So \( rac{a + b}{a - b} = rac{3 \pm \sqrt{5}}{2} \). Letâs take one root (both valid due to symmetry). Set:
["# Solving So ( \dfrac{a + b}{a - b} = \dfrac{3 \pm \sqrt{5}}{2} ): A Step-by-Step Algebraic Exploration", "When solving equations involving ratios like ( \dfrac{a + b}{a - b} ), particularly those leading to expressions involving radicals, a common challenge is translating the algebraic form into meaningful values or relationships. One elegant form arises when this expression equals ( \dfrac{3 \pm \sqrt{5}}{2} ). In this article, we explore how to manipulate this equation, derive key relationships between ( a ) and ( b ), and uncover the mathematical structure behind such ratios.", "---", "## What Does the Equation ( \dfrac{a + b}{a - b} = \dfrac{3 \pm \sqrt{5}}{2} ) Represent?", "The expression ( \dfrac{a + b}{a - b} ) represents the slope of a linear relationship between variables ( a ) and ( b ), assuming ( a <br/>\ne b ) to avoid division by zero. When this expression equals one of the roots of the quadratic form—( \dfrac{3 + \sqrt{5}}{2} ) or ( \dfrac{3 - \sqrt{5}}{2} )—we are effectively describing two distinct proportional relationships between ( a ) and ( b ) that preserve symmetry and algebraic consistency.", "Both roots are valid due to symmetry, and choosing either will yield structurally equivalent solutions. For clarity, we’ll compute one root using substitution and express ( a ) in terms of ( b ) (or vice versa), leading to a clean algebraic pathway.", "---", "## Step 1: Set Up the Equation", "We begin with:", "[\n\dfrac{a + b}{a - b} = \dfrac{3 + \sqrt{5}}{2}\n]", "Let ( r = \dfrac{3 + \sqrt{5}}{2} ). So,", "[\n\dfrac{a + b}{a - b} = r\n]", "Multiply both sides by ( a - b ):", "[\na + b = r(a - b)\n]", "Expand the right-hand side:", "[\na + b = ra - rb\n]", "Bring all terms to one side:", "[\na - ra + b + rb = 0\n]", "Factor:", "[\na(1 - r) + b(1 + r) = 0\n]", "---", "## Step 2: Solve for ( \dfrac{a}{b} )", "To find a meaningful ratio between ( a ) and ( b ), solve for ( \dfrac{a}{b} ):", "[\na(1 - r) = -b(1 + r)\n]", "Divide both sides by ( b(1 - r) ) (assuming ( b <br/>\ne 0 ) and ( r <br/>\ne 1 ), which holds for our values):", "[\n\dfrac{a}{b} = -\dfrac{1 + r}{1 - r}\n]", "Now substitute ( r = \dfrac{3 + \sqrt{5}}{2} ):", "First compute numerator and denominator:", "[\n1 + r = 1 + \dfrac{3 + \sqrt{5}}{2} = \dfrac{2 + 3 + \sqrt{5}}{2} = \dfrac{5 + \sqrt{5}}{2}\n]", "[\n1 - r = 1 - \dfrac{3 + \sqrt{5}}{2} = \dfrac{2 - 3 - \sqrt{5}}{2} = \dfrac{-1 - \sqrt{5}}{2}\n]", "So:", "[\n\dfrac{a}{b} = -\dfrac{ \dfrac{5 + \sqrt{5}}{2} }{ \dfrac{-1 - \sqrt{5}}{2} } = -\dfrac{5 + \sqrt{5}}{-1 - \sqrt{5}} = \dfrac{5 + \sqrt{5}}{1 + \sqrt{5}}\n]", "---", "## Step 3: Rationalize the Fraction", "To simplify ( \dfrac{5 + \sqrt{5}}{1 + \sqrt{5}} ), multiply numerator and denominator by the conjugate ( 1 - \sqrt{5} ):", "[\n\dfrac{(5 + \sqrt{5})(1 - \sqrt{5})}{(1 + \sqrt{5})(1 - \sqrt{5})}\n]", "Denominator:", "[\n(1 + \sqrt{5})(1 - \sqrt{5}) = 1 - (\sqrt{5})^2 = 1 - 5 = -4\n]", "Numerator:", "[\n(5 + \sqrt{5})(1 - \sqrt{5}) = 5 \cdot 1 - 5\sqrt{5} + \sqrt{5} \cdot 1 - \sqrt{5} \cdot \sqrt{5} = 5 - 5\sqrt{5} + \sqrt{5} - 5 = (5 - 5) + (-5\sqrt{5} + \sqrt{5}) = -4\sqrt{5}\n]", "Thus:", "[\n\dfrac{a}{b} = \dfrac{-4\sqrt{5}}{-4} = \sqrt{5}\n]", "---", "## Step 4: Interpret the Result", "We find:", "[\n\dfrac{a}{b} = \sqrt{5}\n]", "This implies ( a = b\sqrt{5} ), a simple and elegant relationship. Due to symmetry, choosing the other root ( \dfrac{3 - \sqrt{5}}{2} ) would yield ( \dfrac{a}{b} = -\sqrt{5} ), or equivalently, ( a = -b\sqrt{5} ). Both satisfy the original equation.", "---", "## Step 5: Verify the Solution", "Let’s verify ( a = b\sqrt{5} ) in the original expression:", "[\n\dfrac{a + b}{a - b} = \dfrac{b\sqrt{5} + b}{b\sqrt{5} - b} = \dfrac{b(\sqrt{5} + 1)}{b(\sqrt{5} - 1)} = \dfrac{\sqrt{5} + 1}{\sqrt{5} - 1}\n]", "Rationalize:", "[\n= \dfrac{(\sqrt{5} + 1)^2}{(\sqrt{5})^2 - 1^2} = \dfrac{(5 + 2\sqrt{5} + 1)}{5 - 1} = \dfrac{6 + 2\sqrt{5}}{4} = \dfrac{3 + \sqrt{5}}{2}\n]", "Verified.", "---", "## Step 6: Final Observations", "The equation ( \dfrac{a + b}{a - b} = \dfrac{3 \pm \sqrt{5}}{2} ) yields a symmetric family of proportional relationships, expressible as ( a = \pm b\sqrt{5} ). These solutions arise naturally when linear ratios involve constructed algebraic constants tied to quadratic equating.", "Understanding such relations strengthens problem-solving agility in algebra, particularly when symmetrical forms hint at deeper geometric or number-theoretic interpretations.", "---", "## Conclusion", "The equation ( \dfrac{a + b}{a - b} = \dfrac{3 \pm \sqrt{5}}{2} ) elegantly constrains the ratio ( \dfrac{a}{b} ) to ( \pm\sqrt{5} ), revealing two valid and symmetric solutions. Through rational manipulation and rationalization, we uncover the proportionality behind the expression—proof that even abstract ratios stem from concrete algebra.", "For students and enthusiasts of mathematics, exploring these forms deepens intuition about symmetry, radicals, and rational equations. Whether in competition problems or real-world modeling, such insights empower precise and elegant reasoning.", "---", "## Key Takeaways", "- The expression ( \dfrac{a + b}{a - b} ) models a key ratio.\n- Solving for ( a/b ) leads to simplified relation via substitution.\n- One valid solution is ( \dfrac{a}{b} = \sqrt{5} ); the symmetric solution is ( -\sqrt{5} ).\n- Verification confirms correctness.\n- This insight strengthens algebraic fluency and problem-solving versatility.", "---", "# Related Searches\n- Solving ( \dfrac{a + b}{a - b} = k )\n- Algebraic simplification of radical expressions\n- Symmetric rational equations and their solutions\n- Proportional relationships in algebra\n- Deriving ( \sqrt{5} ) from linear ratios", "---", "*Keywords: ( \dfrac{a + b}{a - b} = \dfrac{3 \pm \sqrt{5}}{2} ), rational equation, proportionality, ( a/b = \sqrt{5} ), algebraic simplification, mathematical derivation, symmetry in ratios."]
