Polynomial Equations of Higher Degree: Very Difficult Problems with Solutions

Solution:
Rewrite the equation as [tex]x^4=2x^2+400x+9999[/tex] and add [tex]2x^2+1[/tex] to both sides, so that the left side becomes a perfect square:
[tex]x^4+2x^2+1=4x^2+400x+10000\iff\left(x^2+1\right)^2=\left(2x+100\right)^2[/tex]
Hence
[tex]\left(x^2+1-2x-100\right)\left(x^2+1+2x+100\right)=0\iff\left(x^2-2x-99\right)\left(x^2+2x+101\right)=0[/tex]
The first factor gives [tex]x^2-2x-99=(x-11)(x+9)=0[/tex], so [tex]x=11[/tex] or [tex]x=-9[/tex]. The second factor has no real zeros, because [tex]x^2+2x+101=(x+1)^2+100\gt 0[/tex]. Check:
[tex]11^4-2\times 11^2-400\times 11=14641-242-4400=9999[/tex]
[tex](-9)^4-2\times (-9)^2-400\times (-9)=6561-162+3600=9999[/tex]
Therefore [tex]x\in\left\{-9, 11\right\}[/tex].

Solution:
Denote [tex]c=\sqrt{5}[/tex], so that [tex]5=c^{2}[/tex], and consider the left side as a quadratic in [tex]c[/tex]:
[tex]c^{2}-\left(2x^{2}+1\right)c+x^{4}+x[/tex]
Its discriminant is a perfect square:
[tex]D=\left(2x^{2}+1\right)^{2}-4\left(x^{4}+x\right)=4x^{2}-4x+1=(2x-1)^{2}[/tex]
so its roots with respect to [tex]c[/tex] are
[tex]c_{1,2}=\frac{2x^{2}+1\pm(2x-1)}{2},\qquad c_{1}=x^{2}+x,\qquad c_{2}=x^{2}-x+1[/tex]
Hence [tex]c^{2}-\left(2x^{2}+1\right)c+x^{4}+x=\left(c-x^{2}-x\right)\left(c-x^{2}+x-1\right)[/tex] (indeed [tex]\left(x^{2}+x\right)\left(x^{2}-x+1\right)=x^{4}+x[/tex]), and the equation is equivalent to
[tex]\left(x^{2}+x-\sqrt{5}\right)\left(x^{2}-x+1-\sqrt{5}\right)=0[/tex]
The equation [tex]x^{2}+x-\sqrt{5}=0[/tex] has discriminant [tex]1+4\sqrt{5}\gt 0[/tex], so it has two real roots. The equation [tex]x^{2}-x+1-\sqrt{5}=0[/tex] has discriminant [tex]1-4\left(1-\sqrt{5}\right)=4\sqrt{5}-3\gt 0[/tex], so it also has two real roots.
A common root would satisfy [tex]x^{2}+x=x^{2}-x+1[/tex], i.e. [tex]x=\frac{1}{2}[/tex], but then [tex]x^{2}+x=\frac{3}{4}\ne\sqrt{5}[/tex]. Therefore the four roots are different, and the equation has [tex]4[/tex] real roots.

Solution:
Let [tex]x_{1},x_{2},x_{3},x_{4}[/tex] be the roots, with [tex]x_{1}x_{2}=-32[/tex]. By Vieta's formulas
[tex]x_{1}+x_{2}+x_{3}+x_{4}=18,\quad \sum_{i\lt j}x_{i}x_{j}=k,\quad \sum_{i\lt j\lt l}x_{i}x_{j}x_{l}=-200,\quad x_{1}x_{2}x_{3}x_{4}=-1984[/tex]
From the last formula [tex]x_{3}x_{4}=\frac{-1984}{-32}=62[/tex]. Put [tex]s=x_{1}+x_{2}[/tex] and [tex]t=x_{3}+x_{4}[/tex], so [tex]s+t=18[/tex]. The sum of the products of three roots is
[tex]x_{1}x_{2}\left(x_{3}+x_{4}\right)+x_{3}x_{4}\left(x_{1}+x_{2}\right)=-32t+62s=-200[/tex]
Substituting [tex]t=18-s[/tex]:
[tex]-576+32s+62s=-200\Rightarrow 94s=376\Rightarrow s=4,\quad t=14[/tex]
Finally,
[tex]k=x_{1}x_{2}+x_{3}x_{4}+\left(x_{1}+x_{2}\right)\left(x_{3}+x_{4}\right)=-32+62+4\times 14=86[/tex]
Check: [tex]x^{4}-18x^{3}+86x^{2}+200x-1984=\left(x^{2}-4x-32\right)\left(x^{2}-14x+62\right)[/tex], and the roots [tex]8[/tex] and [tex]-4[/tex] of the first factor have product [tex]-32[/tex].

