Solving a 'Stanford' University entrance exam | a²+b²+c²=?

[@hamishbindrinkin]

Given bc is double ab c must be equal to 2a. Sub into 3rd equation for 2a² = 3000. a² = 1500. c² must be 4 times the size = 4000. b²c² = 2000². Solve for b² and add them all up = 8166.66...

[@AsadInternationalAcademy]

okay dear

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[@gilbertogarbi4479]

Much simpler.

[@joaowilsonvieira6196]

Socorro curió!

[@pedrojose392]

Let the equations be (i), (ii) and(iii) in this order. (ii):(i) ==> c=2a e (iii) ==> a^2=1.500 e c^2= 6.000 (iii):(i) ==> c=3b e (ii) ==> 3b^2=2.000 ==> b^2=2.000/3 a^2+b^2+c^2=1.500 +6.000+ 2.000/3= 24.500/3

[@ricardom.a8486]

Correto

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ok

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[@quinty.support]

given data: ab = 1000 bc = 2000 ca = 3000 we need to find a² + b² + c² = ? lets observe an interesting pattern of (a² + b² + c²)'s whole square... therefore... (a² + b² + c²)² = a⁴ + b⁴ + c⁴ + 2(a²b² + b²c² + c²a²) a²b² = (ab)² = 1000^2 = 1000000 b²c² = (bc)² = 2000^2 = 4000000 c²a² = (ca)² = 3000^2 = 9000000 find the sum of (a²b² + b²c² + c²a²) = 1000000 + 4000000 + 9000000 = 14000000 now, a⁴ can be written as → [a²b² • c²a²]/b²c²... → putting all the values... [1000000 • 9000000]/4000000 = 225000 now, b⁴ can be written as → [a²b² • b²c²]/c²a²... → putting all the values.... [1000000 • 4000000]/9000000 = 444444.44444444 now, c⁴ can be be written as → [b²c² • c²a²]/a²b².... → putting all the values... [4000000 • 9000000]/1000000 = 36000000 (a² + b² + c²)² = a⁴ + b⁴ + c⁴ + 2(a²b² + b²c² + c²a²) → 225000 + 444444.44444444 + 36000000 + 2(1000000 + 4000000 + 9000000) → 36669444.444444 + 2(14000000) → 36669444.444444 + 28000000 = 64669444.444444 therefore, (a² + b² + c²) = √(64669444.444444) → 8401.7314332452 ≈ 8402

[@ricardom.a8486]

Não. 8500

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ok

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[@eliandpardo1652]

La respuesta es 4.

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no

[@bardoinutil2587]

ca.ab/bc+ab.bc/ca+bc.ca/ab

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ok

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