Portugal l A Nice Algebra Problem l Math Olympiad

[@quinty.support]

(x + 9)⁴ + (x + 11)⁴ = 706
Let, y = x + 10
=> x + 10 - y = 0 => - x = 10 - y => x = y - 10
(x + 9) = (y - 10 + 9) = (y - 1)
(x + 11) = (y - 10 + 11) = (y + 1)
(y - 1)⁴ + (y + 1)⁴ = 706
=> y⁴ + 1 - 4y³ + 6y² - 4y + y⁴ + 1 + 4y³ + 6y² + 4y = 706 → [Consideration of (a + b)⁴ expansion → ]
Combine the terms:
=> 2y⁴ + 12y² + 2 = 706
=> 2y⁴ + 12y² + 2 - 706 = 0
=> 2y⁴ + 12y² - 704 = 0
[Divide both sides by 2 to simplify further]
=> y⁴ + 6y² - 352 = 0
Let, z = y²
=> z² + 6z - 352 = 0
(a = 1, b = 6, c = -352)
∆ = b² - 4ac = 6² - 4(1)(-352) = 36 + 1408 = 1444
z = (-b ± √∆)/2a = (-6 ± √1444)/2 = (-6 ± 38)/2 = (-6 - 38)/2 or (-6 + 38)/2 = -44/2 or 32/2 = -22 or 16
[Recall → z = y²]
y² = -22 or y² = 16
=> y = ±√-22 or y = ±√16
=> y = ±i√22 or y = ±4
[Recall → y = x + 10 => x = y - 10].... Substituting this value:
y = 4, x = 4 - 10 = -6
y = -4, x = -4 - 10 = -(4 + 10) = -14
y = ±i√22 then x = -10 ± i√22 or -10 + i√22 and -10 - i√22

[@Rocio62154]

EXCELLENTONIO!

[@vijaymaths5483]

Thanks a lot 🎈

[@Quest3669]

5^4+3^4= 706
Comparing well gives
X= -6; -14

[@vijaymaths5483]

👍

[@adribber]

(x + 9)⁴ + (x + 11)⁴ = 706
put, t = x + 10
(t - 1)⁴ + (t + 1)⁴ = 706
(t² - 2t + 1)² + (t² + 2t + 1)² = 706
t⁴ + 4t² + 1 + 2(-2t³ - 2t + t²) + t⁴ + 4t² + 1 + 2(2t³ + 2t + t²) = 706
2t⁴ + 12t² + 2 = 706
t⁴ + 6t² + 1 = 353
t⁴ + 6t² - 352 = 0
put, y = t² (>= 0)
y² + 6y - 352 = 0
D = 36 + 4*352 = 4(9 + 352) = 2²*19²
y = (- 6 ± √D) /2*1 = (- 6 ± 2*19) /2 = - 3 ± 19
y = 16 , y = - 22
∴ y = 16 (∵ y = t² >= 0)
t² = 16
∴ t = ± 4
Case 1 : t = 4
t = x + 10
x = t - 10
∴ x = 4 - 10 = - 6
verifying
(-6 + 9)⁴ + (- 6 + 11)⁴ =? 706
(3)⁴ + (5)⁴ =? 706
81 + 625 =? 706
706 = 706 Pass
Case 2 : t = - 4
t = x + 10
x = t - 10
∴ x = - 4 - 10 = - 14
verifying
(-14 + 9)⁴ + (- 14 + 11)⁴ =? 706
(-3)⁴ + (-5)⁴ =? 706
81 + 625 =? 706
706 = 706 Pass
∴ x = - 6 , x = - 14

[@key_board_x]

(x + 9)⁴ + (x + 11)⁴ = 706
(x + 9)².(x + 9)² + (x + 11)².(x + 11)² = 706
(x² + 18x + 81).(x² + 18x + 81) + (x² + 22x + 121).(x² + 22x + 121) = 706
(x⁴ + 18x³ + 81x² + 18x³ + 324x² + 1458x + 81x² + 1458x + 6561) + (x⁴ + 22x³ + 121x² + 22x³ + 484x² + 2662x + 121x² + 2662x + 14641) = 706
(x⁴ + 36x³ + 486x² + 2916x + 6561) + (x⁴ + 44x³ + 726x² + 5324x + 14641) = 706
2x⁴ + 80x³ + 1212x² + 8240x + 20496 = 0
x⁴ + 40x³ + 606x² + 4120x + 10248 = 0 → the aim, if we are to continue effectively, is to eliminate terms to the 3rd power
Let: x = z - (b/4a) → where:
b is the coefficient for x³, in our case: 40
a is the coefficient for x⁴, in our case: 1
x⁴ + 40x³ + 606x² + 4120x + 10248 = 0 → let: x = z - (40/4) → x = z - 10
(z - 10)⁴ + 40.(z - 10)³ + 606.(z - 10)² + 4120.(z - 10) + 10248 = 0
(z - 10)².(z - 10)² + 40.(z - 10)².(z - 10) + 606.(z² - 20z + 100) + 4120z - 41200 + 10248 = 0
(z² - 20z + 100).(z² - 20z + 100) + 40.(z² - 20z + 100).(z - 10) + 606z² - 12120z + 60600 + 4120z - 30952 = 0
(z⁴ - 20z³ + 100z² - 20z³ + 400z² - 2000z + 100z² - 2000z + 10000) + 40.(z³ - 10z² - 20z² + 200z + 100z - 1000) + 606z² - 8000z + 29648 = 0
(z⁴ - 40z³ + 600z² - 4000z + 10000) + 40.(z³ - 30z² + 300z - 1000) + 606z² - 8000z + 29648 = 0
z⁴ - 40z³ + 600z² - 4000z + 10000 + 40z³ - 1200z² + 12000z - 40000 + 606z² - 8000z + 29648 = 0
z⁴ + 6z² - 352 = 0 ← no more item to the power 3
z⁴ + 6z² = 352
z⁴ + 6z² + 9 = 352 + 9
(z² + 3)² = 19
z² + 3 = ± 19
First case: z² + 3 = 19
z² = 16
z = ± 4 → recall: x = z - 10
First solution: x = - 14
Second solution: x = - 6
Second case: z² + 3 = - 19
z² = - 22
z² = 22i²
z = ± i√22 → recall: x = z - 10
Third solution: x = - 10 + i√22
Fourth solution: x = - 10 - i√22