A very tricky Question from Oxford University Entrance Exam | Find the Value of x=?

[@MrGeorge1896]

Nice problem but the solution seems to be too complicated. Put ((x+1)/(x-1))² on the RHS and multiply by x² (x - 1)²: (x + 1)² (x - 1)² = x² (x - 1)² + x² (x + 1)² x⁴ - 2x² + 1 = x² (2x² + 2) x⁴ + 4x² - 1 = 0 solve for x²: x² = -2 ± √5 now we get x = ±√( -2 ±√5)

[@9허공]

totally agreed!

[@木下雄哉]

nice🎉

[@jlp8573]

did the same. But Super Academy seems to like to overcomplicate his solutions...

[@souzasilva5471]

Resolvi de maneira bem mais simples. Desenvolvi e cheguei em x^4 + 4x^2 = 1, adicionei 4 aos dois lados, fatorei e calculei as raízes. (I solved it in a much simpler way. I developed and arrived at x^4 + 4x^2 = 1, I added 4 to both sides, factored and calculated the roots.)

[@icebear771]

I did the same, it tooks minutes five minutes. But why make it simple when you can make it complicated😂😂

[@anestismoutafidis4575]

(-0,5+1/-0,5)^2 - (-0,5+1/-0,5-1)^2= (0,25/0,25)-(0,25/2,25)=1-0,1=0,9 (-0,485+1/-0,485)^2-(-0,485+1/-0,485-1)^2= (0,265/0,235)-(0,265/2,205)= 1,127-0,120=1,00 x=-0,485

[@maamouhinda7722]

Your way it's ok but too hard and long. This is a short solution: (x+1/x)^2*(1-(x/x+1)^2*(x+1/x-1)^2)=1 (x+1/x)^2*(1-(x/(x-1))^2)=1 1-(x/(x-1))^2=(x/(x+1))^2 (x/(x-1))^2+(x/(x+1))^2=1 With commun dominator : x^2*(2x^2+2)/(x^2-1)^2=1 And put y=x^2 y^2+4y-1=0 y1= √5-2 et y2= -√5-2 Only y1 so X = +-√(√5-2)

[@superacademy247]

Thanks for the nice method 😎😍

[@quinty.support]

[(x+1)/x]² - [(x+1)/(x-1)]² = 1 Let, [(x+1)/x]² be A and [(x+1)/(x-1)]² be B A = [(x+1)/x]² = (x+1)²/x² = (x²+2x+1)/x² B = [(x+1)/(x-1)]² = (x+1)²/(x-1)² = (x²+2x+1)/(x²-2x+1) A-B = 1 => (x²+2x+1)/x² - (x²+2x+1)/(x²-2x+1) = 1 => [(x²+2x+1)(x²-2x+1) - (x²+2x+1)x²]/[x²(x²-2x+1)] = 1 L.H.S = N/D N = (x²+2x+1)(x²-2x+1) - x²(x²+2x+1) = x⁴ - 2x³ + x² + 2x³ - 4x² + 2x + x² - 2x + 1 = x⁴ - 2x³ + 1 D = x²(x²+2x+1) = x⁴ + 2x³ + x² N/D = [x⁴-2x²+1-(x⁴+2x³+x²)]/[x²(x²-2x+1)] = 1 => x²(x²-2x+1) = x⁴-2x²+1-x⁴-2x³-x² => x⁴-2x³+x³+2x³+3x²-1=0 => x⁴ + 4x² - 1 = 0 (suppose: y = x²) (a=1, b=4, c =-1) ∆ = b²-4ac = (4)²-4(1)(-1) = 16 -(-4) = 16+4 = 20 y = (-b±√∆)/2a = (-4±√20)/2 = (-4 ± 2√5)/2 = -2±√5 [Recall → y² = x] Case 1: x² = -2 - √5 => x² = -(2 + √5) => x = i²(2 + √5) => x= ±i√(2+√5)) Case 2: x² = -2 + √5 => x = ±√(-2 + √5)

[@superacademy247]

Nice method 😎💯🥰

[@didierblaizeau1194]

Il y a en efffet beaucoup plus simple !

[@key_board_x]

[(x + 1)/x]² - [(x + 1)/(x - 1)]² = 1 [(x + 1)²/x²] - [(x + 1)²/(x - 1)²] = 1 [(x + 1)².(x - 1)² - x².(x + 1)²] / [x².(x - 1)²] = 1 (x + 1)².(x - 1)² - x².(x + 1)² = x².(x - 1)² (x + 1)².(x - 1)² - x².(x + 1)² - x².(x - 1)² = 0 (x + 1)².(x - 1)² - x².[(x + 1)² + (x - 1)²] = 0 (x² + 2x + 1).(x² - 2x + 1) - x².[x² + 2x + 1 + x² - 2x + 1] = 0 (x⁴ - 2x³ + x² + 2x³ - 4x² + 2x + x² - 2x + 1) - x².(2x² + 2) = 0 (x⁴ - 2x² + 1) - 2x⁴ - 2x² = 0 x⁴ - 2x² + 1 - 2x⁴ - 2x² = 0 - x⁴ - 4x² + 1 = 0 x⁴ + 4x² - 1 = 0 Δ = 4² - (4 * - 1) = 16 + 4 = 20 x² = (- 4 ± √20)/2 x² = (- 4 ± 2√5)/2 x² = - 2 ± √5 First case: x² = - 2 + √5 x² = - 2 + √5 x = ± √(- 2 + √5) Second case: x² = - 2 - √5 x² = - (2 + √5) x² = i².(2 + √5) x = ± i.√(√5 + 2)