MHB Can You Solve This Quartic Equation with Roots Involving Square-Free Integers?

[anemone]

Here is this week's POTW:

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Given $x=p+\sqrt{q}+\sqrt{r}+\sqrt{s}$ (where $q,\,r,\,s$ are square-free integers) is one root of the equation $x^4-4x^3-16x^2-8x+4=0$, find the values for $p,\,q,\,r$ and $s$.

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[anemone]

Congratulations to the following members for their correct solution:

1. greg1313
2. kaliprasad
3. Opalg

Solution from Opalg:

Spoiler

Start by looking for a factorisation of the form $$x^4 - 4x^3 - 16x^2 - 8x + 4 = (x^2 + \alpha x + 2)(x^2 + \beta x + 2).$$ Comparing the coefficients of the powers of $x$ on each side, you see that this is possible provided that $\alpha + \beta = -4$, $\alpha\beta = -20$.

Thus $\alpha$ and $\beta$ are the solutions of the quadratic equation $\lambda^2 + 4\lambda - 20 = 0$. These solutions are $\lambda = -2(1\pm\sqrt6\,).$

So if $x$ is a solution of the given quartic equation, it must satisfy $x^2 - 2(1\pm\sqrt6\,)x + 2 = 0$. Taking the equation with the positive sign for $\sqrt6$ gives the solution $x = 1 + \sqrt6 + \sqrt{(1+\sqrt6\,)^2 - 2}.$ But $(1+\sqrt6\,)^2 - 2 = 5 + 2\sqrt6 = 2 + 3 + 2\sqrt2\sqrt3 = (\sqrt2 + \sqrt3\,)^2.$ This gives the solution $x = 1 + \sqrt6 + \sqrt2 + \sqrt3$.

Therefore $p=1$ and $\{q,r,s\} = \{6,2,3\}.$

The other solutions to the quartic equation are $$1 + \sqrt6 - \sqrt2 - \sqrt3,$$ $$1 - \sqrt6 + \sqrt2 - \sqrt3,$$ $$1 - \sqrt6 - \sqrt2 + \sqrt3.$$