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If two roots of the equation $(a-1)(x^2+x+1)^2-(a+1)(x^4+x^2+1)=0$ are real and distinct, then find the interval in which $a$ lies.

My Approach:
I have expanded the equation to obtain a quartic equation which I am not able to factorize: $$ x^4+(1-a)x^3+(2-a)x^2-ax+1=0 $$

If only I could factorize it I would get two quadratic equations, one of which should have real roots. I know how to proceed further in order to find the interval in which a lies.

But as for now I don't know how to proceed further. It would be great if I could get a hint to move forward.

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  • $\begingroup$ Just to be sure, as your question seems to suggest it; is $a$ assumed to be real? $\endgroup$ – Servaes Dec 31 '16 at 3:34
  • $\begingroup$ @Servaes yes thats what the options in the mcq indicate too $\endgroup$ – Osheen Sachdev Dec 31 '16 at 3:35
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Hint: divide by $x^2 \ne 0$ (since $x=0$ is obviously not a root) and rewrite the equation as:

$$(a-1)(x+\frac{1}{x}+1)^2-(a+1)(x^2+\frac{1}{x^2}+1)=0$$

Let $t = x+\cfrac{1}{x}$ then the equation becomes a quadratic in $t$ which then factors by inspection:

$$(a-1)(t+1)^2-(a+1)(t^2-1)=0$$

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The term $x^4 + x^2 + 1$ easily gets factorised as follows:

$$ x^4 + x^2 + 1 = (x^2 + x + 1)(x^2 - x + 1) $$

Applying this in the original equation,

$$ (x^2 + x + 1)[(a - 1)(x^2 + x + 1) + (a + 1)(x^2 - x + 1)] = 0. $$

Here $x^2 + x + 1 = 0$ is the quadratic with the complex roots. Focusing on the other quadratic, it can be simplified to:

$$ ax^2 - x + a = 0 $$

For this to have real and distinct roots, $1 - 4a^2 > 0 \implies a^2 < 1/4$ which simply gives us

$$ a \in \left (- \frac{1}{2}, \frac{1}{2}\right ). $$

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    $\begingroup$ Good answer, but $a$ cannot be zero because then your quadratic would not remain quadratic and discriminant would not be defined. $\endgroup$ – akhmeteni Sep 23 '17 at 8:10
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HINT: Note that $$x^4+x^2+1=(x^2+x+1)(x^2-x+1).$$

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