There are comments that the given equation is quintic at the beginning.Remember the degree of the polynomial is defined only in the standard form and hence quartic.The technique used to solve is very clever.I intially thought why two variables while a single suffice in the substitution.It does the magic.
@@SyberMath yea but WHY ..I thought of replacing each term with u and v but then i didn't think to substract v from u to get 1..why on warth would anyone see that kr think of that..i don't think they would so why do that?? Why not do difference of two fifth powers at that point like you do differnce of two cubes..
Along lines that others have mentioned: you could just put 'y = x - 3'. Then 'y^5' and the const term drop out, leaving: '5y^4 + 10y^3 + 10y^2 + 5y' = 0. hence 'y(y^3 + 2y^2 + 2y +1) = 0', y(y + 1) (y^2 + y + 1) = 0, so' y = 0, y = -1' plus complex sol'ns, and so ' x = 3, x = 2'. However, I am quite partial to Syber's treks along meandering paths just to enjoy the sights!
(x-2)^5- (x-3)^5=1 1^5+(x-3)^5=(x-2)^5 By Fermat last theorem a^5+b^5=/=c^5 a,b,c are positive integers But one of a , b is 0then Other number is equal to c a is 1so b is equal to 0 (x-3)^5=0 x-3=0 x=3 Now (x-2)^5+(3-x)^5=1 Here we take (x-2)^5=0 x-2=0 x=2 x=(2,3)
I realize this is like a year late, but by shifting by 2.5, i.e. letting y = x - 5/2, it becomes a quadratic equation in y^2, which will solve for y^2 = 1/4 and y^2 = - 3/4, giving the two real solutions x = 2, 3 and the complex solutions 2.5 +- sqrt(3)/2 i
x - 2 = 1 and x - 3 = 0 is the first obvious solution. Once I worked through the resulting cubic, I found the other obvious solution: x - 2 = 0 and x - 3 =-1. The remainder quadratic gave me a pair of irrational, complex conjugate solutions.
@@jingyuanzhou9061 What I did now may differ from what I did ten months ago. But to make life easier, I let u = x - 2 and u-1 = x - 3. Then I factored out the product u(u-1) from u^5 - (u-1)^5 - 1. Assuming no mistakes, the polynomial factor was u^2 - u + 1. The quantity under the square root in the quadratic formula is a negative number, -3. The square root is imaginary irrational. The plus-minus makes the two solutions complex-conjugate.
I did the same, and actually think this approach requires less work, although the one in the video is "cleverer". Writing u= x-3: (u+1)^5 - u^5 = 1 = 5u^4 + 10u^3 + 10u^2 + 5u + 1 5u^4 + 10u^3 + 10u^2 + 5u = 0 5u(u^3+2u^2+2u+1) = 0 So u=0 or u^3+2u^2+2u+1 = 0 . From the coefficients, that looks very much like (u+1) is a factor - you get (u+1)(u^2+u+1) = 0 So u=-1 or u^2+u+1. That quadratic has complex solutions at u = (-1 +/- i.sqrt(3))/2 Since x = u+3, we have x = 3 or 2 or (5 +/- i.sqrt(3))/2
Right in the beginning if we substitute x=2 or x=3, equation is satified. So they are real solutions for this equation. Why go further.and even if you want to find out complex solution use remainder theorem find the the remaining factors.
Isn't this overkill, at least when looking for integer solutions? Because if the difference of two quintics is 1 then one of them is 0^5 and the other 1^5 or (-1)^5.
@@angelmendez-rivera351 I hadnt watched the whole video yet sorry. But it is a quintic equation at the beginning because it's a difference two fifth powers. So you could accurately call it quintic also.
@@leif1075 Well, by similar logic, couldn't you also claim that x^2 - 1 = 0 is a degree 10 polynomial? Just multiply everything by x^8 to get: x^10 - x^8 = 0. Surely you can see the problem with this. I could say that I am maybe 7.5 feet tall because that's how high I can reach my hand. But, we have standards for these things that at least most of us agree to follow in order to be able to communicate clearly and meaningfully. As I understand it, we must first put the expression into standard polynomial form to determine its degree. I may be wrong somehow, though. I welcome constructive criticism.
