A Very Nice Math Olympiad Problem | Solve for x | Algebra

[@MrPandaJJ]

At the beginning of the 2nd page (~5:18), you know x^4 - 4x^3 - 6x^2 - 4x + 1 = 0 Observe that flipping the sign of 6x^2 gives you (1-x)^4, so we can in fact rewrite the equation as (x^4 - 4x^3 + 6x^2 - 4x + 1) - 12x^2 = 0 i.e. (1-x)^4 - 12x^2 = 0 This is the difference of two squares, factorize it and you would get two quadratic equations, which give you the 4 solutions of x. It would save you some times compare with the t substitution.

[@dan-florinchereches4892]

This is the solution I saw but later I also noticed this is a palindromic equation so dividing by x^2 can also work for it

[@SpencersAcademy]

Great job 👏

[@師太滅絕]

(continue) [(1-x)^2 - sqrt(12x)][(1-x)^2 + sqrt(12x)] = 0 case A (1-x)^2 - sqrt(12x) = 0 case B(1-x)^2 + sqrt(12x) = 0

[@MrPandaJJ]

@@師太滅絕 not sqrt(12x) but sqrt(12)x therefore much easier than you think

[@ernestdecsi5913]

A very nice solution indeed.

[@MrPandaJJ]

At the end you say all 4 solutions are real not complex, however it is not true. x_3 and x_4 are complex, as [3 - 2*sqrt(3)] is, in fact, negative.

[@gintautasruksenas6129]

Ačiū!

[@SpencersAcademy]

My pleasure, sir 🙏 Thanks for the Super Thanks 😊 I'm indeed grateful 🙏

[@giuseppemalaguti435]

Risolvo la quartica..2+2x^4=1+6x+4x^2+6x^3+x^4..dopo i calcoli risulta (x^2-3x+1)^2-15x^2=0..poi è semplice .

[@JoseManuel-zn8lm]

Hola Las soluciones son todas válidas al hacer la comprobación? Gracias

[@SpencersAcademy]

Yes, they are.

[@key_board_x]

(1 + x⁴)/(1 + x)⁴ = 1/2 2.(1 + x⁴) = (1 + x)⁴ 2 + 2x⁴ = (1 + x)².(1 + x)² 2 + 2x⁴ = (1 + 2x + x²).(1 + 2x + x²) 2 + 2x⁴ = 1 + 2x + x² + 2x + 4x² + 2x³ + x² + 2x³ + x⁴ x⁴ - 4x³ - 6x² - 4x + 1 = 0 → the aim, if we are to continue effectively, is to eliminate terms to the 3rd power x⁴ - 4x³ - 6x² - 4x + 1 = 0 → let: x = z - (b/4a) → where: b is the coefficient for x³, in our case: - 4 a is the coefficient for x⁴, in our case: 1 x⁴ - 4x³ - 6x² - 4x + 1 = 0 → let: x = z - (- 4/4) → x = z + 1 (z + 1)⁴ - 4.(z + 1)³ - 6.(z + 1)² - 4.(z + 1) + 1 = 0 (z + 1)².(z + 1)² - 4.(z + 1)².(z + 1) - 6.(z² + 2z + 1) - 4z - 4 + 1 = 0 (z² + 2z + 1).(z² + 2z + 1) - 4.(z² + 2z + 1).(z + 1) - 6z² - 12z - 6 - 4z - 4 + 1 = 0 (z⁴ + 2z³ + z² + 2z³ + 4z² + 2z + z² + 2z + 1) - 4.(z³ + z² + 2z² + 2z + z + 1) - 6z² - 12z - 6 - 4z - 4 + 1 = 0 (z⁴ + 4z³ + 6z² + 4z + 1) - 4.(z³ + 3z² + 3z + 1) - 6z² - 16z - 9 = 0 z⁴ + 4z³ + 6z² + 4z + 1 - 4z³ - 12z² - 12z - 4 - 6z² - 16z - 9 = 0 z⁴ - 12z² - 24z - 12 = 0 z⁴ - (12z² + 24z + 12) = 0 z⁴ - 12.(z² + 2z + 1) = 0 z⁴ - 12.(z + 1)² = 0 z⁴ - [4.(z + 1)² * 3] = 0 z⁴ - [2².(z + 1)² * (√3)²] = 0 (z²)² - [2.(z + 1).√3]² = 0 → recall: a² - b² = (a + b).(a - b) [z² + 2.(z + 1).√3].[z² - 2.(z + 1).√3] = 0 First case: [z² + 2.(z + 1).√3] = 0 z² + 2.(z + 1).√3 = 0 z² + 2z√3 + 2√3 = 0 Δ = (2√3)² - (4 * 2√3) = 12 - 8√3 ← it's negative → complex number Δ = 12 - 8√3 Δ = - (8√3 - 12) Δ = i².(8√3 - 12) Δ = 4i².(2√3 - 3) z = [- 2√3 ± 2i√(2√3 - 3)]/2 z = - √3 ± i√(2√3 - 3) Second case: [z² - 2.(z + 1).√3] = 0 z² - 2.(z + 1).√3 = 0 z² - 2z√3 - 2√3 = 0 Δ = (- 2√3)² - (4 * - 2√3) = 12 + 8√3 = 4.(3 + 2√3) z = [2√3 ± 2√(3 + 2√3)]/2 z = √3 ± √(3 + 2√3) Recall: x = z + 1 When: z = - √3 ± i√(2√3 - 3) → x = 1 - √3 ± i√(2√3 - 3) When: z = √3 ± √(3 + 2√3) → x = 1 + √3 ± √(3 + 2√3)

[@SpencersAcademy]

Incredible!!! 👏

[@souzasilva5471]

3-2V3 < 0 Logo sqrt(3-2sqrt(3) i

[@jimwinchester339]

Why not begin by cancelling out one of the powers of (x + 1) in the denominator with the numerator, then flipping over both sides of the equation, giving (x+1)^ 3 = 2? Surely that would be a lot simpler.

[@jeanluchenry2152]

Pour x3,x4, c'est 2 nombres complexes car 3-2racine3 est négatif

[@RealQinnMalloryu4]

{1+1 ➖ }+{x^4+x^4 ➖ }/4x^4={2+x^8}/4x^4=2x^8/4x^4 2x^2^3/2^2x^2^2 1x^1^1^1/1^1x^1^2 x^1^2 (x ➖ 2x+1).

[@Lemda_gtr]

👍🏻🫡 waw great

[@milencenov6421]

4 real solutions ??? BS !!! Note that √(3 - 2√3) is NOT real, because 3 - 2√3 is less than 0. Also note that (x - 1)^4 expands very similarly to (x + 1)^4. All even powers are the same and the odd powers of x have minuses. After you cross multiplied and rearranged, you could have noticed that: (x - 1)^4 = 12 * x^2 Thus: (x - 1)^2 = |x * 2√3| (you may also write plus minus and normal brackets instead of the modulus) Now you have the same 2 cases you arrived at. x^2 - 2x + 1 = -x * 2√3 x^2 - 2(1 - √3)x + 1 = 0 Or x^2 - 2x + 1 = x * 2√3 x^2 - 2(1 + √3)x + 1 = 0 Note that when B is equal to -2(1 - √3), the discriminant is negative, because B^2 < 4AC.