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Question :

$x^4 + px^3 + qx^2 + px + 1 =0 $ has real roots. Then what is the minimum value of $ p^2 +q^2 $ .

How I started ?

I started by dividing the whole equation by $x^2$ then we get $ (x + \frac{1}{x} ) ^2 + p (x + \frac{1}{x} ) + q - 2 = 0 $ Then put $(x + \frac{1}{x} ) = t$. Then discriminant should be greater than equal to zero. But now the problem arises that $t$ does not belong to $(-2,2)$ , so taking care of that part leads to solving inequality which I am unable to do .

Have I started the right way? One more thing to notice is that the sum of roots of the equation is equal to the sum of reciprocal of the roots . How to proceed further ?

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marked as duplicate by Saad, Michael Rozenberg algebra-precalculus Mar 3 '18 at 4:40

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

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I don't know how much this is going to help, but if you use this you can find the roots

\begin{eqnarray} x_1&=& -\frac{1}{4} \sqrt{p^2-4 q+8}-\frac{1}{2} \sqrt{\frac{p^2}{2}-\frac{-p^3+4 p q-8 p}{2 \sqrt{p^2-4 q+8}}-q-2}-\frac{p}{4},\\ x_2&=& -\frac{1}{4} \sqrt{p^2-4 q+8}+\frac{1}{2} \sqrt{\frac{p^2}{2}-\frac{-p^3+4 p q-8 p}{2 \sqrt{p^2-4 q+8}}-q-2}-\frac{p}{4},\\ x_3&=& \frac{1}{4} \sqrt{p^2-4 q+8}-\frac{1}{2} \sqrt{\frac{p^2}{2}+\frac{-p^3+4 p q-8 p}{2 \sqrt{p^2-4 q+8}}-q-2}-\frac{p}{4}, \\ x_4&=& \frac{1}{4} \sqrt{p^2-4 q+8}+\frac{1}{2} \sqrt{\frac{p^2}{2}+\frac{-p^3+4 p q-8 p}{2 \sqrt{p^2-4 q+8}}-q-2}-\frac{p}{4} \end{eqnarray}

And from here the constraint

$$ p^2 - 4q+8 > 0 \tag{1} $$

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Not a full solution, but could be a direction to go with: $$t = x + \frac{1}{x} \in (-\infty, -2] \cup [2, +\infty) \text{ for } x \in \mathbb R$$

We want $t^2 + pt + q -2 = 0$ to have solution for $t$ as above:

$$p^2-4(q-2) \ge 0 \text{ and }$$ $$\frac{-p+\sqrt{p^2-4(q-2)}}{2} \ge 2 \text{ or } \frac{-p-\sqrt{p^2-4(q-2)}}{2} \le -2$$

Then we could get relations about $p$ and $q$, and get the min of $p^2 + q^2$.

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