We may easily see that the polynomial: p(x) = (x-2^5 - (x-3)^5 - 1 has 2 zeros: x=2 and x=3. Therefore, we may factorize : p(x) in the following way: p(x) = (x-2) (x-3) (4x^2-13x+31) The equation: p(x)=0 has therefore 2 real solutions: x_1=2 , x_2=3 and 2 complex ones: x_3,4 = (13 ± i√327) / 8
@@SyberMath Well, your method isn't necessarily more complicated. For some people, it may be helpful to use a change in variables, since this simplifies the equations visually. But for those who appreciate a more straightforward method, doing it like this is completely doable.
What if you had (x-2)^5 + (x-3)^5 = 1? One obvious solution is x=3. Now x=2 is not a solution any more. Do we have other real solutions? This is a more difficult problem to consider.
@@SyberMath I used calculus... the function f(x) = (x-2)^5 + (x-3)^5 - 1 is strictly increasing, as its derivative is positive (sum of 4-th powers) and can not have more than one root, the obvious root x=3
There are comments that the given equation is quintic at the beginning.Remember the degree of the polynomial is defined only in the standard form and hence quartic.The technique used to solve is very clever.I intially thought why two variables while a single suffice in the substitution.It does the magic.
Good thinking! Thank you for your comment.
@@SyberMath DAMNYOUARE SOOOO AWESOME.
@@SyberMath yea but WHY ..I thought of replacing each term with u and v but then i didn't think to substract v from u to get 1..why on warth would anyone see that kr think of that..i don't think they would so why do that?? Why not do difference of two fifth powers at that point like you do differnce of two cubes..
Along lines that others have mentioned: you could just put 'y = x - 3'.
Then 'y^5' and the const term drop out, leaving:
'5y^4 + 10y^3 + 10y^2 + 5y' = 0. hence 'y(y^3 + 2y^2 + 2y +1) = 0',
y(y + 1) (y^2 + y + 1) = 0, so' y = 0, y = -1' plus complex sol'ns, and so ' x = 3, x = 2'.
However, I am quite partial to Syber's treks along meandering paths just to enjoy the sights!
set y = x-3, use y to rewrite the equation
(x-2)^5- (x-3)^5=1
1^5+(x-3)^5=(x-2)^5
By Fermat last theorem
a^5+b^5=/=c^5
a,b,c are positive integers
But one of a , b is 0then
Other number is equal to c
a is 1so b is equal to 0
(x-3)^5=0
x-3=0
x=3
Now
(x-2)^5+(3-x)^5=1
Here we take
(x-2)^5=0
x-2=0
x=2
x=(2,3)
別解 x-3=tとおくと、5t(t^3+2t ^2+2t+1)=0 、変形すると、5t(t+1)(t^2+t+1)=0 、これより、x-3=0 、 x---3= ---1 、x-3=(-1±√3i)/2 、解として、x=3 、x=2 、x= (5±√3i)/2
I realize this is like a year late, but by shifting by 2.5, i.e. letting y = x - 5/2, it becomes a quadratic equation in y^2, which will solve for y^2 = 1/4 and y^2 = - 3/4, giving the two real solutions x = 2, 3 and the complex solutions 2.5 +- sqrt(3)/2 i
I solved in similar way.
Let t/2=x-5/2.
((t+1)/2)^5-((t-1)/2)^5=1
(t+1)^5-(t-1)^5=32
t^4+2t^2-3=0
t^2=1, t^2=-3.
x - 2 = 1 and x - 3 = 0 is the first obvious solution. Once I worked through the resulting cubic, I found the other obvious solution: x - 2 = 0 and x - 3 =-1. The remainder quadratic gave me a pair of irrational, complex conjugate solutions.
In your method, how do you confirm "The remainder quadratic gave me a pair of irrational, complex conjugate solutions"?
@@jingyuanzhou9061 What I did now may differ from what I did ten months ago. But to make life easier, I let u = x - 2 and u-1 = x - 3. Then I factored out the product u(u-1) from u^5 - (u-1)^5 - 1. Assuming no mistakes, the polynomial factor was u^2 - u + 1. The quantity under the square root in the quadratic formula is a negative number, -3. The square root is imaginary irrational. The plus-minus makes the two solutions complex-conjugate.
nice approach, i did using u and u+1...
same 2 solutions of x = { 2 , 3 }.
i also got the complex solutions, x = 1/2 (5 +/- i.sqrt(3))...
