Want to confirm the answer of the following question

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gamaz321

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The question is as follows
If (x + 1/x) = sqrt(2) then find the value of (x^8n + 1/x^8n)
I got the answer as 2 which I would like to confirm if it is correct.
Thank you.
 
gamaz321 said:
(x^8n + 1/x^8n)

I got the answer as 2
Click to expand...
You probably meant

x^(8n) + 1/x^(8n)

Have you been asked to demonstrate that the expression is 2 for all n?

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[imath]\;[/imath]
 
n is any positive integer. I was experimenting with a problem that was somewhat similar but was a bit easy Here, I wanted to see that for any positive integer n the value of the expression is 2 which I think it is. However, I wanted to check if there is any condition this is not. Thank you.
 
For real [imath]x[/imath] we have [imath]\left|x+\frac{1}{x}\right|\geq 2 > \sqrt{2}[/imath], and I suspect the equation is not solvable for complex [imath]x[/imath] either.
 
blamocur said:
For real [imath]x[/imath] we have [imath]\left|x+\frac{1}{x}\right|\geq 2 > \sqrt{2}[/imath], and I suspect the equation is not solvable for complex [imath]x[/imath] either.
Click to expand...
\(\displaystyle x+\frac{1}{x}=\sqrt{2} \implies x^2-\sqrt{2}x+1=0\)

Find the roots.
 
gamaz321 said:
The question is as follows
If (x + 1/x) = sqrt(2) then find the value of (x^8n + 1/x^8n)
I got the answer as 2 which I would like to confirm if it is correct.
Thank you.
Click to expand...
I agree with your answer (assuming that the speculations in #3 are correct); but it's far better if you show us your work, rather than make us go through the work of figuring out what the problem means and then solving it with no context from you.

I can only guess that you did it the way I did, by repeatedly squaring the equation, and then using the results to prove the claim.
 
I have a problem here.
\(\displaystyle x + \dfrac{1}{x} =\sqrt2\)
Then \(\displaystyle (x + \dfrac{1}{x})^2 =2\)
This yields that \(\displaystyle x^2 +\dfrac{1}{x^2}=0\)
This last equation is never true.
If the premise is not true......
 
Steven G said:
I have a problem here.
\(\displaystyle x + \dfrac{1}{x} =\sqrt2\)
Then \(\displaystyle (x + \dfrac{1}{x})^2 =2\)
This yields that \(\displaystyle x^2 +\dfrac{1}{x^2}=0\)
This last equation is never true.
If the premise is not true......
Click to expand...
The reality is that the problem is more complex than you think--puns intended.

You're a blind man and need to look for that black cat which isn't there. Once you've done that, you'll transcend into a mathematician.

If you need help finding the black cat, see post#5 or contact @Harry_the_cat & @Otis. They might know one. Meow...?‍
 
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BigBeachBanana said:
The reality is that the problem is more complex than you think--puns intended.

You're a blind man and need to look for that black cat which isn't there. Once you've done that, you'll transcend into a mathematician.

If you need help finding the black cat, see post#5 or contact @Harry_the_cat & @Otis. They might know one. Meow...?‍
Click to expand...
Yes, I did think about complex solutions but I also felt that this problem was from intermediate algebra and was not about complex roots. I guess that I was mistaken.
 
Steven G said:
I have a problem here.
\(\displaystyle x + \dfrac{1}{x} =\sqrt2\)
Then \(\displaystyle (x + \dfrac{1}{x})^2 =2\)
This yields that \(\displaystyle x^2 +\dfrac{1}{x^2}=0\)
This last equation is never true.
If the premise is not true......
Click to expand...
This is an important thing to at least think about. I have occasionally seen a problem of this general sort, where you can easily show that if a given equation is true, then some other thing is true; but when you dig into it, you find that in fact the condition is never satisfied (for any kind of number), so the nice answer is meaningless -- though still valid!
Steven G said:
Yes, I did think about complex solutions but I also felt that this problem was from intermediate algebra and was not about complex roots. I guess that I was mistaken.
Click to expand...
But the problem doesn't ask you to find x; and it is probably assigned at a level where complex numbers are at least known to exist (though proving the conclusion for the actual values of x might be a lot of work). It's an exercise in a different kind of thinking than that. (And proving it true for all n implies some level of sophistication, too.)
 
blamocur said:
For real [imath]x[/imath] we have [imath]\left|x+\frac{1}{x}\right|\geq 2 > \sqrt{2}[/imath], and I suspect the equation is not solvable for complex [imath]x[/imath] either.
Click to expand...
Ooops again: there is, of course, a solution in complex numbers. And the answer for [imath]x^8 + \frac{1}{x^8}[/imath] is 2 indeed.
 
gamaz321 said:
OK, here is the attachment of my attempt.
Click to expand...
Good work.

I'd had actually gone far past the key equation there, because I was focused on a more general technique for solving such problems. As you found, although you didn't have to actually calculate x, you could easily do so along the way. It's easier than I realized.
 
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