You made a math error in the 1st method since the right hand side " = 1" got changed to " = 0" by mistake. Thus, the constant term on the left hand side should simplify to +84 and not +85. It doesn't matter much since you stopped trying to solve the problem at that point and switched to the 2nd method, but it would be a big deal if you had continued solving the 1st method.
Do you know what? After following you for some time now, I immediately went for the substitution you used. Method 2. As you say, how powerful! And to be still learning and improving at 81 is great. Thank you so much. Fittingly, my age is 3^4 and in a few more days it will be 2025, which is of course a perfect square. Better still, 2025 = (5^2) x (3^4).
I'm also 81 and have been using the substitution method almost since I was in diapers. I always look for ways to symmetrize and cancel out extra terms.
The brute force method is not so formidable. The equation can be simplified to: x^4 + 8x^3 + 30.5x^2 + 58x + 42 = 0. Now with repeated application of the rational root theorem, we first find the 2 coincident real roots at x = (negative)2 and then the 2 complex roots: x = (negative)2 +/- i.sqrt(26)/2
Substitute x=y-2 into the given equation and revise to 2[y^4+6y^2+1]+y^2-2=2y^4+13y^2=0. Thus y=0 twice and y=±i√26/2 It follows that x=y-2=-2 twice and x=±i√26/2-2.
This time i m a bit lazy x making and solving this interseing exercise.. Tomorrow better. By the way.. Give you a very honest thx x all your lessons, wish you spend a good and happy Christmas 😊 dear teacher
You made a math error in the 1st method since the right hand side " = 1" got changed to " = 0" by mistake. Thus, the constant term on the left hand side should simplify to +84 and not +85. It doesn't matter much since you stopped trying to solve the problem at that point and switched to the 2nd method, but it would be a big deal if you had continued solving the 1st method.
Do you know what? After following you for some time now, I immediately went for the substitution you used. Method 2. As you say, how powerful! And to be still learning and improving at 81 is great. Thank you so much.
Fittingly, my age is 3^4 and in a few more days it will be 2025, which is of course a perfect square. Better still, 2025 = (5^2) x (3^4).
I'm also 81 and have been using the substitution method almost since I was in diapers. I always look for ways to symmetrize and cancel out extra terms.
The brute force method is not so formidable. The equation can be simplified to: x^4 + 8x^3 + 30.5x^2 + 58x + 42 = 0. Now with repeated application of the rational root theorem, we first find the 2 coincident real roots at x = (negative)2 and then the 2 complex roots: x = (negative)2 +/- i.sqrt(26)/2
Agreed! Though I just factored it right from scratch without testing the roots.
Substitute x=y-2 into the given equation and revise to 2[y^4+6y^2+1]+y^2-2=2y^4+13y^2=0. Thus y=0 twice and y=±i√26/2
It follows that x=y-2=-2 twice and x=±i√26/2-2.
I set u=x-2 and solved the resulting biquadratic. The only real solution, x=-2, is a double root.
I also got -2 as the only real solution.
aplusbi😊💯💯💯👍👍🦾💪💥💥
*_1st:_* The ^most^ brute force method would be to graph the function and solve by binary (or root2) chop, or try every possible integer.
This time i m a bit lazy x making and solving this interseing exercise.. Tomorrow better. By the way..
Give you a very honest thx x all your lessons, wish you spend a good and happy Christmas 😊 dear teacher
Merry Christmas and Happy New Year to you too! 🎄
Set y = x + 2
(y - 1)⁴ + (y + 1)⁴ = -y² + 2
2y⁴ + 12y² + 2 = -y² + 2
2y⁴ + 13y² = 0
Set u = y²
2u² + 13u = 0
u(2u + 13) = 0
u = 0 or u = -13/2
Then, x = -2 or x = -2 ± sqrt(-13/2)i
x = -2
An Interesting Equation From Math Olympiads:
(x + 1)⁴ + (x + 3)⁴ = 1 - (x + 1)(x + 3); x =?
Let: y = x + 2, x + 1 = y - 1, x + 3 = y + 1; (y - 1)⁴ + (y + 1)⁴ = 1 - (y - 1)(y + 1)
(y ± 1)⁴ = y⁴ ± 4y³ + 6y² ± 4y + 1, (y - 1)⁴ + (y + 1)⁴ = 2(y⁴ + 6y² + 1) = 1 - (y² - 1)
2(y⁴ + 6y² + 1) = 1 - (y² - 1) = 2 - y², 2y⁴ + 13y² = 0, y²(2y² + 13) = 0
y = 0; Double root or 2y² + 13 = 0, y = ± i√(13/2) = ± (i√26)/2
y = x + 2 = 0, x = - 2 or y = x + 2 = ± i√(13/2), x = - 2 ± (i√26)/2 = (- 4 ± i√26)/2
Answer check:
x = - 2
(x + 1)⁴ + (x + 3)⁴ = (- 1)⁴ + 1⁴ = 2, 1 - (x + 1)(x + 3) = - (- 1) = (1); Confirmed
x = (- 4 ± i√26)/2, y = x + 2; y² = - 13/2: (y - 1)⁴ + (y + 1)⁴ = 2 - y²
(y - 1)⁴ + (y + 1)⁴ = 2(y⁴ + 6y² + 1) = 2(y²)(y² + 6) + 2
= 2(- 13/2)(- 13/2 + 6) + 2 = 2 + 13/2 = 2 - (- 13/2) = 2 - y²; Confirmed
Final answer:
x = - 2, Double root; x = (- 4 + i√26)/2 or x = (- 4 - i√26)/2
Two complex value roots, if acceptable
(x+3)^4+(x+1)(x+3)=1-(x+1)^4...(x+3)(x^3+9x^2+28x+28)=(-2x-x^2)(x^2+2x+2)..(x+3)(x+2)(x^2+7x+14)=-x(x+2)(x^2+2x+2).. 1 soluzione x=-2...(x+3)(x^2+7x+14)=-x(x^2+2x+2)..2x^3+12x^2+37x+42=0..(x+2)(2x^2+8x+21)=0.. 2 soluzione x=-2...x=(-4+i√26)/2..x=(-4-i√26)/2
problem
(x+1)⁴ +(x+3)⁴ = 1-(x+1)(x+3)
Let
a = x + 1
x = a - 1
a⁴ +(a+2)[ (a+2)³ + a ] - 1 = 0
Expand.
2a⁴ + 8 a³ + 25 a² + 34 a + 15 = 0
Multiply by 8 to get 16a⁴ leading term.
16 a⁴ + 64 a³ + 200 a² + 272 a + 120 = 0
Let
a = b/2
b⁴ +8b³ + 50 b² + 136 b + 120 = 0
To remove the cubic term let
b = y-2
y⁴ + 26 y² = 0
Factor.
y² ( y² + 26 ) = 0
Apply 0 product property.
First term:
y = 0
Second term:
y = ± i√26
Roots in y:
0, - i√26, i√26
Back substitute.
Roots in b = y-2
-2, -2- i√26, -2 + i√26
Roots in a = b / 2
-1, -1 - i√26 / 2, -1 + i√26 / 2
Roots in x = a - 1
-2, -2 - i√26 / 2, -2 + i√26 / 2
answer
x ∈ { -2, -2 - i√26/2 , -2 + i√26 /2 }