You probably meant(x^8n + 1/x^8n)
I got the answer as 2
\(\displaystyle x+\frac{1}{x}=\sqrt{2} \implies x^2-\sqrt{2}x+1=0\)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.
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.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.
The reality is that the problem is more complex than you think--puns intended.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......
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.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...?
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!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......
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.)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.
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.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.
Good work.OK, here is the attachment of my attempt.