Harvard University Pure Mathematics Entrance Exam Tricks

[@nikolayplatnov5148]

The equation can be,simply reduced to a squate form by means of substitution (a/b)^2 =x. (a+b)/(a-b)=(a/b+1)/(a/b-1). Swuare,equation is x^2+1=10x

[@ShriH-d1o]

(a^2+b^2)^2=a^4+ b^4+2a^2*b^2;=>12a^2*b^2=> a^2+ b^2= ab√12; (a+b)^2=a^2+b^2+2ab=(√12+2)ab;=(a-b)^2=> (√12-2)ab; =>(a+b)^2/(a-b)^2 =(√12+2)ab/(√12-2)ab=(√3+1)/(√3-1)=(√3+1)(√3+1)/(3-1)= (√3+1)^2/2; =>(a+b)/(a-b)= ±(√3+1)/√2= ±(√6+√2)/2

[@key_board_x]

a⁴ + b⁴ = 10a²b² (a⁴ + b⁴)/a²b² = 10 (a⁴/a²b²) + (b⁴/a²b²) = 10 (a²/b²) + (b²/a²) = 10 (a²/b²) + [1/(a²/b²)] = 10 → let: x = a²/b² x + (1/x) = 10 (x² + 1)/x = 10 x² + 1 = 10x x² - 10x + 1 = 0 Δ = (- 10)² - 4 = 96 = 16 * 6 x = (10 ± 4√6)/2 x = 5 ± 2√6 x = 2 ± 2√6 + 3 x = [(√2)² ± 2.(√2 * √3) + (√3)²] x = (√2 ± √3)² → x = a²/b² a²/b² = (√2 ± √3)² (a/b)² = (√2 ± √3)² First case: (a/b)² = (√2 + √3)² → a/b = ± (√2 + √3) First possibility: a/b = √2 + √3 → a = (√2 + √3).b = (a + b)/(a - b) = [(√2 + √3).b + b] / [(√2 + √3).b - b] = [(√2 + √3) + 1] / [(√2 + √3) - 1] = [(√2 + √3) + 1]² / { [(√2 + √3) - 1].[(√2 + √3) + 1] } = [(√2 + √3) + 1]² / [(√2 + √3)² - 1²] = [(√2 + √3)² + 2.(√2 + √3) + 1²] / [2 + 2√6 + 3 - 1] = [2 + 2√6 + 3 + 2√2 + 2√3 + 1] / [4 + 2√6] = [6 + 2√6 + 2√2 + 2√3] / [4 + 2√6] = [3 + √6 + √2 + √3] / [2 + √6] = [(3 + √6 + √2 + √3).(2 - √6)] / [(2 + √6).(2 - √6)] = [6 - 3√6 + 2√6 - 6 + 2√2 - √12 + 2√3 - √18] / [4 - 6] = [6 - 3√6 + 2√6 - 6 + 2√2 - 2√3 + 2√3 - 3√2] / [- 2] = [- √6 + 2√2 - 3√2] / [- 2] = (√6 - √2)/2 Second possibility: a/b = - √2 - √3 → a = - (√2 + √3).b = (a + b)/(a - b) = [- (√2 + √3).b + b] / [- (√2 + √3).b - b] = [- (√2 + √3) + 1] / [- (√2 + √3) - 1] = [(√2 + √3) - 1] / [(√2 + √3) + 1] = [(√2 + √3) - 1]² / { [(√2 + √3) + 1].[(√2 + √3) - 1] } = [(√2 + √3)² + 2.(√2 + √3) + 1²] / [(√2 + √3)² - 1²] = [2 + 2√6 + 3 + 2√2 + 2√3 + 1] / [2 + 2√6 + 3 - 1] = [6 + 2√6 + 2√2 + 2√3] / [4 + 2√6] = [3 + √6 + √2 + √3] / [2 + √6] = [(3 + √6 + √2 + √3).(2 - √6)] / [(2 + √6).(2 - √6)] = [6 - 3√6 + 2√6 - 6 + 2√2 - √12 + 2√3 - √18] / [4 - 6] = [- √6 - √2] / [4 - 6] = (√6 + √2)/2 Second case: (a/b)² = (√2 - √3)² → a/b = ± (√2 - √3) First possibility: a/b = √2 - √3 → a = (√2 - √3).b = (a + b)/(a - b) = [(√2 - √3).b + b] / [(√2 - √3).b - b] = [(√2 - √3) + 1] / [(√2 - √3) - 1] = [(√2 - √3) + 1]² / { [(√2 - √3) - 1].[(√2 - √3) + 1] } = [(√2 - √3) + 1]² / [(√2 - √3)² - 1²] = [(√2 - √3)² + 2.(√2 - √3) + 1²] / [2 - 2√6 + 3 - 1] = [2 - 2√6 + 3 + 2√2 - 2√3 + 1] / [4 - 2√6] = [6 - 2√6 + 2√2 - 2√3] / [4 - 2√6] = [3 - √6 + √2 - √3] / [2 - √6] = [(3 - √6 + √2 - √3).(2 + √6)] / [(2 - √6).(2 + √6)] = [6 + 3√6 - 2√6 - 6 + 2√2 + √12 - 2√3 - √18] / [4 - 6] = [3√6 - 2√6 + 2√2 + 2√3 - 2√3 - 3√2] / [4 - 6] = [√6 - √2] / [4 - 6] = - (√6 - √2)/2 Second possibility: a/b = - (√2 - √3) → a = (√3 - √2).b = (a + b)/(a - b) = [(√3 - √2).b + b] / [(√3 - √2).b - b] = [(√3 - √2) + 1] / [(√3 - √2) - 1] = [(√3 - √2) + 1]² / { [(√3 - √2) - 1].[(√3 - √2) + 1] } = [(√3 - √2) + 1]² / [(√3 - √2)² - 1²] = [(√3 - √2)² + 2.(√3 - √2) + 1²] / [3 - 2√6 + 2 - 1] = [3 - 2√6 + 2 + 2√3 - 2√2 + 1] / [4 - 2√6] = [6 - 2√6 + 2√3 - 2√2] / [4 - 2√6] = [3 - √6 + √3 - √2] / [2 - √6] = [(3 - √6 + √3 - √2).(2 + √6)]/[(2 - √6).(2 + √6)] = [6 + 3√6 - 2√6 - 6 + 2√3 + √18 - 2√2 - √12]/[4 - 6] = [√6 + 2√3 + 3√2 - 2√2 - 2√3]/[4 - 6] = [√6 + √2]/[4 - 6] = - (√6 + √2)/2 Conclusion: (a + b)/(a - b) = (√6 ± √2)/2 (a + b)/(a - b) = - (√6 ± √2)/2

[@superacademy247]

Thanks for detailed explanation 🙏💕💯🥰✅