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This is a question from an undergraduate course on Galois theory:

Find all complex numbers which are roots of $P(T)=T^4+2T^2-\sqrt{6}T+\frac{3}{4}$

Can we use Galois theory to solves this?

Or do we just do soem long calculations to get quadratic polynomials $A(T)$ and $B(T)$ such that $P(T)=A(T)B(T)$ and then use quadratic formula?


$P(T)=T^4+2T^2-\sqrt{6}T+\frac{3}{4}=(T^2+aT+b)(T^2+cT+d)=T^4+(c+a)T^3+(d+ac+b)T^2+(ad+bc)T+bd$

$\implies $

$c+a=0 \implies a=-c$

$ d+ac+b=2 \implies d-c^2+b=2 \implies c=\sqrt{d+b-2}$

$ad+bc=-\sqrt{6} \implies -d\sqrt{d+b-2}+b\sqrt{d+b-2}=-\sqrt{6}$

$bd=\frac{3}{4} \implies b=\frac{3}{4d}$

This leads to: $-d\sqrt{d+\frac{3}{4d}-2}+\frac{3}{4d}\sqrt{d+\frac{3}{4d}-2}=-\sqrt{6}=(\frac{3}{4d}-d)\sqrt{d+\frac{3}{4d}-2}$

Squaring both sides gives: $(\frac{3}{4d}-d)^2({d+\frac{3}{4d}-2})=6$

Then we have a cubic in $d$ which I can not solve

So this would us a value of $d$ and , in turn, a value of $b$, $c$ and $a$, then we could use the quadratic formula to find complex roots.

This is very lengthy (not ideal under exam timed conditions)


Is there a more efficient way to solve this problem?

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  • $\begingroup$ With the $ad+bc=-\sqrt 6$ I would be tempted to multiply through by $d$ to get $-cd^2+cbd =-d\sqrt 6\to -c(d^2+\frac 34)=-d\sqrt 6$ and then solve for $d$ in terms of $c$... This doesn't alleviate the overall time issue much. $\endgroup$
    – abiessu
    Apr 19, 2016 at 22:57
  • $\begingroup$ OK thanks. This is a recurring question on Galois Theory papers of my course and I have calculated I can spend 7 minutes on it..... This method takes me much longer tho $\endgroup$
    – amiz9
    Apr 19, 2016 at 23:25

2 Answers 2

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(Too long for a comment.)

Just some heuristics, which happen to pay off in this case, though not necessarily always.

  • Calculations are always easier/faster in integers. With $\,z = \sqrt{6}\, T\,$ the equation becomes: $$ z^4 + 12 z^2 + 36 z + 27 = 0 $$

  • A quadratic factorization can be written as a sum or difference of squares, which may sometimes be easier to spot than the factorization itself. Here, remembering the original equation, it is expected that at least one of the squares may have a multiple of $6$ multiplier, which narrows down the guessing range somewhat. With a stroke of luck, one may find: $$ \begin{align} z^4 + 12 z^2 + 36 z + 27 &= (z^2-3)^2 + 18 (z+1)^2 \\ &= \left(z^2 + 3 \sqrt{2}\,i\,z - 3 + 3 \sqrt{2}\,i\right)\left(z^2 - 3 \sqrt{2}\,i\,z - 3 - 3 \sqrt{2}\,i\right) \end{align} $$

  • However, the calculations would be easier with real quadratics, and such factorization must exist. With more stroke of luck, one may find: $$ \begin{align} z^4 + 12 z^2 + 36 z + 27 &= (z^2 + 9)^2 - 6(z-3)^2 \\ &= \left(z^2 + \sqrt{6} z + 9 - 3\sqrt{6}\right)\left(z^2 - \sqrt{6} z + 9 + 3\sqrt{6}\right) \end{align} $$ From here, it's just a matter of calculating the roots and reversing the substitution back to $\,T\,$.

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  • $\begingroup$ Thanks, so much! Comments removed in a few seconds. $\endgroup$ Jul 8, 2023 at 8:46
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If $\alpha$ is a root of $P$ then $$\frac{\alpha^4 + 2 \alpha^2 + 3/4}{\alpha} = \sqrt 6 \in \mathbb Q(\alpha)$$ so $\alpha$ is quadratic over $\mathbb Q(\sqrt 6)$ and $P$ factors as $$P(x) = (x^2 + a x + b)(x^2 - a x + c)$$ over $\mathbb Q(\sqrt 6)$. In particular $a^2 > 0$, since $\mathbb Q(\sqrt 6) \subset \mathbb R$. Expanding the product above and some algebraic manipulation leads to $$(a^2 + 2)^2 - \frac 6 {a^2} = 3.$$ Together with $a^2 > 0$ this shows that $a = \pm1$ and the factorization can be completed from there.

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