This is an interesting question that tests a lot of concepts! Hope you are all well Playlist to watch all videos on Learncommunolizer • Maths Olympiad And • Maths Subscribe: / @learncommunolizer
Пікірлер: 24
@ashkanshekarchi77538 күн бұрын
Explain the solving strategy and ideas rather than a long confusing mechanical calculation
@user-fl1go2lr3c5 күн бұрын
Is here anybody, who automatically calculated that x=0😂😂 Being honest, I didn't watch till the end, because I already thought that 2^4+1^4 equals to 17 And second x equals to -3
@laoxian8995 күн бұрын
Yes, at the first glance one can get the two results, since both parentheses must be 1/-1 and 2/-2.
@abbasmasum16333 күн бұрын
The goal is to learn how to apply several rules of algebra to find the solutions. This channel is the best there is and PLEASE stop offering guess and check solutions. He's connecting the problem to so many profound concepts. I truly like his methods and if they are too long, then at least offer a better way.
@professorsargeanthikesclim92933 күн бұрын
Did you also find the complex solutions?
@is77288 күн бұрын
Let h be x + 1. (h + 1)^4 + h^4 = 17 h^4 + 4h^3 + 6h^2 + 4h + 1 + h^4 = 17 2h^4 + 4h^3 + 6h^2 + 4h - 16 = 0 h^4 + 2h^3 + 3h^2 + 2h - 8 = 0 When h = 1, the eqn is satisfied. ∴ x = 0 h^3 + 3h^2 + 6h + 8 = 0 When h = -2, the eqn is satisfied. ∴ x = -3 h^2 + h + 4 = 0 h = ( -1 +- √(1 - 16) ) / 2 = -0.5 +- √15 i / 2 ∴ x = -1.5 +- √15 i / 2
I made the same variable change. You could have developped faster by using the binomial expansion method: (a+b)^4=a⁴+4a³b+6a²b²+4ab³+b⁴ It was even possible to skip the calculation of the terms in y and y³ by noticing that if y is solution then -y is also a solution, so only the even powers of y would remain. Best method in my opinion.
@walterwen29758 күн бұрын
Math Olympiad: (x + 2)⁴ + (x + 1)⁴ = 17; x = ? Let: y = x + 2; y⁴ + (y - 1)⁴ = 2y⁴ - 4y³ + 6y² - 4y + 1 = 17 2y⁴ - 4y³ + 6y² - 4y = 16, y⁴ - 2y³ + 3y² - 2y - 8 = 0 (y⁴ - 2y³ + y²) + (2y² - 2y) - 8 = 0, (y² - y)² + 2(y² - y) - 8 = 0 (y² - y - 2)(y² - y + 4) = (y + 1)(y - 2)(y² - y + 4) = 0 y + 1 = 0, y = - 1; y - 2 = 0, y = 2 or y² - y + 4 = 0, y = (1 ± i√15)/2 x = y - 2: x = - 3; x = 0 or x = (1 ± i√15)/2 - 2 = (- 3 ± i√15)/2 Answer check: x = - 3: (x + 2)⁴ + (x + 1)⁴ = (- 1)⁴ + (- 2)⁴ = 1 + 16 = 17; Confirmed x = 0: 2⁴ + 1⁴ = 16 + 1 = 17; Confirmed x = (- 3 ± i√15)/2: y = x + 2, y² - y + 4 = 0, y² - y = - 4 (x + 2)⁴ + (x + 1)⁴ = 2(y⁴ - 2y³ + 3y² - 2y) + 1 = 2[(y² - y)² + 2(y² - y)] + 1 = 2[(- 4)² + 2(- 4)] + 1 = 2(16 - 8) + 1 = 17; Confirmed Final answer: x = - 3, x = 0, Two complex value roots, if acceptable; x = (- 3 + i√15)/2 or x = (- 3 - i√15)/2
@NNaween6 күн бұрын
Nice explanation @naweenraaj
@is77286 күн бұрын
Maybe this better? Let h be x + 1. (h + 1)^4 + h^4 = 17 h^4 + 4h^3 + 6h^2 + 4h + 1 + h^4 = 17 2h^4 + 4h^3 + 6h^2 + 4h - 16 = 0 h^4 + 2h^3 + 3h^2 + 2h - 8 = 0 When h = 1, the eqn is satisfied. ∴ x = 0 h^3 + 3h^2 + 6h + 8 = 0 When h = -2, the eqn is satisfied. ∴ x = -3 h^2 + h + 4 = 0 h = ( -1 +- √(1 - 16) ) / 2 = -0.5 +- √15 i / 2 ∴ x = -1.5 +- √15 i / 2