An Interesting Equation From Math Olympiads

[@timothybohdan7415]

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.

[@robyzr7421]

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

[@SyberMath]

Merry Christmas and Happy New Year to you too! 🎄

[@AntimatterBeam8954]

Hope you had a good Christmas Sybermath! Happy new year for tonight! From London, UK

[@wes9627]

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.

[@dwm1943]

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).

[@wes9627]

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.

[@AntimatterBeam8954]

Learning from maths videos is great though. I enjoy watching them to learn stuff. If I already know the methods it's still great fun. I sometimes binge watch maths videos. My age is only a semiprime: 5×7

[@dwm1943]

@@AntimatterBeam8954 This is the best of the maths sites that I have found, so far. 35 years old! You're square next year, and in the prime of life the year after that. Me too, if I'm still around, at 83. And is 2027 a prime year? It might be the year I understand Lambert.

[@AntimatterBeam8954]

@dwm1943 hehe yes I love the properties of numbers. I was playing around with a challenge earlier on. Oh the Lambert W function? I think I understand it, I do make silly mistakes with it a lot though.

[@InnocentNeuron]

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

[@philipfoy7117]

Agreed! Though I just factored it right from scratch without testing the roots.

[@neuralwarp]

*_1st:_* The ^most^ brute force method would be to graph the function and solve by binary (or root2) chop, or try every possible integer.

[@yoav613]

aplusbi😊💯💯💯👍👍🦾💪💥💥

[@scottleung9587]

I also got -2 as the only real solution.

[@bobbyheffley4955]

I set u=x-2 and solved the resulting biquadratic. The only real solution, x=-2, is a double root.

[@damiennortier8942]

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

[@rakenzarnsworld2]

x = -2

[@walterwen2975]

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

[@giuseppemalaguti435]

(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