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Home » between $(0,0,0)$ and $(0,1,1)$: $\sqrt2$, - Appfinity Technologies

between $(0,0,0)$ and $(0,1,1)$: $\sqrt2$, - Appfinity Technologies

📅 March 1, 2026 👤 scraface

Mar 01, 2026

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📌 Now both are negative — still problem. But ratio of third to first: if $ a + d > a - 3d $, then third > first, but denominator negative implies first negative, so magnitude issue.

📌 But in symmetric AP, if $ d > 0 $, $ a - 3d $ could be negative. To make sense, assume $ a $ large. But both signs give negative — so only one is positive? Try numerically: $ \sqrt{7} \approx 2.6458 $. Then $ a - 3d \approx a - 5.29 $, $ a + 3d \approx a + 7.97 $, $ a + d \approx a + 2.6458 $. For first term positive, $ a > 5.29 $; third term positive. Then $ \frac{a + d}{a - 3d}

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From \( f(-1) = 4 \):
-a + b - c + d = 4 \quad \text{(1)}
From \( f(0) = 1 \):
d = 1 \quad \text{(2)}
From \( f(2) = 10 \):
8a + 4b + 2c + d = 10 \quad \text{(3)}
From \( f(3) = 19 \):
27a + 9b + 3c + d = 19 \quad \text{(4)}
Substitute \( d = 1 \) into (1), (3), and (4):
Now solve the system:

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