Stanford University Admission Interview | Find x+y=?

[@key_board_x]

x² - y² = 64 → given: xy = 8 → y = 8/x x² - (8/x)² = 64 x² - (64/x²) = 64 x⁴ - 64 = 64x² x⁴ - 64x² - 64 = 0 → let: X = x² ← where X ≥ 0 X² - 64X - 64 = 0 Δ = (- 64)² - (4 * - 64) = (64 * 64) + (4 * 64) = 68 * 64 = 17 * 4 * 64 = 17 * 2² * 8² = 17 * 16² X = (64 ± 16√17)/2 X = 32 ± 8√17 → we keep only the positive value (recall: X ≥ 0) X = 32 + 8√17 ← memorize this result x + y = x + (8/x) ← because xy = 8 → y = 8/x x + y = (x² + 8)/x (x + y)² = (x² + 8)²/x² → recall: x² = X (x + y)² = (X + 8)²/X → recall:the memorized result (x + y)² = (32 + 8√17 + 8)² / (32 + 8√17) (x + y)² = (40 + 8√17)² / (32 + 8√17) (x + y)² = [8 * (5 + √17)]² / [8 * (4 + √17)] (x + y)² = [8² * (5 + √17)²] / [8 * (4 + √17)] (x + y)² = [8 * (5 + √17)²] / (4 + √17) (x + y)² = [8 * (25 + 10√17 + 17)] / (4 + √17) (x + y)² = [8 * (42 + 10√17)] / (4 + √17) (x + y)² = [16 * (21 + 5√17)] / (4 + √17) (x + y)² = 16 * [(21 + 5√17) * (4 - √17)] / [(4 + √17) * (4 - √17)] (x + y)² = 16 * [84 - 21√17 + 20√17 - 85] / [16 - 17] (x + y)² = 16 * [- 1 - √17] / [- 1] (x + y)² = 16 * (1 + √17) x + y = ± 4√(1 + √17)

[@superacademy247]

Thanks for sharing detailed and resourceful explanation 🙏💡✅✅✅

[@benyasir423]

Que pensez vous de la demarche suivante? Objectif: former un système à 2 équations à 2 inconues connaissant leur somme et leur produit. on pose m = x^2 et n = - y^2. on a xy = 8 donc (x^2)(y^2) = 64 avec xy > 0 soit x^2(- y^2) = - 64 d'ou le système m+n = 64 = S et mn = -64 = P donc m et n sont solutions de l'équation Z^2 - SZ + P = 0 avec m > 0 et n < 0 Z^2 - 64Z - 64 = 0. le reste est simple, respectez les signes et l'ordre des grandeurs de x et y. vous avez x^2 - y^2 < 0 donc |x| > |y| et xy > 0. Merci.!

[@superacademy247]

Is the quadratic equation formed in terms of Z solvable?

[@ai.oekaki-v3w]

let x+y=A, x-y=B x^2 - y^2 = 64 (x+y)(x-y) = 64 AB = 64 2x = (x+y)+(x-y)=A+B 2y = (x+y)-(x-y)=A-B 4xy = (A+B)(A-B) (A+B)(A-B)=4*8=32 A^2 - B^2 = 32 A^4 - (AB)^2 = 32A^2 A^4 - 64^2 = 32A^2 let A^2=C, then 0