MHB Polynomial Challenge V: Real Solution Implies $p^2+q^2\ge 8$

[anemone]

Show that if $x^4+px^3+2x^2+qx+1$ has a real solution, then $p^2+q^2\ge 8$.

 

[Opalg]

Show that if $x^4+px^3+2x^2+qx+1 = 0$ has a real solution, then $p^2+q^2\ge 8$.

[sp]Divide through by $x^2$: $(x+x^{-1})^2 + px+qx^{-1} = 0$. Now let $y = x+x^{-1}$, so that the equation becomes $y^2 = -(px+qx^{-1}).$

By the Cauchy–Schwarz inequality, $(px+qx^{-1})^2 \leqslant (p^2+q^2)(x^2+x^{-2}) = (p^2+q^2)(y^2 - 2).$ If $y^2 = -(px+qx^{-1})$, then $y^4 = (px+qx^{-1})^2 \leqslant (p^2+q^2)(y^2 - 2)$, so that $y^4 - (p^2+q^2)y^2 + 2(p^2+q^2) \leqslant0$. The discriminant of that quadratic expression in $y^2$ is $(p^2+q^2)^2 - 8(p^2+q^2) = (p^2+q^2)(p^2+q^2-8).$ If $p^2+q^2<8$, the discriminant is negative and so the quadratic expression will have no real zeros. There would then be no real solutions for $y$ and therefore no real solutions for $x$.[/sp]

 

[kaliprasad]

[sp]Divide through by $x^2$: $(x+x^{-1})^2 + px+qx^{-1} = 0$. Now let $y = x+x^{-1}$, so that the equation becomes $y^2 = -(px+qx^{-1}).$

By the Cauchy–Schwarz inequality, $(px+qx^{-1})^2 \leqslant (p^2+q^2)(x^2+x^{-2}) = (p^2+q^2)(y^2 - 2).$ If $y^2 = -(px+qx^{-1})$, then $y^4 = (px+qx^{-1})^2 \leqslant (p^2+q^2)(y^2 - 2)$, so that $y^4 - (p^2+q^2)y^2 + 2(p^2+q^2) \leqslant0$. The discriminant of that quadratic expression in $y^2$ is $(p^2+q^2)^2 - 8(p^2+q^2) = (p^2+q^2)(p^2+q^2-8).$ If $p^2+q^2<8$, the discriminant is negative and so the quadratic expression will have no real zeros. There would then be no real solutions for $y$ and therefore no real solutions for $x$.[/sp]

Spoiler

you need to show that y does not lie between -2 and 2 ( or $y^2 \ge 4 $) else x becomes complex)

 

[Opalg]

Spoiler

you need to show that y does not lie between -2 and 2 ( or $y^2 \ge 4 $) else x becomes complex)

[sp]I was expecting to have to do that. But when I came to write out the solution, I saw that if $p^2+q^2<8$ then there is no real solution for $y$ (never mind whether it lies between $-2$ and $2$ or not), and therefore no real solution for $x$.[/sp]

 

[anemone]

[sp]Divide through by $x^2$: $(x+x^{-1})^2 + px+qx^{-1} = 0$. Now let $y = x+x^{-1}$, so that the equation becomes $y^2 = -(px+qx^{-1}).$

By the Cauchy–Schwarz inequality, $(px+qx^{-1})^2 \leqslant (p^2+q^2)(x^2+x^{-2}) = (p^2+q^2)(y^2 - 2).$ If $y^2 = -(px+qx^{-1})$, then $y^4 = (px+qx^{-1})^2 \leqslant (p^2+q^2)(y^2 - 2)$, so that $y^4 - (p^2+q^2)y^2 + 2(p^2+q^2) \leqslant0$. The discriminant of that quadratic expression in $y^2$ is $(p^2+q^2)^2 - 8(p^2+q^2) = (p^2+q^2)(p^2+q^2-8).$ If $p^2+q^2<8$, the discriminant is negative and so the quadratic expression will have no real zeros. There would then be no real solutions for $y$ and therefore no real solutions for $x$.[/sp]

Well done, Opalg! It never occurred to be to approach this problem using your method! Thank you Opalg for your neat and easy-to-follow solution and thanks for participating as well.

A solution that I saw online somewhere:

Spoiler

Note that the degree of the polynomial is even, if there exists one real root, then there must exist either two roots of multiplicity one or one root of multiplicity two. Therefore we may write:

$x^4+px^3+2x^2+qx+1=(x^2+ax\pm1)(x^2+bx\pm1)$ for some real $a,\,b$.

We now split the problem into two cases:

Case I: $x^4+px^3+2x^2+qx+1=(x^2+ax+1)(x^2+bx+1)$

The coefficient of $x^2$ from the RHS equals to $2+ab$, setting this equals to 2 gives $ab=0$. WLOG, let $a=0$, then the factorization boils down to $(x^2+1)(x^2+bx+1)$. Since $x^2+1$ has no real rots, $x^2+bx+1$ must have one, so $b^2\ge 4$. Finally, expanding out the RHS and comparing coefficients gives $p=q=b$, so $p^2+q^2=2b^2\ge2(4)=8$, as desired.

Case II: $x^4+px^3+2x^2+qx+1=(x^2+ax-1)(x^2+bx-1)$

The coefficient of $x^2$ from the RHS equals to $-2+ab$, setting this equals to 2 gives $ab=4$. WLOG, let $p=a+b$, and $p=-(a+b)$. This means that $p^2+q^2=2(a+b)^2=2(a^2+b^2)+4ab=2(a^2+b^2)+16$ and since $a^2+b^2\ge 0$, the entire expression is $\ge 16$, which obviously makes the expression greater than 8, as desired.

Finally, note that when both $x^2$ coefficients are negated, the only aspect of the problem that differs is the sign of the $ax$ and $bx$ coefficients. Running through the steps shown above, it can be seen that this does not affect the value of $p^2+q^2$.

Hence all cases are covered, and so we're done.