Germany | Can you solve this? | Math Olympiad

[@bonilsson6349]

JI don't understand anything... Remove the exponent in both joints so you have X-2=5 => X = 7 ????? ...

[@Syife-n2z]

كما فعلت أنا

[@santer70]

@bonilsson6349 the problem is that you have to find 6 solution for X cause the exponent is ^6

[@waltertonelli2174]

you must add x-2=-5 -> x=-3 but only if you are looking for real solutions

[@Kwalliteht]

5^6=(-5)^6 And for this reason |x-2|=5 first case: x-2=5 | +2 x=7 don't need to demonstrate this, wou already fond this solution second case: 2-x=5 | -2 -x=3 | *(-1) x=-3 demonstate that the x fits: (-3-2)^6=5^6 -5^6=5^6 15625=15625

[@stpat7614]

Slightly different (and easier) method: (x - 2)^6 = 5^6 (x - 2)^(3*2) = 5^(3*2) ([x - 2]^3)^2 = (5^3)^2 Let a = (x - 2)^3, and b = 5^3 ([x - 2]^3)^2 = (5^3)^2 => a^2 = b^2 => a^2 - b^2 = b^2 - b^2 => a^2 - b^2 = 0 => (a - b)(a + b) = 0 => ([x - 2]^3 - 5^3)([x - 2]^3 + 5^3) = 0 Let a = x - 2, and b = 5 ([x - 2]^3 - 5^3)([x - 2]^3 + 5^3) = 0 => (a^3 - b^3)(a^3 + b^3) = 0 => (a - b)(a^2 + ab + b^2)(a + b)(a^2 - ab + b^2) = 0 => (a - b)(1 * a^2 + b * a + b^2)(a + b)(1 * a^2 - b * a + b^2) = 0 Suppose a - b = 0 a - b = 0 Remember, a = x - 2, and b = 5 (x - 2) - 5 = 0 x - 2 - 5 = 0 x - 7 = 0 x - 7 + 7 = 0 + 7 x = 7 x1 = 7 Suppose 1 * a^2 + b * a + b^2 = 0 1 * a^2 + b * a + b^2 = 0 a = (-b +/- sqrt[b^2 - 4 * 1 * b^2]) / (2 * 1) a = (-b +/- sqrt[b^2 * (1 - 4)]) / (2) a = (-b +/- sqrt[b^2 * (-3)]) / 2 a = (-b +/- sqrt[b^2 * 3 * (-1)]) / 2 a = (-b +/- sqrt[b^2] * sqrt[3] * sqrt[-1]) / 2 a = (-b +/- b * sqrt[3] * i) / 2 Remember, a = x - 2, and b = 5 x - 2 = (-5 +/- 5 * sqrt[3] * i) / 2 x - 2 + 2 = 2 + (-5 +/- 5 * sqrt[3] * i) / 2 x = 2 + (-5 +/- 5 * sqrt[3] * i) / 2 x = 2 * 2 / 2 + (-5 +/- 5 * sqrt[3] * i) / 2 x = 4 / 2 + (-5 +/- 5 * sqrt[3] * i) / 2 x = (4 - 5 +/- 5 * sqrt[3] * i) / 2 x = ([4 - 5] +/- 5 * sqrt[3] * i) / 2 x = (-1 +/- 5 * sqrt[3] * i) / 2 x = (-1 + 5 * sqrt[3] * i) / 2, or x = (-1 - 5 * sqrt[3] * i) / 2 x = -(1 - 5 * sqrt[3] * i) / 2, or x = -(1 + 5 * sqrt[3] * i) / 2 x2 = -(1 - 5 * sqrt[3] * i) / 2 x3 = -(1 + 5 * sqrt[3] * i) / 2 Suppose a + b = 0 a + b = 0 Remember, a = x - 2, and b = 5 (x - 2) + 5 = 0 x - 2 + 5 = 0 x + 3 = 0 x + 3 - 3 = 0 - 3 x = -3 x4 = -3 Suppose 1 * a^2 - b * a + b^2 = 0 1 * a^2 - b * a + b^2 = 0 a = (-[-b] +/- sqrt[(-b)^2 - 4 * 1 * b^2]) / (2 * 1) a = (b +/- sqrt[(-1)^2 * b^2 - 4 * b^2]) / (2) a = (b +/- sqrt[1 * b^2 - 4 * b^2]) / 2 a = (b +/- sqrt[b^2 * 1 - b^2 * 4]) / 2 a = (b +/- sqrt[b^2 * (1 - 4)]) / 2 a = (b +/- sqrt[b^2 * (-3)]) / 2 a = (b +/- sqrt[b^2 * 3 * (-1)]) / 2 a = (b +/- sqrt[b^2] * sqrt[3] * sqrt[-1]) / 2 a = (b +/- b * sqrt[3] * i) / 2 Remember, a = x - 2, and b = 5 x - 2 = (5 +/- 5 * sqrt[3] * i) / 2 x - 2 + 2 = 2 + (5 +/- 5 * sqrt[3] * i) / 2 x = 2 + (5 +/- 5 * sqrt[3] * i) / 2 x = 2 * 2 / 2 + (5 +/- 5 * sqrt[3] * i) / 2 x = 4 / 2 + (5 +/- 5 * sqrt[3] * i) / 2 x = (4 + 5 +/- sqrt[3] * i) / 2 x = ([4 + 5] +/- 5 * sqrt[3] * i) / 2 x = (9 +/- 5 * sqrt[3] * i) / 2 x = (9 + 5 * sqrt[3] * i) / 2, or x = (9 - 5 * sqrt[3] * i) / 2 x5 = (9 + 5 * sqrt[3] * i) / 2 x6 = (9 - 5 * sqrt[3] * i) / 2 {x1, x2, x3, x4, x5, x6} = {7, -(1 - 5 * sqrt[3] * i) / 2, -(1 + 5 * sqrt[3] * i) / 2, -3, (9 + 5 * sqrt[3] * i) / 2, (9 - 5 * sqrt[3] * i) / 2}

[@MohammadrezaAyat]

x_2 =+5 or-5 x= 7 or -3

[@ruedizzle317]

I would have failed this one because I removed the powers since they were the same to get : (x - 2) = 5 , x=5 + 2, x = 7. Wouldn't even have thought to solve anything else 😁

[@MasterTMathsClass]

😊

[@gopinathsamji4324]

X=7 is correct Why this lengthy procedure?

[@stanislavshunin4265]

x=7 it is only one of the solutions. There must be 6 solutions, so to find all of them we need this length procedure

[@SelinaKhatun-vt7nk]

I think, it's solution will be: (x-2)⁶=5⁶ or, ⁶√(x-2)⁶=⁶√5⁶ or, (x-2)=5 or, x=5+2 x=7(Ans)

[@Kwalliteht]

(x-2)^6=5^6 |x-2|=5 x1=x-2=5 x1=7 x2=2-x=5 x2=-3

[@kamdeoray3573]

For those saying the answer is seven youre not wrong but there are 6 possible answers here

[@MatteoButtazzo-yi4in]

x^6+6•(-2x^5)+15•(x^4)•4+20x^3•-8+15x^2•16+6x•-32+64=5^6 ---> 15625 x^6-12x^5+60x^4-160x^3+240x^2-192x+ ( -15561)=0 😱 Boh???? Non ricordo più tutte le altre regole e come si raggruppa😅🤷🏻‍♂️ ☺️ Mi sono impegnato un po'... Po' po'... 🤏🏻... 👋🏻

[@rupertkirchgasser2048]

warum einfach, wenn es kompliziert auch geht!?

[@JagoKhan-t4o]

X=7 is correct 😡😡