Two Paths to the Same Answer — Isolate or Factor the Radical | Math Olympiad

Two solution methods, radical factorization, olympiad algebra — the same problem yields one answer whether you isolate √x first or factor it out as a common term. Both approaches reveal why squaring creates extraneous roots. √(5x)−√x = 8. Method 1 isolates the larger radical, moves the smaller across, then squares twice to eliminate all radicals. That produces a quadratic with two candidates: x = 24+8√5 and x = 24−8√5. Checking both in the original equation reveals that only x = 24+8√5 works — the second is extraneous, introduced by squaring. Method 2 recognises that √(5x) = √5·√x, so factoring √x from both terms gives √x(√5−1) = 8. Isolate √x instantly: √x = 8/(√5−1). Rationalise by multiplying by (√5+1)/(√5+1), getting √x = 2(√5+1). Square once to find x = 24+8√5 directly. No extraneous roots appear because there's no squaring until the very end, and the factorisation locks down √x uniquely. The second method is faster — one isolation, one rationalisation, one squaring. The first method is more mechanical and requires checking extraneous roots at the end. Both are valid, but Method 2 demonstrates the elegance of factoring before squaring. Students who don't recognise √(5x) = √5·√x miss the factorisation entirely and default to Method 1. Spotting the common factor halves the algebra and eliminates extraneous solutions by construction. This factorisation-before-squaring technique applies to any radical equation where one radicand is a multiple of another. Factor out the common radical, isolate, then square once instead of twice. Fewer squarings mean fewer extraneous roots to check. 🔔 Daily olympiad problems — algebra, number theory, geometry, combinatorics. #OlympiadMath #TwoMethods #RadicalEquation #FactorCommon #AlgebraTrick #CompetitionMath #AMCPrep #AIМЕPrep #MathOlympiad #RationaliseDenominator #ExtraneousRoots #OlympiadAlgebra