11) = 0

Then, b = 9 or b = 11

So, when a = (-25), a < b for b = 9 and a < b for b = 11

And when a = (-16), a < b for b = 9 and a < b for b = 11

∴ we can observe that a < b.

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