Let's Solve A Nice Polynomial Equation | Math Olympiads

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[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

[wes]

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

[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.

[allanmarder]

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.

[gdmathguy]

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

[maximusmeridius]

Whoever came up with this substitution is a genius

[WahranRai]

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

[goldfing]

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.

[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

[peterburkhard]

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

[artandata]

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