Let's Solve A Nice Polynomial Equation | Math Olympiads

[@NadiehFan]

As has already been noted here, the best approach seems to be to create symmetry. We have (x + (2 − x))/2 = 2/2 = 1 so the quantities x and 2 − x are equidistant from 1, which is their average. Half their difference is (x − (2 − x))/2 = (2x − 2)/2 = x − 1 so we have x = 1 + (x − 1) and 2 − x = 1 − (x − 1) and we can therefore set x − 1 = z which gives x = 1 + z and 2 − x = 1 − z so the equation becomes (1 + z)⁵ + (1 − z)⁵ = 82 which gives z⁴ + 2z² − 8 = 0 which is a quadratic in z² that is easily solved. (z² − 2)(z² + 4) = 0 z = √2 ∨ z = −√2 ∨ z = 2i ∨ z = −2i and since x = 1 + z this gives x = 1 + √2 ∨ x = 1 − √2 ∨ x = 1 + 2i ∨ x = 1 − 2i

[@bobbyheffley4955]

1+sqrt 2 is the silver ratio.

[@wes9627]

Substitute x=y+1 into the given equation and rearrange to (y+1)^5-(y-1)^5-82=0. Pascal's triangle: 1 5 10 10 5 1 The odd powers of y cancel out leaving 2(5y^4+10y^2+1)-82=0 or y^4+2y^2-8=0 Thus, y^2=(-2±6)/2=2 or -4 and y=±√2 or ±2i. It follows that x=y+1=1±√2 or 1±2i

[@molodoy564]

Nice , I also decided

[@MrGeorge1896]

Yes, I took the same path. It is much more straight forward and less error prone.

[@NadiehFan]

Regarding your first method: the quartic equation x⁴ − 4x³ + 8x² − 8x − 5 = 0 is easily solved by completing the square twice. You should really have done that because it is quite instructive. First complete the square with respect to the quartic and cubic terms, which gives us (x² − 2x)² + 4x² − 8x − 5 = 0 Now observe that we can factor out 4 from the quadratic and linear terms which gives (x² − 2x)² + 4(x² − 2x) − 5 = 0 Now again complete the square with respect to the quadratic and linear terms which gives ((x² − 2x) + 2)² − 4 − 5 = 0 (x² − 2x + 2)² − 3² = 0 And yes, we have a difference of two squares at the left hand side so we get (x² − 2x + 2 − 3)(x² − 2x + 2 + 3) = 0 (x² − 2x − 1)(x² − 2x + 5) = 0 x² − 2x − 1 = 0 ∨ x² − 2x + 5 = 0 (x − 1)² = 2 ∨ (x − 1)² = −4 x = 1 + √2 ∨ x = 1 − √2 ∨ x = 1 + 2i ∨ x = 1 − 2i and we are done.

[@maximusmeridius2060]

Whoever came up with this substitution is a genius

[@gdmathguy]

That looks like such a satisfying solution. I love this problem

[@SyberMath]

Thanks!

[@mcwulf25]

Good use of substitutions 👍

[@allanmarder456]

If you assume the quartic x^4 - 4x^3 +8x^2 -8x -5 can be factored as 2 quadratics with integer coefficients it's not that bad. (x^2 +bx +c)(x^2 +dx +e) gives the following coefficients b+d=-4 c +e +bd =8 ce =-5. Fortunately 5 is prime so if you assume integer coefficients you have (c,e) = 1,-5 -5,1 -1,5 5,-1 and you have to check 4 cases. It turned out -1,5 was a good choice. That gave the system b+d =-4 4 +bd =8 or b+d =-4 and bd=4. b=-2 d=-2 works and also c=-1 and e=5. The coefficient for x is be +cd = -8 and (5)(-2) + (-1)(-2) =-8 so (x^4 - 4x^3 + 8x^2 - 8x -5) factors as (x^2 - 2x -1)(x^2 - 2x + 5). Now just apply the quadratic formula.

[@farhansadik5423]

nice! I loved this one! though, is it safe to not assign a coefficient to x^4 in this case? I mean i can see that the coefficient will be 1, so it's not needed, but does that ever cause problems?

[@allanmarder456]

@@farhansadik5423 Thanks for the comment. I've never experienced a problem like that. But I always check afterward to make sure the factors check out.

[@farhansadik5423]

@@allanmarder456 yeah, that's what I think should happen, i'm still learning. Thanks for the knowledge!

[@goldfing5898]

x^5 + (2 - x)^5 = 82 x^5 + ((-(x - 2))^5 = 82 x^5 - (x - 2)^5 = 82 x^5 = (x - 2)^5 + 82 In such cases, it is often advantegeous to substitute the average of the two terms: t = x - 1 So we can replace x = t + 1 x - 2 = t - 1 We now get the equation (t + 1)^5 = (t - 1)^5 + 82 t^5 + 5t^4 + 10t^3 + 10t^2 + 5t + 1 = = t^5 - 5t^4 + 10t^3 - 10t^2 + 5t - 1 + 82 We subtract t^5 and 10t^3 and 5t: 5t^4 + 10t^2 + 1 = - 5t^4 - 10t^2 + 81 10t^4 + 20t^2 - 80 = 0 t^4 + 2t^2 - 8 = 0 A very was to solve biquadeatic: (t^2 + 4)(t^2 - 2) = 0 Either t^2 = -4 t = +- 2i x = 1 +- 2i or t^2 = 2 t = +- sqrt(2) x = 1 +- sqrt(2) Two real and two complex solutions.

[@WahranRai]

If y is a solution then 2- y is also a solution (equation doesn't change if replace x by 2-x )

[@gametimewitharyan6665]

y^5 + (2-y)^5 does not equal (2-y)^5 + (-y)^5 Because y^5 does not equal (-y)^5

[@filipeoliveira7001]

@@gametimewitharyan6665but (2-(2-y)) = (2-2+y) = y, so he is correct

[@gametimewitharyan6665]

@@filipeoliveira7001 Oh my bad, thanks

[@filipeoliveira7001]

@@gametimewitharyan6665 np homie!

[@hktundra]

if you are looking for roots in complex numbers, then there must be exactly five roots for the original equation, if counted with their multiplicities

[@MrGeorge1896]

So you think x⁴ - x⁴ + x = 1 has more solutions than just x = 1?

[@artandata]

I've got a question: at time= 8:50 isn't it x²-2x +1 = 0 ? thanks!

[@peterburkhard8700]

Set u=1-x. All odd powers of u are eliminated. Bi quadratic in u. 4 steps to solution

[@scottleung9587]

Cool!

[@jcfgykjtdk]

z=x-1

[@dwaindean1730]

🤪 "promosm"