Math Olympiad | A Nice Algebra Problem | Find the values of X
Vložit
- čas přidán 27. 06. 2024
- 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
Explain the solving strategy and ideas rather than a long confusing mechanical calculation
Very confusing procedure.
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
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
2^4 + 1^4 = 17
(-1)^4 + (-2)^4 = 17
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
What did we learn?
X=0 et.....
Real Number
X = 0 and (-3)
16 + 1 = 17
Immediately guessing, x=0.
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
Yes, at the first glance one can get the two results, since both parentheses must be 1/-1 and 2/-2.
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.
Did you also find the complex solutions?
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
Nice explanation
@naweenraaj
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
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)
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.
I'm so blown away. This is amazing ����