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Home » \(a^2 = 4\), \(b^2 = 21\) is not possible - Nelissen Grade advocaten

\(a^2 = 4\), \(b^2 = 21\) is not possible - Nelissen Grade advocaten

📅 March 1, 2026 👤 scraface

Mar 01, 2026

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📌 Now find \( f(g(5)) \):

📌 g(5) = \sqrt{5+3} = \sqrt{8} = 2\sqrt{2}

📌 Substitute into \( f(x) \):

📌 f(g(5)) = (2\sqrt{2})^2 + 2(2\sqrt{2}) + 1 = 8 + 4\sqrt{2} + 1 = 9 + 4\sqrt{2}

📌 \boxed{9 + 4\sqrt{2}}

📌 Question:** A philosopher of science is analyzing a relationship where \( a + b = 10 \) and \( a^2 + b^2 = 58 \). Determine \( a^3 + b^3 \).

📌 Start by using the identity for the sum of cubes:

📌 a^3 + b^3 = (a+b)(a^2 - ab + b^2)

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