Determine the irrational numbers x such that both x and x 3 − 6 x are rational numbers

Question

Determine the irrational numbers   x such that both   x and       x    3    −  6  x are rational numbers

metodikkf6z · Accepted Answer

Suppose   q  =      x    2    +  2  x is rational.Then, by the quadratic formula,   x  =            −      2      ±              4        +        4        q              2    =  −  1  ±      α  , where   α  =  q  +  1 can be any rational number that is not a perfect square (since x is irrational).Now,      x    3    −  6  x  =  (  −  1  ±      α        )    3    −  6  (  −  1  ±      α    )  =  ±      α          3              /            2        −  3  α  ∓  3      α    +  5will be rational if and only if       α          3              /            2        −  3      α   is (since the other terms are always rational). But this is equal to       α    (  α  −  3  ). Since   α  −  3 is rational and       α   irrational, the entire quantity will be rational iff it is zero: that is, iff   α  =  3.Thus the numbers   x that satisfy the condition are   −  1  ±      3

pighead73283r · Accepted Answer

Suppose       x    2    +  2  x  =      q    1   and       x    3    −  6  x  =      q    2   with       q    1    ,      q    2    ∈      Q  . Clearly   [      Q    (  x  )  :      Q    ]  =  2. Thus the polynomial       t    2    +  2  t  −      q    1   divides       t    3    −  6  t  −      q    2  ; that is, there exists some   α such that      t    3    −  6  t  −      q    2    =  (      t    2    +  2  t  +      q    1    )  (  t  −  α  )  =      t    3    +  (  2  −  α  )      t    2    +  (      q    1    −  2  α  )  t  −  α      q    1    .Equating corresponding coefficients of       t    n   gives   α  =  2,       q    1    =  −  2  , and       q    2    =  4. Thus both polynomials are rational iff       x    2    +  2  x  +  2  =  0; that is, iff   x  =  −  1  ±      3  .

Determine the irrational numbers x such that both x^2+2x and x^3−6x are rational numbers

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