Solving a friendly algebra problem.

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[kkanden]

"strong vietas formula vibes" is one the best things ive heard recently

[leonhardeuler]

Let them be a, b, c. Then abc=1 and d=a+b+c=ab+bc+ac, so they are the roots of the cubic x^3-dx^2+dx-1. Hence one of them is 1. The only way this can happen is x=2.

[bentationfunkiloglio]

Really interesting! Love seeing equation solving strategies.

[aidenmcdonald]

just so you know the thumbnail is missing a ^2 on the x in sqrt(x-1)/x^2 and x^2/sqrt(x-1)

[fedoslozben]

Really cool trick with main symmetrical polynoms!

[brinzanalexandru]

For the quartic equation we can use Ferrari method:so start by rewriting the initial equation as (x²+1)²=2x²+x, now introduce a constant t such that (x²+1+t)²=2x²+x+t²+2t(x²+1), here notice that if we would expand lhs then nothing will change, now in the lhs we have a perfect square and in the rhs we have a quadratic polynomial which we can rewrite as (2+2t)x²+x+t²+2t now this will be a perfect square only if the discriminant is 0 so after all we get a cubic in t which can be solved by cardano but the only trouble is that we'll have cubic roots of complex numbers, but they can be transformed into reals using demoivres theorem and some trick but it will just be a mess with a bunch of ugly stuff😅.

[heldercomp]

x >= 1. It’s then pretty obvious
x^4 >= x^2 > x-1
since there’s no solution in the first case.

[chaoticoli]

This makes me wonder about solving the more general problem of sums of functions equaling sums of their reciprocals. Is there anything general that can be gleaned?

[emmanuellaurens]

7:10 or you can do it directly just looking at √(x-1)/x² = 1
x is real, so x² is real and the square root also has to be since it is equal to x², so we must have x>=1.
Now for x>=1 we have x²>=x>=√x
and x > x--1 so √x>√(x-1)
putting both together x²>√(x-1) and they are both positive so for x>=1 sqrt(x-1)/x² <1 and there are no real solutions.

[roberttelarket]

It's a good place to stop being unfriendly to this problem!

[kehrierg]

I tried the substitution u = sqrt(x-1) and got a gnarly 9th degree polynomial equation, which the rational zero theorem told me could only have roots of +/-1. I thought, if there is a god, 1 will be a root of the thing. It was. After factoring out (u-1) and getting a degree 8 polynomial, I could show there were no more real roots. So the only real solution for x was that corresponding to my u = 1 which was x = 2.

[ScienceTalkwithJimMassa]

x=2 makes sense simply from domain considerations. From the equation, we know that x =/= 0 and has to be >=1. But with the sqrt(x-1) in denominator, x has to be > 1. So domain for any real solutions has to be x>1. x=2 satisfies this requirement.

[robshaw]

I have a funny feeling this might be a repeat problem on the channel

[dkravitz]

For the third equation, all you have to do is say that the left side is above the right side on the y axis, and also that the left side has a bigger derivative for x greater than zero

[zuzaaa]

While watching it I had a very strong feeling that you had done this problem already. I can't find exact episode but I'm sure I have seen this trick with substitution and that this new variables multiply to 1

[zakiabg]

True it's a hard equation but not impossible.

Wolfram Alpha really doesn't think that quadratic polynomial has easily expressible complex roots. A single one takes the better part of my screen

[CTJ]

You’ve done this problem before or one very very similar

[Happy_Abe]

How would one find the complex roots to the quartic?

[YotYot]

Is this a reupload? I think I saw it previously