I solved it in a more brutal way: I rewrite the equation as a polynomial x⁴+2x³-x²-6x-3=0 And by trying I found x⁴+2x³-x²-6x-3= (x²+3x+3)(x²-x-1)=0 Your solution is very elegant! 👏🏻
I also did it this way. To get more specific: As it's a math olympiad question you can assume that you don't need Cardano's formulas. Thus it must be possible to rewrite x^4+2x^3-x^2-3=0 as (x^2+Ax+B)*(x^2+Cx+D) with "nice" A,B,C and D. As B*D=-3, the natural first guesses are (B,D)=(1,-3) or (B,D)=(-1,3). The latter works and results in your equation.
i did it by substitution x+1=u, then multiply both sides by u^2 and put all on left side. We get W(u) =0 where W is polynomial with cooficients being same when being read from left to right and right to left. We have known substitution is such case: u+1/u=y and we get the quadratic polynomial which is trivial.
Fun fact: The real solutions are the golden ratio and negative golden ration plus 1(1.618 and -0.618). phi = (1 + sqrt(5))/2 -phi + 1 = (1 - sqrt(5))/2
Your solution is unnecessarily complicated. If we start by multiplying both sides by (x + 1)² to get rid of the fraction, then we can proceed as follows: x²(x + 1)² + x² = 3(x + 1)² x²(x² + 2x + 2) = 3(x + 1)² ((x² + x + 1) − (x + 1))((x² + x + 1) + (x + 1)) = 3(x + 1)² (x² + x + 1)² − (x + 1)² = 3(x + 1)² (x² + x + 1)² = 4(x + 1)² x² + x + 1 = 2x + 2 ∨ x² + x + 1 = −2x − 2 x² − x − 1 = 0 ∨ x² + 3x + 3 = 0 Now we only need to solve the two quadratics which is easy using the quadratic formula.
I tried to solve it, but i wrote it wrongly. However, i decided to post it 'cause i found interesting completing the squares twice.That's what i solved: x^2+[x^2/(x-1)^2]=3 x^2+[x/(x-1)]^2+2*[x^2/(x-1)]-2*[x^2/(x-1)]=3 [x+x/(x-1)]^2-2*[x^2/(x-1)]=3 [x^2/(x-1)]^2-2*[x^2/(x-1)]=3 [x^2/(x-1)]^2-2*[x^2/(x-1)]+1-1=3 {[x^2/(x-1)]-1}^2=4 case A. [x^2/(x-1)]-1=2 x^2-3x+3=0 x=[3±i*sqrt(3)]/2 case B. [x^2/(x-1)]-1=-2 x^2+x-1=0 x=[-1±sqrt(5)]/2 Bye!
Its long winded usually you work out the brackets first then rearranging using polynomial sequence there must be a easier method than the one you were using.
After going over the steps I could see how it was done but I don't know why. I mean that presented by that equation I wouldn't know to follow that process.
@@garyheron looks confusing because of the double root of the squared term in the denominator. I'll assume you didn't get my joke - do a quick crash course in partial fraction decomposition, then it will make sense whats going on.
@@leif1075 no it is not out of nowhere. from the beginning my thought was to create (x+1)^2 on the left side so that I can write it as (x+1+x/(x+1))^2+something
@@leif1075 I think the last step is obvious ....(x+1+x/(x+1))^2=4 >> x+1+x/(x+1)=+2 or -2 so you get two quadratic equations exactly as in the last minute of the video
2:47 "u" doesn't exist if x=-1 and also all the equation doesn't exist if x=-1, in other words, the domain must be determined before solving any equation and you didn't do that! this is a frequent mistake made by the great learned of our 21st century 😎😊
He didn’t make any mistake lmfao. As long as you go back and check your values in the original equation (in which case the ones he arrived at obviously don’t lead to any issues), then there’s no problem with the solution.
Nice way to convert into degree 2.
I solved it in a more brutal way:
I rewrite the equation as a polynomial
x⁴+2x³-x²-6x-3=0
And by trying I found x⁴+2x³-x²-6x-3=
(x²+3x+3)(x²-x-1)=0
Your solution is very elegant! 👏🏻
I also did it this way. To get more specific: As it's a math olympiad question you can assume that you don't need Cardano's formulas. Thus it must be possible to rewrite x^4+2x^3-x^2-3=0 as (x^2+Ax+B)*(x^2+Cx+D) with "nice" A,B,C and D. As B*D=-3, the natural first guesses are (B,D)=(1,-3) or (B,D)=(-1,3). The latter works and results in your equation.
same
I was amazed by this trick😱 Thanks for showing us!
i did it by substitution x+1=u, then multiply both sides by u^2 and put all on left side. We get W(u) =0 where W is polynomial with cooficients being same when being read from left to right and right to left. We have known substitution is such case: u+1/u=y and we get the quadratic polynomial which is trivial.
I did the exact same!
what?
Fun fact: The real solutions are the golden ratio and negative golden ration plus 1(1.618 and -0.618).
phi = (1 + sqrt(5))/2
-phi + 1 = (1 - sqrt(5))/2
Negative golden ratio +1 is called the golden ratio conjugate and is both equal to 1- phi and -1/phi
Subscribed….and a big thanks so far. Looking forward to more.
