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If $\log _{\frac{1}{3}} (|a|+1)>-1$, then find the domain of

$$f(x)=\sqrt{2x^4+ax^3-6x^2-4ax-8}$$

Using $\log _{\frac{1}{3}} (|a|+1)>-1$, I got $-2<a<2$ but to use it in $\sqrt{2x^4+ax^3-6x^2-4ax-8} \geq0$

Could someone help me with this.

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HINT: $$2x^4+ax^3-6x^2-4ax-8=2x^4-6x^2-8+ax(x^2-4)=\\=2(x^4-3x^2-4)+ax(x^2-4)=2(x^2-4)(x^2+1)+ax(x^2-4)=(x^2-4)(2x^2+ax+2)\\$$ NOTE:For factoring $x^4-3x^2-4$ you can use the change of variables $t=x^2$ and use a quadratic.

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  • $\begingroup$ Brilliant. How did you observe this? Could you share a bit of your thought process? $\endgroup$ Commented Apr 10, 2017 at 22:19
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    $\begingroup$ My thought was that you can't find a domain of a fourth power polynomial and even harder with a parameter,derivatives also won't help much,so the second thought was that when you group terms by $a$ you will get $x^2-4$ now on first look it's possible that the rest of polynomial shares at least the $(x-2)$ factor or in best case whole $x^2-4$ $\endgroup$
    – kingW3
    Commented Apr 10, 2017 at 22:24

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