Very good!
I did the same, and actually think this approach requires less work, although the one in the video is "cleverer". Writing u= x-3:
(u+1)^5 - u^5 = 1 = 5u^4 + 10u^3 + 10u^2 + 5u + 1
5u^4 + 10u^3 + 10u^2 + 5u = 0
5u(u^3+2u^2+2u+1) = 0
So u=0 or u^3+2u^2+2u+1 = 0 . From the coefficients, that looks very much like (u+1) is a factor - you get
(u+1)(u^2+u+1) = 0
So u=-1 or u^2+u+1. That quadratic has complex solutions at u = (-1 +/- i.sqrt(3))/2
Since x = u+3, we have x = 3 or 2 or (5 +/- i.sqrt(3))/2
I expressed 1 as (x-2)-(x-3) then took all terms on lhs , then factored using formulas and identities and I got the same solutions.
Right in the beginning if we substitute x=2 or x=3, equation is satified. So they are real solutions for this equation. Why go further.and even if you want to find out complex solution use remainder theorem find the the remaining factors.
Isn't this overkill, at least when looking for integer solutions? Because if the difference of two quintics is 1 then one of them is 0^5 and the other 1^5 or (-1)^5.
It probably is!
The complex solutions are from taking p as -1 right?
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A great technique is used here! It's not very common to increase the number of variables. Seems like a step backward but it's for a good cause! 😁
The video should say quintic not quartic right?
Good question! If you consider the difference, you will notice that it's actually quartic, not quintic. 😊
@@leif1075 The equation is a quartic, because after you simplify, the resulting polynomial is degree 4.
@@angelmendez-rivera351 I hadnt watched the whole video yet sorry. But it is a quintic equation at the beginning because it's a difference two fifth powers. So you could accurately call it quintic also.
@@leif1075 Well, by similar logic, couldn't you also claim that x^2 - 1 = 0 is a degree 10 polynomial? Just multiply everything by x^8 to get:
x^10 - x^8 = 0. Surely you can see the problem with this. I could say that I am maybe 7.5 feet tall because that's how high I can reach my hand. But, we have standards for these things that at least most of us agree to follow in order to be able to communicate clearly and meaningfully. As I understand it, we must first put the expression into standard polynomial form to determine its degree. I may be wrong somehow, though. I welcome constructive criticism.
why not taking Log for both equation ends?
We may easily see that the polynomial: p(x) = (x-2^5 - (x-3)^5 - 1 has 2 zeros: x=2 and x=3. Therefore, we may factorize : p(x) in the following way:
p(x) = (x-2) (x-3) (4x^2-13x+31)
The equation: p(x)=0 has therefore 2 real solutions: x_1=2 , x_2=3 and 2 complex ones: x_3,4 = (13 ± i√327) / 8
3 is also a valid solution, as it is another real root. Plug 3 in and you get 3 - 2 - 3 + 3 = 1.
Just substitute y = x + 2.5 and you’ll get there so much faster.
