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Home » Since $\|\mathbfx\| = \|\mathbfy\| = 1$, this becomes: - Treasure Valley Movers

Since $\|\mathbfx\| = \|\mathbfy\| = 1$, this becomes: - Treasure Valley Movers

📅 April 15, 2026 👤 scraface

Apr 15, 2026

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6r^2 - 12r + 8 = 96
6r^2 - 12r - 88 = 0
3r^2 - 6r - 44 = 0
r = \frac{6 \pm \sqrt{(-6)^2 - 4 \cdot 3 \cdot (-44)}}{2 \cdot 3} = \f...
\sqrt{564} = \sqrt{4 \cdot 141} = 2\sqrt{141}
r = \frac{6 \pm 2\sqrt{141}}{6} = \frac{3 \pm \sqrt{141}}{3}
Since radius must be positive and greater than 2, we take the positive...
r = \frac{3 + \sqrt{141}}{3}
Check if this is greater than 2:
\(\sqrt{141} \approx 11.87\), so \( r \approx (3 + 11.87)/3 = 14.87/3...

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