A Nice Polynomial Equation

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[MichaelRothwell]

Here's my solution.

Let's solve this equation without restricting to real solutions.

First note that (2x+1)¹⁶=(2x-1)¹⁶ is actually a polynomial of degree 15, as, on expansion, the terms (2x)¹⁶ on each side cancel.

As x=½ is not a solution, 2x-1≠0, and we get
(2x+1)¹⁶=(2x-1)¹⁶
⇔(2x+1)¹⁶/(2x-1)¹⁶=1
⇔[(2x+1)/(2x-1)]¹⁶=1
So (2x+1)/(2x-1) is one of the 16th roots of 1, given by cis(2kπ/16)=cis(kπ/8), k=0 to 15.

In actual fact, as 2x+1≠2x-1, (2x+1)/(2x-1)≠1, so (2x+1)/(2x-1) is one of the other 15 16th roots of 1: one real root (-1) and 7 complex conjugate pairs (cis(±kπ/8), k=1 to 7). This is fine as the polynomial is of degree 15.

(2x+1)/(2x-1)=cis(kπ/8), k=1 to 15
⇔2x+1=cis(kπ/8)(2x-1)
⇔2x+1=2cis(kπ/8)x-cis(kπ/8)
⇔2x(1-cis(kπ/8))=-1-cis(kπ/8)
⇔x=-½(1+cis(kπ/8))/(1-cis(kπ/8)), k=1 to 15
Note that as we excluded the case cis(kπ/8)=1, we have 1-cis(kπ/8)≠0.

So the solutions are:
x=0 [real root, case k=8, when cis(kπ/8)=-1]
or x=-½(1+cis(±kπ/8))/(1-cis(±kπ/8)), k=1 to 7 [7 complex conjugate pairs].

[tylergamingofficial]

Incredible stuff fam loving these math vids

[anuragsoni]

take everything on LHS
(2x+1)^16 -(2x-1)^16=0
now solve this with a^2-b^2= (a+b) . (a-b)
now equate a+b = 0 and a-b = 0 we got 2 sol
now focus on a-b=0 part

no focus on only (2x+1)^8 -(2x-1)^8=0 this can be again re introduced into a sq -b sq formula
keep doing these steps 😘 until you get exponent of 1 ....
and finally you will faa-king get to know that only possible answer is 0 and nothing else 😂😂😂 remaining all solutions will be in complex no

[Daniel-efgg]

Of course, in the complex world there are other 16th roots of unity besides 1 and -1.

[christophecornet]

Uh, yeah. When I see something to the 16 like that. I'm immediately thinking complex numbers

[moeberry]

Lol you can’t have 15 complex solutions there must be another real solution unless x=0 is repeated. You have to find the other 15 solutions Syber. Never mind I just looked at it again this is really a polynomial of degree 15. And since you found one solution there can be 14 complex in other words 7 pairs of conjugates.

[hayaafareen]

in step 3, why is the equation disproving itself?