Math Olympiad | A Nice Algebra Problem

Math Olympiad | A Nice Algebra Problem | Find the values of X

Рет қаралды 17,904

7 күн бұрын

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: 20

@ashkanshekarchi7753 5 күн бұрын

Explain the solving strategy and ideas rather than a long confusing mechanical calculation

@opulence3222 5 күн бұрын

Very confusing procedure.

@is7728 5 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

@maximcoroli8306 3 күн бұрын

17 = 1**4 +2**4 or (-1)**4+(-2)**4 х+2=2, х+1=1 => x=0 x+2=-1, x+1=-2 = x=-3

@is7728 5 күн бұрын

2^4 + 1^4 = 17 (-1)^4 + (-2)^4 = 17

@joiceroosita5317 4 күн бұрын

Mine is more simple. (x+2)⁴ + (x+1)⁴ = 17 [(x+1)+1]⁴ + (x+1)⁴ = 17 example ; y = (x+1) (y+1)⁴ + y⁴ = 17 y⁴ + 4y³ + 6y² + 4y + 1 + y⁴ - 17 = 0 2y⁴ + 4y³ + 6y² + 4y - 16 = 0 2(y⁴ + 2y³ + 3y² +2y - 8) = 0 Lets find the factors of y Using polinomial Factors of y⁴ + 2y³ + 3y² + 2y - 8 = 0 (y - 1) → y = 1 → 1+2+3+2-8 = 0 (ok!) (y + 2) → y = -2 → -8+12-12+8 = 0 (ok.!) The others y = i So, we get y = 1 and y = -2 (y = x+1) y = 1 x+1 = 1 → x = 0 (check ok.!) y = -2 x+1 = -2 → x = -3 (check ok.!) So, x = 0, -3

@thomaslangbein297 2 күн бұрын

What did we learn?

@user-nd7th3hy4l 5 күн бұрын

X=0 et.....

@yiutungwong315 20 сағат бұрын

Real Number X = 0 and (-3) 16 + 1 = 17

@edwardwang7929 3 күн бұрын

Immediately guessing, x=0.

@user-fl1go2lr3c 3 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

@laoxian899 2 күн бұрын

Yes, at the first glance one can get the two results, since both parentheses must be 1/-1 and 2/-2.

@abbasmasum1633 12 сағат 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.

@professorsargeanthikesclim9293 10 сағат бұрын

Did you also find the complex solutions?

@walterwen2975 5 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

@NNaween 4 күн бұрын

Nice explanation @naweenraaj

@is7728 4 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

@E.h.a.b 4 күн бұрын

Let y = x + 3/2 (y + 1/2)^4 + (y - 1/2)^4 = 17 (y^2 + y + 1/4)^2 + (y^2 - y + 1/4)^2 = 17 [ (y^2 + 1/4) + y ]^2 + [ (y^2 + 1/4) - y ]^2 = 17 [ (y^2 + 1/4)^2 + 2 y (y^2 + 1/4) + y^2 ] + [ (y^2 + 1/4)^2 - 2 y (y^2 + 1/4) + y^2 ]=17 2((y^2 + 1/4)^2 + y^2) = 17 2(y^4 + 1/2 y^2 + 1/16 + y^2) = 17 2 y^4 + y^2 + 1/8 + 2 y^2 - 17 = 0 2 y^4 + 3 y^2 - 135/8 = 0 y^2 = (-3 +/- √(9 - 4 * 2 * (- 135/8)))/(2*2) y^2 = (-3 +/- √(9 + 135))/4 y^2 = (-3 +/- √144)/4 y^2 = (-3 +/- 12)/4 y^2 = (-3 + 12)/4 = 9/4 //y^2 = (-3 - 12)/4 < 0 so it is rejected y = +/- 3/2 x = y - 3/2 = +/- 3/2 - 3/2 x = (0, -3)

@Altair705 11 сағат бұрын

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.