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$\begingroup$

I've tried simplifying the term by substituting $y := x^2$ and also to use a certain algorithm that outputs the f(a), but it gets too big for the input r, that is a variable and not a number to begin with (added little example):

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If you have a good idea how to do this, please let me know

  • 2
    Hint: Show that $$(1+\sqrt2)^4-6(1+\sqrt2)^2+1=0$$ – 2017-01-05
  • 0
    @AlexSilva This is talking about solutions of $f(x)=0$. – 2017-01-05
  • 0
    @Did, I get it! Thanks! – 2017-01-05

3 Answers 3

1

Let $y=x^2$ which reduces the equation to $y^2-6qy+q^2=0$

$y=\frac{6q\pm\sqrt{36q^2-4q^2}}{2}$

$x^2=\left(3 \pm 2\sqrt{2}\right)q$

$x^2=\left(\sqrt{2}\pm1\right)^2 q$

$x=\pm\left(\sqrt{2}\pm1\right)\sqrt{q}$

$x=\pm\left(\sqrt{2q}\pm\sqrt{q}\right)$

Q.E.D

  • 0
    Can you explain why $(3 \pm 2\sqrt{2})q = (\sqrt{2} \pm 1)q$ – 2017-01-05
  • 0
    $\left(\sqrt{2}+1\right)^2=\left(\sqrt{2}\right)^2+2\sqrt{2}+1=2+2\sqrt{2}+1$ – 2017-01-05
3

Hint

Completing square (it is equivalent to use Baskara's formula):

$x^4-6qx^2+q^2=(x^2-3q)^2-8q^2=0 \rightarrow (x^2-3q)^2=8q^2 \rightarrow x^2=(3 \pm2\sqrt{2})q=(1\pm \sqrt{2})^2q$

$$x^2=(1\pm \sqrt{2})^2q$$

Can you finish?

2

\begin{align}x^4-6qx^2+q^2&=x^4-6qx^2+9q^2-8q^2\\[0.2cm]&=(x^2-3q)^2-(\sqrt{8}q)^2\\[0.2cm]&=\left(x^2-3q-2\sqrt{2}q\right)\left(x^2-3q+2\sqrt{2}q\right)\end{align} Now, note that $$r^2=\left(\sqrt{q}+\sqrt{q}\sqrt{2}\right)^2=q\left(1+2\sqrt2+2\right)=(3+2\sqrt{2})q$$ to conclude.