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# Can you prove this?

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Why bother with a post grad course - waste of time? 17-10-2016
1. x^5+x=10

How can you prove that the positive root of this equation is irrational?

(hints appreciated )
2. Factorise the LHS completely, leave the RHS side as it is.

No this won't work.
3. Show the root is positive by using the change of sign test.

Now assume the root is rational i.e. x=a/b where a and b are integers and are coprime.

So a^5/b^5+a/b=10
a^5+ab^4=10b^5

a(a^4+b^4)=10b^5

a^5=10b^5-ab^4
a^5=b(10b^4-ab^3)

So a must divide 10 since we assumed a and b are coprime and similarly b must divide 1 (we use the fundamental theorem of arithmetic here).

So x=±1,±2,±5,±10.
You can check none of these are roots of the equation (In fact if you use the change of sign test then you only need check 2 of these). So the root must be irrational.
4. General solution for a quadratic equation. Look at the part in the square root. The square root of a prime number is always irrational, and an irrational number multiplied by a rational number is always irrational
5. (Original post by offhegoes)
General solution for a quadratic equation. Look at the part in the square root. The square root of a prime number is always irrational, and an irrational number multiplied by a rational number is always irrational
That would work if it was a quadratic, but this isn't quite true here. It's x^5.
6. (Original post by Zacken)
That would work if it was a quadratic, but this isn't quite true here. It's x^5.
So it is!

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