Increasing the number of variables is not necessary either. You can actually solve it directly by rewriting the equation as (x - 2)^5 = (x - 3)^5 + 1 = [(x - 3) + 1]·[(x - 3)^4 - (x - 3)^3 + (x - 3)^2 - (x - 3) + 1] = (x - 2)·[(x - 3)^4 - (x - 3)^3 + (x - 3)^2 - (x - 3) + 1]. There is a common factor of x - 2, which indicates that x = 2 solves the equation. To find the other solutions now, simply divide by x - 2, since finding the other solutions implies |x - 2| > 0. Therefore, (x - 2)^4 = (x - 3)^4 - (x - 3)^3 + (x - 3)^2 - (x - 3) + 1. In a similar fashion as in the beginning, this equation can be rearranged for further factorization as (x - 2)^4 - 1 = (x - 3)^4 - (x - 3)^3 + (x - 3)^2 - (x - 3) = (x - 3)·[(x - 3)^3 - (x - 3)^2 + (x - 3) - 1] = [(x - 2) - 1]·[(x - 2) + 1]·[(x - 2)^2 + 1] = (x - 3)·(x - 1)·[(x - 2)^2 + 1]. There is a common factor of x - 3, indicating that x = 3 solves the equation. To find other solution, simply divide by x - 3, since now |x - 3| > 0. Therefore, (x - 1)·[(x - 2)^2 + 1] = (x - 3)^3 - (x - 3)^2 + (x - 3) - 1 = (x - 2)^3 + (x - 2)^2 + (x - 2) + 1. Now the equation can be rearranged yet again as -[(x - 3)^2 - (x - 3) + 1] = (x - 2)^3 - (x - 3)^3 + (x - 2)^2 + (x - 2) + 1 = [(x - 2) - (x - 3)]·[(x - 2)^2 + (x - 2)·(x - 3) + (x - 3)^2] + (x - 2)^2 + (x - 2) + 1 = (x - 2)^2 + (x - 2)·(x - 3) + (x - 3)^2 + (x - 2)^2 + (x - 2) + 1, and rearrange again as (x - 3) - 1 = 2·(x - 2)^2 + (x - 2)·(x - 3) + 2·(x - 3)^2 + (x - 2) + 1, and finally rearrange as 2·(x - 2)^2 + (x - 2)·(x - 3) + 2·(x - 3)^2 + (x - 2) - (x - 3) + 2 = 2·(x - 2)^2 + (x - 2)·(x - 3) + 2·(x - 3)^2 + 3 = 0. Now it is merely a matter of expanding the quadratic polynomials, which is much better than expanding cubic ones, so 2·(x^2 - 4·x + 4) + (x^2 - 5·x + 6) + 2·(x^2 - 6·x + 9) + 3 = 5·x^2 - 25·x + 35 = x^2 - 5·x + 7 = 0. x^2 - 5·x + 7 = x^2 - 5·x + 25/4 + 7 - 25/4 = (x - 5/2)^2 + 3/4 = 0, implying (x - 5/2)^2 = -3/4, implying x - 5/2 = i·sqrt(3)/2 or x - 5/2 = -i·sqrt(3)/2, implying x = [5 + i·sqrt(3)]/2 or x = [5 - i·sqrt(3)]/2. Since this is a quartic equation, the complete set of solutions is {2, 3, [5 - i·sqrt(3)]/2, [5 + i·sqrt(3)]/2}.
You're absolutely right! I like to complicate things! 😄
@@SyberMath Well, your method isn't necessarily more complicated. For some people, it may be helpful to use a change in variables, since this simplifies the equations visually. But for those who appreciate a more straightforward method, doing it like this is completely doable.
that was so cool
The two solutions are obvious. [1-0=1 et 0-(-1)=1]
I propose (x-2)^5-(x-3)^5=11 which have also two real solutions.
Idem for (x-2)^5-(x-3)^5=61 or 151
You dont use p = -1 why ???????????????????
because that doesn't give us real solutions. You can try and see.
Thank u now i get it
@@SyberMath NO WE CAN LOOK FOR COMPLEX SOLUTIONS!!!!!!
putting x-2 = t and expanding (x-3)^5=(t-1)^5 using binomial will make it easier than this one and will easily give the imaginary roots also.
Interesting!
@@SyberMath oh thanks....your videos are more interesting keep going 👍🏻👍🏻
@@zakirreshi6737 Np. Thanks! I will!
@@SyberMath Do you really think anyone would do that crazy manipukarion at 4:08?? I don't see why anyone would rewrite it that way..
I could see at a glance that the equation would work for x=2 and x=3...
What if you had (x-2)^5 + (x-3)^5 = 1? One obvious solution is x=3. Now x=2 is not a solution any more. Do we have other real solutions? This is a more difficult problem to consider.
Not that hard. Replace x with y+5/2 and you'll get a nice equation
@@SyberMath I used calculus... the function f(x) = (x-2)^5 + (x-3)^5 - 1 is strictly increasing, as its derivative is positive (sum of 4-th powers) and can not have more than one root, the obvious root x=3
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