I always try to find simple solutions first. Trying x = 1 gives 0^8 + 3^4 = 3^4, so x = 1 is the first solution. Now I will try to convert the octic to a quartic. Set y = (x - 1)^2, which means that y = 0 for the first solution Then -x^2 + 3x + 1 = x + 2 - (x^2 - 2x + 1) = x + 2 - y Let z = x + 2 and we have y^4 + (z - y)^4 = z^4 Expanding gives y^4 + z^4 - 4z^3y + 6z^2y^2 - 4zy^3 + y^4 = z^4 Canceling like terms and factoring 2y(2y^4 - 4zy^3 + 6z^2y^2- 4z^3y) = 0 Dividing by 2y gives the cubic y^3 - 2zy^2 + 3z^2y - 2z^3 = 0 As noted earlier, the first solution is y = 0, which yields (x - 1)^2 = 0 or x = 1 twice as TWO roots of this equation. One solution to the remaining cubic equation is y = z. Thus (x - 1)^2 = x^2 - 2x + 1 = x + 2 or x^2 - 3x - 1 = 0, which has solutions x = [3 ±√(9 +4)]/2 = (3 ± √13)/2 Applying synthetic division by (y - z) to the above cubic equation yields y^2 - yz + 2z^2 = 0 The quadratic solution gives y = [z ±√(z^2 - 8z^2)]/2 = z(1 ± i√7)/2 where i = √(-1) This yields the equations (x - 1)^2 = (x + 2)(1 ± i√7)/2 Expanding gives x^2 - 2x + 1 = [(1 ± i√7)/2]x + (1 ± i√7) or x^2 - [(5 ± i√7)/2]x ∓ i√7 = 0 I’ll leave it to a complex number calculator to evaluate these four complex roots using the quadratic formula.
Very interesting. Though it is difficult to predict (or notice) in advance, that summing (x-1)² and 3x - x² + 1 on the left gives x + 2 on the right. It is obviously kind of made up example.
Subtracting (x-1)^8 and factoring the difference of squares gives something of the form A^4 = B^2 A^2.
I always try to find simple solutions first.
Trying x = 1 gives 0^8 + 3^4 = 3^4, so x = 1 is the first solution.
Now I will try to convert the octic to a quartic.
Set y = (x - 1)^2, which means that y = 0 for the first solution
Then -x^2 + 3x + 1 = x + 2 - (x^2 - 2x + 1) = x + 2 - y
Let z = x + 2 and we have
y^4 + (z - y)^4 = z^4
Expanding gives
y^4 + z^4 - 4z^3y + 6z^2y^2 - 4zy^3 + y^4 = z^4
Canceling like terms and factoring
2y(2y^4 - 4zy^3 + 6z^2y^2- 4z^3y) = 0
Dividing by 2y gives the cubic
y^3 - 2zy^2 + 3z^2y - 2z^3 = 0
As noted earlier, the first solution is y = 0, which yields (x - 1)^2 = 0 or x = 1 twice as TWO roots of this equation.
One solution to the remaining cubic equation is y = z. Thus
(x - 1)^2 = x^2 - 2x + 1 = x + 2 or
x^2 - 3x - 1 = 0, which has solutions
x = [3 ±√(9 +4)]/2 = (3 ± √13)/2
Applying synthetic division by (y - z) to the above cubic equation yields
y^2 - yz + 2z^2 = 0
The quadratic solution gives
y = [z ±√(z^2 - 8z^2)]/2 = z(1 ± i√7)/2 where i = √(-1)
This yields the equations
(x - 1)^2 = (x + 2)(1 ± i√7)/2
Expanding gives
x^2 - 2x + 1 = [(1 ± i√7)/2]x + (1 ± i√7)
or
x^2 - [(5 ± i√7)/2]x ∓ i√7 = 0
I’ll leave it to a complex number calculator to evaluate these four complex roots using the quadratic formula.
Solving quartic for reaching to the complex.
x^4 - 5x^3 + 8x^2 + 7x + 7 = 0 (for k = (-3 +- i√7)/4)
substitute x = y + 5/4
then,
y^4 - (11/8)y^2 + (91/8)y + (5357/256) = 0 ...(A)
the resolvent cubic for this
z^3 - (11/4)z^2 - (1309/16)z - (8281/64) = 0
one of solution z = 9/4 + √79 (positive value)
so equation (A) is factorized to two quadratic
y^2 - y√(9/4 + √79) + (7 + 8√79)/16 + 91/(16√(9/4 + √79)) = 0
and
y^2 + y√(9/4 + √79) + (7 + 8√79)/16 - 91/(16√(9/4 + √79)) = 0
Solving these, we get 4 complex.
y = (1/4)√(9 + 4√79) +- i(1/2)√[{623 + 14√79 + 91√(9 + 4√79)}/(18 + 8√79)]
y = -(1/4)√(9 + 4√79) +- i(1/2)√[{(623 + 14√79 - 91√(9 + 4√79)}/(18 + 8√79)]
Yes.
Excelente
Huey Lewis said it's hip 2b².
Very interesting. Though it is difficult to predict (or notice) in advance, that summing (x-1)² and 3x - x² + 1 on the left gives x + 2 on the right. It is obviously kind of made up example.
So Interesting 👍👍🙏🏻🙏🙏🫡🫡🫡🇮🇳
2a^2+3ab+2b^2=(a+b)^2 + (a+b/2)^2 + 3b^2/4 >0 since both a and B cannot be zero simultaneously. So no real roots from this factor.
❤❤❤❤❤❤