One of the solutions is the GOLDEN RATIO.
Your solution is unnecessarily complicated. If we start by multiplying both sides by (x + 1)² to get rid of the fraction, then we can proceed as follows:
x²(x + 1)² + x² = 3(x + 1)²
x²(x² + 2x + 2) = 3(x + 1)²
((x² + x + 1) − (x + 1))((x² + x + 1) + (x + 1)) = 3(x + 1)²
(x² + x + 1)² − (x + 1)² = 3(x + 1)²
(x² + x + 1)² = 4(x + 1)²
x² + x + 1 = 2x + 2 ∨ x² + x + 1 = −2x − 2
x² − x − 1 = 0 ∨ x² + 3x + 3 = 0
Now we only need to solve the two quadratics which is easy using the quadratic formula.
Awesome manupulation❤️
I tried to solve it, but i wrote it wrongly.
However, i decided to post it 'cause i found interesting completing the squares twice.That's what i solved:
x^2+[x^2/(x-1)^2]=3
x^2+[x/(x-1)]^2+2*[x^2/(x-1)]-2*[x^2/(x-1)]=3
[x+x/(x-1)]^2-2*[x^2/(x-1)]=3
[x^2/(x-1)]^2-2*[x^2/(x-1)]=3
[x^2/(x-1)]^2-2*[x^2/(x-1)]+1-1=3
{[x^2/(x-1)]-1}^2=4
case A.
[x^2/(x-1)]-1=2
x^2-3x+3=0
x=[3±i*sqrt(3)]/2
case B.
[x^2/(x-1)]-1=-2
x^2+x-1=0
x=[-1±sqrt(5)]/2
Bye!
I have another simple way to approach it. Let y=x+1, then we will have the following easily: (y+1/y)^2-2(y+1/y)-3=0. The rest are just routines.
The solution can be more easily if first change z=x+1, then we have (z-1)^2+(z-1)^2/z^2=3 => (z+1/z)^2 -2*(z+1/z)=3.
I’m not convinced that arriving at the final equation with your substitution is any easier than what he did in the video.
@@Deathranger999 Dawg literally found final equation in 2 steps.
@@fuji_films How do you get from the first step to the second?
For real answers, this is the same as solving for a right angle triangle with sides x, x/(x+1) and hypotenuse √3.
square root of 3
@@MATHSTHEMATIQUES correct! I have edited accordingly. Thanks!
@@MrLidless👍 don't mention it
How does that help you solve it though?
@@leif1075 I didn’t say it did. What it does do though is give the solver many more tools at their disposal.
Very nice solution. Thank you for sharing 👍
Nice question and excellent explanation! 👍👍
Great solution
Its long winded usually you work out the brackets first then rearranging using polynomial sequence there must be a easier method than the one you were using.
This literally decomposed my brain. I understood a fraction of it, but only partially.
After going over the steps I could see how it was done but I don't know why. I mean that presented by that equation I wouldn't know to follow that process.
@@garyheron looks confusing because of the double root of the squared term in the denominator. I'll assume you didn't get my joke - do a quick crash course in partial fraction decomposition, then it will make sense whats going on.
You can simple take x^2 to the right hand side as 3 - x^2 then cross multiply to get a quadratic 🙄
That’s not true - you get a quartic
@@rupertmillard o yes! thanks
Unexpected golden ratio.
Beautiful
You can substitute u=x+1
Somehow the golden ratio pops up
Nice intuition
Nice solution
Wait how do we know that x+1 is non zero?? Cus we can't cancel then
I guess that it's assumed in the equation
Hi, you just forgot first to set that x shall be different from -1. 😉
good idea thanks
add 2x+1 to both sides. then you have (x+1-x/(x+1))^2+2x = 4+2x and done!
WHY would anyone thinknfonthwt all..don't youbthink that's random and outnof nowhere??
A d you don't explain after that what tondo since now youbhave a quadratic in the denominator..
@@leif1075 no it is not out of nowhere. from the beginning my thought was to create (x+1)^2 on the left side so that I can write it as (x+1+x/(x+1))^2+something
@@leif1075 I think the last step is obvious ....(x+1+x/(x+1))^2=4 >> x+1+x/(x+1)=+2 or -2 so you get two quadratic equations exactly as in the last minute of the video
@@amirb715 right i thought of that too but that's not st all what you said before younsaid add 2x plus 1
А не проще было бы преобразовать левую сторону как сумму квадратов ипосле этого замена очевидная ?
nice solution!
Ну такое решали в Советских яслях...
The answer is the Golden Ratio :)
2:47 "u" doesn't exist if x=-1 and also all the equation doesn't exist if x=-1, in other words, the domain must be determined before solving any equation and you didn't do that! this is a frequent mistake made by the great learned of our 21st century 😎😊
Nobody works like that. We just remember it and reject it if it arises as a solution down the line. This is just being pedantic for no reason
He didn’t make any mistake lmfao. As long as you go back and check your values in the original equation (in which case the ones he arrived at obviously don’t lead to any issues), then there’s no problem with the solution.
Your true brother maybe this question is not that imp but when you work on harder question you often regret not checking the domain
@@jumblefumble 👍
Nice
easy가 어딧지..
어캐했누