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Easy Problem? Surely not binomial expansion... watch

1. Hey guys, my brother is doing GCSEs and asked me to help him with this problem his teacher set him. I did Furhter Maths last year, but looking at this it looked a bit like binomial expansion, which is way out of GCSE knowledge. But my maths is rusty, and i'm sure i've just missed the easy method. Any help? Thanks

Given that x + 1/x = 5

Find the value of x in:
x^5 = 1/(x^5)
2. (Original post by Guaglio)
Find the value of x in:
x^5 = 1/(x^5)
Take the 5th root of both sides.
3. (Original post by Kolya)
Take the 5th root of both sides.
and?
4. (Original post by Guaglio)
and?
That gives you x = 1/x. If you have x=1/x and x + 1/x = 5, it is easy to solve for x.
5. (Original post by Guaglio)
Hey guys, my brother is doing GCSEs and asked me to help him with this problem his teacher set him. I did Furhter Maths last year, but looking at this it looked a bit like binomial expansion, which is way out of GCSE knowledge. But my maths is rusty, and i'm sure i've just missed the easy method. Any help? Thanks

Given that x + 1/x = 5

Find the value of x in:
x^5 = 1/(x^5)
it would help if you used latex.
6. (Original post by Guaglio)
Hey guys, my brother is doing GCSEs and asked me to help him with this problem his teacher set him. I did Furhter Maths last year, but looking at this it looked a bit like binomial expansion, which is way out of GCSE knowledge. But my maths is rusty, and i'm sure i've just missed the easy method. Any help? Thanks

Given that x + 1/x = 5

Find the value of x in:
x^5 = 1/(x^5)
Do you mean which is what you have written, or
7. This doesn't make sense to me.

Thought you could just multiply both sides of to get a quadratic. But that doesn't work numerically.

Kolya's method gives 2.5 which doesn't seem to work.
8. I don't believe the question has been posted accurately.

If x^5 = 1/x^5, then x^5 = +/-1 and so x = +/-1 (assuming x real). Which has no connection to "Given that x+1/x=5" (and indeed if x+1/x = 5, it is impossible that x^5 = 1/x^5).
9. (Original post by miml)
This doesn't make sense to me.

Thought you could just multiply both sides of to get a quadratic. But that doesn't work numerically.

Kolya's method gives 2.5 which doesn't seem to work.
Yeah, you're right. The question's wrong (or a trick). x+1/x=5 has two irrational solutions. x^5=1/x^5 has 10 complex solutions, only two of which are real (which are +-1). In other words, there are values of x which satisfy both equations.
10. Kolya: Did you mean "there are no values of x which satisfy both equations"?
11. LOL. Nvm
12. (Original post by Kolya)
Yeah, you're right. The question's wrong (or a trick). x+1/x=5 has two irrational solutions. x^5=1/x^5 has 10 complex solutions, only two of which are real (which are +-1). In other words, there are values of x which satisfy both equations.
This is a GCSE question.
13. this question is weird!!!
14. (Original post by DFranklin)
Kolya: Did you mean "there are no values of x which satisfy both equations"?
Yeah, typo. Sorry.

Can you see where the flaw is in the method I posted right at the beginning, Dave? It's deceptively "legit", even though it doesn't actually work. i.e. the reasoning is:

Note that if x is s.t. x^5 = 1/x^5, then x is also s.t. x = 1/x. (This loses solutions, but it still preserves truth.)
Now, from the first equation, we have that 1/x = 5-x.
Therefore, combining the two equations gives us x = 5-x.
So x = 5/2 is a solution.
15. What's wrong is that if x = 5/2, then x does NOT equal 1/x. The two equations are inconsistent.

(I kind of realise you knew that, but if they're inconsistent, you can't expect substituting one into the other to yield anything useful).

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