Theorem: If a cubic has any constructible root then it has at least one real root.
Proof:
Suppose that the in cubic:
3 + 2 + + = 0
there are constructible roots but none in ℚ (the set of rational numbers) and there is some value such that the field 𝐹( ) is the closest field to the rational number field in which a root exists. I.e. that the root is = + √𝑘 for , , 𝑘 from a field lower than 𝐹( ) , 𝐹( − 1) and √𝑘 is in 𝐹( ) .
Then we put = + √𝑘 into the cubic:
We can create many fields from the field of rational numbers
( + √𝑘) 3 + ( + √𝑘) 2 + ( + √𝑘) + = 0
If we expand the brackets we get:
( + 3 + 2 𝑘 + 2 + 2 𝑘 + ) + (3 2 + 3 𝑘 + 2 + )√𝑘 = 0
which we can simplify as :
+ √𝑘 = 0
So either √𝑘 = −/ or = = 0 . Because −/ is in 𝐹( − 1) and √𝑘 is not, this case cannot be true; therefore S=T=0 must be the true case.
Therefore we ask: is = − √𝑘 a root?
( − √𝑘) 3 + ( − √𝑘) 2 + ( − √𝑘) + = 0
( + 3 + 2 𝑘 + 2 + 2 𝑘 + ) − (3 2 + 3 𝑘 + 2 + )√𝑘 = 0
∴ − √𝑘 = 0
Because = = 0 , this is also true. However, a cubic cannot have just 2 real roots (unless the roots are repeated). Let +√𝑘 = 1 , − √𝑘 = 2 , and the third (either repeated or not repeated) be 3 . I.e.:
( − 1
)( − 2
)( − 3
) = 0
3 + ( 1
) 2 + ⋯ = 0
+ 2
+ 3
Comparing with the original cubic, 3 + 2 + + = 0 , we can say that:
) = −( + + 3
) = −2 − 3
= ( 1
+ 2
+ 3
3
= −2 −
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