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Home » Evaluate at $ y = -1 $: - AMAZONAWS

Evaluate at $ y = -1 $: - AMAZONAWS

📅 March 5, 2026 👤 scraface

Mar 05, 2026

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📌 \[ n^2 = 141 \]

📌 \[ n = \sqrt{141} \approx 11.87 \]

📌 Wait — 141 is not a perfect square. Try checking if the middle term is \( n \), so use:

📌 Let the middle integer be \( n \), then:

📌 \[ 3n^2 + 2 = 425 \Rightarrow 3n^2 = 423 \Rightarrow n^2 = 141 \Rightarrow n = \sqrt{141} \]

📌 But 141 is not a perfect square — check if 137, 138, etc., work. Try actual integers.

📌 Try \( n = 11 \): numbers 10, 11, 12

📌 Sum of squares: \( 100 + 121 + 144 = 365 \) → too low

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S = rac{2(1 + (-1))}{1 - (-1)} = 0, ext{ but } a^2 - b^2 = 1 - (-1)...
Another example: $ a = 1 $, $ b = e^{i2\pi/3} $:
a^2 = 1, b^2 = e^{i4\pi/3}, a^2 - b^2 = 1 - e^{i4\pi/3}, a^2 + b^2 = 1...
Numerator: $ 2(1 + e^{i4\pi/3}) $, denominator: $ 1 - e^{i4\pi/3} $, s...
S = 2i \cdot rac{e^{i2\pi/3} + e^{i2\pi/3}}{e^{i2\pi/3} - e^{-i2\pi/3...
But from earlier general form $ S = rac{2(a^2 + b^2)}{a^2 - b^2} $, a...
eq b $, and $ |a| = |b| = 1 $, the only consistent value from examples...
oxed{0}
Question: Find the quadratic polynomial $ p(x) $ such that $ p(1) = 4...
Solution: Let $ p(x) = ax^2 + bx + c $. Using the given values:

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