another solution: notice that 8^x and 2^x are both strictly increasing functions, so 8^x+2^x is strictly increasing. This means that there is at most 1 solution to 8^x + 2^x = 68. since x=2 works, we are done.
t^3+t-68=0 t^3-64+t-4=0 Difference of the cubes is equal to: a^3-b^3=(a-b)(a^2+ab+b^2). Then, we get: (t-4)(t^2+4t+16)+t-4=0 Here, t-4 is common factor, then: (t-4)(t^2+4t+17)=0 As we can see, the equation is more rational here. My question is will it be mistake if we don’t use the most rational solution in the case of school olympiad testing? In my country if we don’t use the most rational solution method we can loose points. Anyway, as you said here is a lot of approaches, and this is just one of them. Thank you for your videos.
I would have found the quadratic by synthetic division after finding the first root. Could also show other 2 roots are complex using Descartes’s rule of signs.
another solution: notice that 8^x and 2^x are both strictly increasing functions, so 8^x+2^x is strictly increasing. This means that there is at most 1 solution to 8^x + 2^x = 68. since x=2 works, we are done.
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t^3+t-68=0
t^3-64+t-4=0
Difference of the cubes is equal to: a^3-b^3=(a-b)(a^2+ab+b^2).
Then, we get:
(t-4)(t^2+4t+16)+t-4=0
Here, t-4 is common factor, then:
(t-4)(t^2+4t+17)=0
As we can see, the equation is more rational here. My question is will it be mistake if we don’t use the most rational solution in the case of school olympiad testing? In my country if we don’t use the most rational solution method we can loose points.
Anyway, as you said here is a lot of approaches, and this is just one of them.
Thank you for your videos.
I would have found the quadratic by synthetic division after finding the first root.
Could also show other 2 roots are complex using Descartes’s rule of signs.
Just by looking at this equation it’s quite clear that x=2. Would this whole proof be needed to get points in the question?
what if you were asked 3 values of x
By inspection x = 2. Difficult to believe this is a World International Maths Olympiad question.
Or ... this way:
8^x + 2^x = 68
2^3^x + 2^x = 68
2^x^3 + 2^x = 68
X = 2^x
X^3 + X = 68
X^3 + X - 68 = 0
based on X³ + Xp + q = 0 model with:
• p = 1
• q = -68
Cardano/Tartaglia formula application:
X = [-q/2 + √(q²/4 + p³/27)]^(1/3) + [-q/2 - √(q²/4 + p³/27)]^(1/3)
X = [-(-68)/2 + √((-68)²/4 + (1)³/27)]^(1/3) +
[-(-68)/2 - √((-68)²/4 + (1)³/27)]^(1/3)
X = 4.081665 + (-0.081665) = 4
2^x = X = 4 => x = 2
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| x = 2 |
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🙂
I don’t think that you need a 10 minute solution to get X=2. It’s incredibly obvious from just looking at it.
Amazing way, the way that i see it is 68 = 64 + 4 which mean 8² + 2²😂
❤
this is why I got 2 in my 11th-grade finals. (10%) 😅🥲