Can you Solve Pure Mathematics ? | Oxford University's Entrance for Software Development.

[@souzasilva5471]

Resolvi dividindo todos os termos por 10x e fiz x/10 -1/x = a . (I solved it by dividing all terms by 10x and made x/10 -1/x = a)

[@superacademy247]

Great strategy, using substitution is a powerful tool in solving equations! 💯🙏

[@raghvendrasingh1289]

(x-10/x-1)^2+ (x-10/x+1)^2 = 100/10
Now (A+B)°2+ (A-B)^2 = 2(A^2+B^2) hence
(x-10/x)^2+1= 5
(x^2-10)/x = 2 , - 2
x^2-2x-10 = 0 , x^2+2x-10 = 0
rest follows at once

[@key_board_x]

[(x² - x - 10)/10x]² - [(x² + x - 10)/10x]² = 1/10 → where: x ≠ 0 because the denominators
[(x² - x - 10)²/100x²] - [(x² + x - 10)²/100x²] = 1/10
[(x² - x - 10)² - (x² + x - 10)²]/100x² = 1/10
(x² - x - 10)² - (x² + x - 10)² = 100x²/10
(x² - x - 10)² - (x² + x - 10)² = 10x²
(x⁴ - x³ - 10x² - x³ + x² + 10x - 10x² + 10x + 100) - (x⁴ + x³ - 10x² + x³ + x² - 10x - 10x² - 10x + 100) = 10x²
(x⁴ - 2x³ - 19x² + 20x + 100) - (x⁴ + 2x³ - 19x² - 20x + 100) = 10x²
x⁴ - 2x³ - 19x² + 20x + 100 - x⁴ - 2x³ + 19x² + 20x - 100 = 10x²
- 4x³ - 10x² + 40x = 0
2x³ + 5x² - 20x = 0
x.(2x² + 5x - 20) = 0 → recall: x ≠ 0
2x² + 5x - 20 = 0
Δ = 5² - 4.(2 * - 20) = 25 + 160 = 185
x = (- 5 ± √185)/4