The Hardest Quintic Functional Equation 1% Solved

Ajouté : 2026-05-05

[00:00:01]This is a classic invert the expressions functional equations. How do you solve it?

[00:00:10]We are given a quintic domain inserted in this functions f of x^ 5 + 1 / x^ 5 = x. Our goal is to find f of x.

[00:00:26]How do you solve this problem? You can drop the answer in the comment box and let's start by providing the solution to this beautiful functional equations.

[00:00:37]So here is method. So if you note that this domain x^ 5 can also be written as x^ 5 + x^ of - 5 and if I let u to be equ= 1 / x + x or you can say x + 1 / x any of these two it works. And after considering this, let's find from this expression x^ 2 + 1 / x^ 2. So this is the same thing as when we have x + 1 / x all^ 2 - 2 with x^ 2 + 1 / x^ 2. Since we are given u to be = x + 1 /x, this becomes u ^2 - 2.

[00:01:42]And again we can also find the solution to x^ 3 + 1 / x^ 3. This gives us x + 1 / x all^ 3 - 3 into bracket of x + 1 / x.

[00:02:09]Then x + 1 /x is given as u. then raised to power 3 - 3 * u.

[00:02:22]Moving on, let's find the power of 5.

[00:02:27]If we have x^ 5 + 1 / x^ 5.

[00:02:34]So this is the same thing as when we have x^ 2 + 1 / x 2.

[00:02:43]then multiply by x^ 3 + 1 / x^ 3. So we multiply these two expression together and we subtract x + 1 / x from it. We are going to get x^ 5 + 1 / x^ 5. And after obtaining all this so we should note that x^ of 2 + 1 x^2 this is given as u 2 - 2 and again x^ 3 + 1 / x^ 3 that one is given as u^ 3 - 3 u then minus u.

[00:03:32]So considering all this we are going to expand u^ 2 - 2 with u^ 3 - 3 uh which everything at this side is given as x^ 5 + 1 / x^ 5. Let's expand. So we multiply u power of 2 with u^ 3 - 3 u. So we have u power 2 bracket u^ 3 - 3 u then again we have - 2 power u^ 3 - 3 u then minus u again u power 2 * u^ 3 this gives us u^ 5 following the rule of indices then what - u power 2 * - 3 U that gives us - 3 U^ 3 then - 2 * U keep that is - 2 U^ 3 then - then 2 * 3 6 and here is U then - U from here we have u power of 5 this is power of three and this is power of 3 - 3 - 2 that is - 5 then U^ 3 plus here is 6 U. This is U. This gives us 5 U.

[00:05:06]And the of this expression is for U^ 5 + 1 / X^ 5 + 1 / X^ 5.

[00:05:17]Then if the whole of this assume I let the whole of this to be equals t because x^ 5 + 1 / x^ 5 is this that is the doain and if I change everything to t that is our new doain because becomes u^ 5 - 5 u^ 3 + 5 u and let's solve for that.

[00:05:44]So this is what we have. But let's call that from the beginning we let u to be equ= x + 1 / x.

[00:05:56]Then if I cross multiply we are going to have x² - ux + 1 = 0.

[00:06:09]So with this we are going to solve for the value of x and x always satisfy these equations.

[00:06:19]So solving for x quadratically we have x = u + or minus the square root of u ^ 2 - 4 / 2.

[00:06:32]And again since what we have from the beginning f of t equals x.

[00:06:43]So because we have already replace the whole of domain with t then we can have f of t = x and we have t here= u^ 5 - 5 u^ 3 + 5 u. So the t of u is a degree of five in u and is not simply expressed in radical form from t. So if this is what we have then we are going to conclude that the value of f of x by taking the highest power of x. So we are going to have it as f of t equals we have it as u + the square root of u ^ 2 we have it as f of t + t ^ 2 - 4 / 2 then all raised to power of 1 / 5.

[00:07:50]So by changing this t to x we have f of x and from here t is greater than or equals 2. So therefore f of x is x + roo<unk> of x^2 - 4 / 2 then all^ of 1.

[00:08:14]solve with s greater than or equals 2.

[00:08:19]This is how I solve this problem. Though we still have another way of solving this or more than one way. So you can just drop it in the comment box if there's any other ideas on how to solve this problem. See you in the next video and don't forget to subscribe to this channel for more exciting functional equation problem. Bye-bye.