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Problem Of The Week #415 May 4th, 2020

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anemone

MHB POTW Director
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Feb 14, 2012
3,522
Here is this week's POTW:

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If $q>0$ and $p$ is a real number, prove that the polynomial $x^4-px^3+qx^2-\sqrt{q}x+1=0$ has no real roots.

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Remember to read the POTW submission guidelines to find out how to submit your answers!
 
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anemone

MHB POTW Director
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Feb 14, 2012
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Hello all!

I apologize to the community for posting a question that is misphrase.:oops:

The question should indeed read:

If $q>0$ and $p$ is a real number, prove that the polynomial $x^4−px^3+qx^2−\sqrt{q}x+1=0$ cannot have all real roots.

Thanks to Opalg for catching it!
 
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anemone

MHB POTW Director
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Feb 14, 2012
3,522
Congratulations to Opalg for his correct solution:cool:, which you can find below!

Solution from Opalg :
If all the roots are real then the polynomial can be factorised as \[(1) \qquad x^4 - px^3 + qx^2 - \sqrt qx + 1 = (x^2 + ax + c)(x^2 + bx + \frac1c),\] and both quadratic factors have real roots. The conditions for that are \[(2)\qquad a^2\geqslant 4c, \qquad b^2\geqslant \frac4c.\] Comparing coefficients of $x$ and $x^2$ in (1), \[(3)\qquad bc + \frac ac = -\sqrt q,\qquad c+ab + \frac1c = q.\] Therefore (using (2)) \[(4)\qquad c+ab + \frac1c = (bc + \frac ac)^2 = b^2c^2 + 2ab + \frac{a^2}{c^2} \geqslant 4c + 2ab + \frac4c,\] from which $3c + ab + \frac3c \leqslant0.$

But from (3), $c+ab + \frac1c = q >0$. It follows that $3c + ab + \frac3c \leqslant0 < c+ab + \frac1c$, from which $c + \frac1c <0$ and $ab>0$. But from (4), \[ (5)\qquad c + \frac1c =b^2c^2 + ab + \frac{a^2}{c^2}.\] However, the left side of (5) is negative and everything on the right side of (5) is positive. That contradiction shows that the original assumption (that all the roots of the polynomial are real) is false. So the polynomial cannot have all real roots.

Alternate solution from other:
Assume that the polynomial has 4 non-zero real roots.

We then let $y_1, y_2, y_3$ and $y_4$ be the inverses of those four real roots and they form another polynomial of the form f $y^4-\sqrt{q}y+qy^2-py+1=0$.

Vieta's formulas say we then have

$(y_1+y_2+y_3+y_4)^2=\sqrt{q}^2=y_1y_2+y_1y_3+y_1y_4+y_2y_3+y_2y_4+y_3y_4$

From the identity $(y_1+y_2+y_3+y_4)^2=y_1^2+y_2^2+y_3^2+y_4^2+2(y_1y_2+y_1y_3+y_1y_4+y_2y_3+y_2y_4+y_3y_4)$, we see that

$y_1^2+y_2^2+y_3^2+y_4^2=q-2q=-q<0$ (since $q>0$)

where we have reached a contradiction so our assumption was false.

Therefore, the polynomial cannot have all real roots.
 
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