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    x^5+x=10

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

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

    No this won't work.
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    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.
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    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
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    (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.
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    (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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