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Home » The time \(t\) to reach the highest point is when the velocity becomes zero: - Treasure Valley Movers

The time \(t\) to reach the highest point is when the velocity becomes zero: - Treasure Valley Movers

📅 April 17, 2026 👤 scraface

Apr 17, 2026

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📌 5Question: Solve for $ x $ in the equation $ \sqrt{x} + 3x = 4 $, inspired by philosophical considerations of root solutions and existence.

📌 Solution: Let $ u = \sqrt{x} $, so $ u \geq 0 $. Substituting, the equation becomes $ u + 3u^2 = 4 $. Rearranging: $ 3u^2 + u - 4 = 0 $. Using the quadratic formula: $ u = \frac{-1 \pm \sqrt{1 + 48}}{6} = \frac{-1 \pm 7}{6} $. Valid solution: $ u = 1 $ (since $ u \geq 0 $). Thus, $ \sqrt{x} = 1 \Rightarrow x = 1 $. \boxed{1}

📌 Question: Find the remainder when $ x^5 - 2x^3 + x^2 + 1 $ is divided by $ x^2 - 3x + 2 $, focusing on polynomial division techniques.

📌 Solution: Factor the divisor: $ x^2 - 3x + 2 = (x-1)(x-2) $. Use the Remainder Theorem: Let $ R(x) = ax + b $ be the remainder. Evaluate $ f(1) = 1 - 2 + 1 + 1 = 1 = a(1) + b $. Evaluate $ f(2) = 32 - 16 + 4 + 1 = 21 = a(2) + b $. Solve: $ a + b = 1 $, $ 2a + b = 21 $. Subtract: $ a = 20 $, then $ b = -19 $. Remainder: $ 20x - 19 $. \boxed{20x - 19}

📌 Question: If $ h(2y + 1) = y^3 - 5y + 6 $, find $ h(y^2 - 1) $.

📌 Solution: Let $ t = 2y + 1 \Rightarrow y = \frac{t - 1}{2} $. Substitute into $ h(t) $: $ h(t) = \left(\frac{t - 1}{2}\right)^3 - 5\left(\frac{t - 1}{2}\right) + 6 $. Simplify: $ h(t) = \frac{(t-1)^3}{8} - \frac{5(t-1)}{2} + 6 $. Now replace $ t $ with $ y^2 - 1 $: $ h(y^2 - 1) = \frac{(y^2 - 2)^3}{8} - \frac{5(y^2 - 2)}{2} + 6 $. Expand and combine terms (if required). \boxed{\frac{(y^2 - 2)^3}{8} - \frac{5(y^2 - 2)}{2} + 6}

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