# Polynomial Doubt [duplicate]

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 ?

## marked as duplicate by Saad, Michael Rozenberg algebra-precalculus StackExchange.ready(function() { if (StackExchange.options.isMobile) return; $('.dupe-hammer-message-hover:not(.hover-bound)').each(function() { var$hover = $(this).addClass('hover-bound'),$msg = $hover.siblings('.dupe-hammer-message');$hover.hover( function() { $hover.showInfoMessage('', { messageElement:$msg.clone().show(), transient: false, position: { my: 'bottom left', at: 'top center', offsetTop: -7 }, dismissable: false, relativeToBody: true }); }, function() { StackExchange.helpers.removeMessages(); } ); }); }); Mar 3 '18 at 4:40

• Do you mean "all roots are real" or "at least one root is real"? – John Wayland Bales Mar 3 '18 at 0:22
• If $q=0$ and either $p=1$ or $p=-1$ then $p^2+q^2=1$ and the polynomial has a real root. So the minimum is less than or equal to $1$. – John Wayland Bales Mar 3 '18 at 0:51
• – kingW3 Mar 3 '18 at 1:36

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

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

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 $p^2 + q^2$.