Math Olympiad | A Nice Rational Equation | 95% Failed to solve!

[@assiya3023]

تمرين صعب شكرا أستاذ

[@nasrullahhusnan2289]

Multiply by x⁵: 16x⁴-8x³+4x²-2x+1=32x⁵ 1-2x+(2x)²-(2x)³+(2x)⁴-(2x)⁵=0 LHS is sum of 5 elements of a geometric series with 1st element 1 and ratio (-2x). Thus [1-(-2x)⁶]/[1-(-2x)]=0 --> [1+(2x)⁶]/(1+2x)=0 1-(2x)⁶=0 --> [1+(2x)³][1-(2x)³]=0 If x is real: • 1+(2x)³=0 --> x=-½ • 1-(2x)³=0 --> x=½ Therefore x=±½ If x can be imaginary number then • 1+(2x)³=0 (1+2x)[1-(2x)+(2x)²]=0 The imaginary x is from 1-(2x)+(2x)²=0 --> 2x=½[1±isqrt(3)] x=¼[1±isqrt(3)] • Similarly for 1-(2x)³=0 (1-2x)[1+(2x)+(2x)²=0 The imaginary x is from 1+(2x)+(2x)²=0 --> 2x=½[-1±isqrt(3)] x=¼[-1±isqrt(3)]

[@kareolaussen819]

With y=-2x the equation can be written y^5+y^4+y^3+y^2+y+1=0, or (y^6-1)/(y-1)=0 The five solutions are all solutions of y^6=1 except y=1. I.e., x = (1/2)*exp(iπn/3), for n= -2, -1, 0, 1, 2.

[@davidshen5916]

Y=1/（2X）， Y-Y^2+Y^3-Y^4+Y^5=1，（Y-1）（Y^4+Y^2+1）=0，Y=1， X=1/2

[@key_board_x]

(16/x) - (8/x²) + (4/x³) - (2/x⁴) + (1/x⁵) = 32 (16x⁴ - 8x³ + 4x² - 2x + 1)/x⁵ = 32 16x⁴ - 8x³ + 4x² - 2x + 1 = 32x⁵ 32x⁵ + 8x³ + 2x - 16x⁴ - 4x² - 1 = 0 (32x⁵ + 8x³ + 2x) - (16x⁴ + 4x² + 1) = 0 2x.(16x⁴ + 4x² + 1) - (16x⁴ + 4x² + 1) = 0 (2x - 1).(16x⁴ + 4x² + 1) = 0 First case: (2x - 1) = 0 2x - 1 = 0 2x = 1 → x = 1/2 Second case: (16x⁴ + 4x² + 1) = 0 16x⁴ + 4x² + 1 = 0 16x⁴ + 4x² = - 1 (4x²) + 2.[4x² * (1/2)] = - 1 (4x²) + 2.[4x² * (1/2)] + (1/2)² = - 1 + (1/2)² [4x² + (1/2)]² = - 3/4 [4x² + (1/2)]² = 3i²/4 4x² + (1/2) = ± (i√3)/2 4x² = - (1/2) ± (i√3)/2 4x² = (- 1 ± i√3)/2 x² = (- 1 ± i√3)/8 x² = [- 2 ± 2i√3]/16 x² = [- (3 - 1) ± 2i√3]/16 x² = [- 3 + 1 ± 2i√3]/16 x² = [1 ± 2i√3 - 3]/16 x² = [1 ± 2i√3 + 3i²]/16 x² = [1 ± 2i√3 + (i√3)²]/16 x² = [1 ± i√3]²/4² x² = [(1 ± i√3)/4]² x = ± (1 ± i√3)/4 First case: x = (1 ± i√3)/4 → x = (1 + i√3)/4 → x = (1 - i√3)/4 Second case: x = - (1 ± i√3)/4 → x = (- 1 + i√3)/4 → x = (- 1 - i√3)/4

[@mchang7901]

here is an easier solution: Let a=1/x, and move 32 to the other side, and rewrite the equation with high power to the low power: (a**5 - 2*a**4 + 4*a**3) - (8*a**2 +16*a - 32) =a**3 *(a**2-2*a+4) - 8*(a**2-2*a+4) =(a**2-2*a+4) (a**3-8) =(a-2)**2 * (a**3-8) so a=2 for real solution

[@-basicmaths862]

Other 4ruts are complex number.

[@-basicmaths862]

x=1/2

[@Rocio62154]

SENSATIONAL!

[@RealQinnMalloryu4]

(16)^2/(x)^2 ➖ (8)^2/(x^2)^2={256/x^2 ➖ 64/x^4}=192/x^2 +{4+4 ➖ }/}x^3+x^3 ➖ }={192/x^2+8/x^6}=200/x^8 ➖ (2))^2/(x^4)^2={200/x^8 ➖ 4/x^16}=196/x^8+{1+1 ➖ }/{x^5+x^5 ➖ }={196/x^8+2/x^10}=198/x^18=11 (x ➖ 11x+1).

[@ganeshdas3174]

x = 1/2