Solution:
Let [tex]f(x)=9x^3-36x^2+45x-17[/tex]. We have
[tex]f(0)=-17\lt 0,\quad f(1)=1\gt 0,\quad f\left(\frac{3}{2}\right)=-\frac{1}{8}\lt 0,\quad f(2)=1\gt 0[/tex]
so the cubic equation has three real roots [tex]x_1\in\left(0, 1\right)[/tex], [tex]x_2\in\left(1, \frac{3}{2}\right)[/tex], [tex]x_3\in\left(\frac{3}{2}, 2\right)[/tex], all of them positive. By Vieta's formulas [tex]x_1+x_2+x_3=\frac{36}{9}=4[/tex], so the semi-perimeter is [tex]p=2[/tex]. Each root is less than [tex]2=p[/tex], i.e. each side is less than the sum of the other two, so such a triangle exists.
By Heron's formula [tex]S^2=p(p-x_1)(p-x_2)(p-x_3)[/tex]. Since [tex]f(x)=9(x-x_1)(x-x_2)(x-x_3)[/tex], we have
[tex]9(2-x_1)(2-x_2)(2-x_3)=f(2)=72-144+90-17=1\Rightarrow (2-x_1)(2-x_2)(2-x_3)=\frac{1}{9}[/tex]
Hence
[tex]S^2=2\times \frac{1}{9}=\frac{2}{9}\Rightarrow S=\frac{\sqrt{2}}{3}[/tex]

Solution:
The equation is [tex]x\left(25x^4+25(a-1)x^2-4(a-7)\right)=0[/tex], so [tex]x=0[/tex] is a root and the other roots satisfy the biquadratic equation [tex]25x^4+25(a-1)x^2-4(a-7)=0[/tex]. Its non-zero roots come in pairs [tex]\pm x[/tex], so the five distinct roots must be [tex]0[/tex], [tex]\pm p[/tex], [tex]\pm q[/tex] with [tex]0\lt p\lt q[/tex], i.e. the equation
[tex]25t^2+25(a-1)t-4(a-7)=0,\qquad t=x^2[/tex]
has two distinct positive roots [tex]p^2[/tex] and [tex]q^2[/tex]. The ordered roots [tex]-q, -p, 0, p, q[/tex] form an arithmetic progression exactly when [tex]p-0=q-p[/tex], i.e. [tex]q=2p[/tex]. Then the roots in [tex]t[/tex] are [tex]p^2[/tex] and [tex]4p^2[/tex], and by Vieta's formulas
[tex]p^2+4p^2=5p^2=1-a,\qquad p^2\times 4p^2=4p^4=\frac{4(7-a)}{25}[/tex]
From the first equality [tex]p^4=\frac{(1-a)^2}{25}[/tex], hence
[tex](1-a)^2=7-a\iff a^2-a-6=0\iff a=3\text{ or }a=-2[/tex]
Since [tex]p^2=\frac{1-a}{5}\gt 0[/tex], we need [tex]a\lt 1[/tex], so [tex]a=3[/tex] is rejected (for [tex]a=3[/tex] the equation in [tex]t[/tex] is [tex]25t^2+50t+16=0[/tex] with negative roots, and [tex]x=0[/tex] is the only real root). For [tex]a=-2[/tex] the equation in [tex]t[/tex] is [tex]25t^2-75t+36=0[/tex] with roots [tex]\frac{3}{5}[/tex] and [tex]\frac{12}{5}=4\times \frac{3}{5}[/tex], and the roots of the given equation are
[tex]-\frac{2\sqrt{15}}{5}, -\frac{\sqrt{15}}{5}, 0, \frac{\sqrt{15}}{5}, \frac{2\sqrt{15}}{5}[/tex]
They form an arithmetic progression with common difference [tex]\frac{\sqrt{15}}{5}[/tex]. Hence [tex]a=-2[/tex].

Solution:
Assume the contrary: all six roots [tex]x_{1},x_{2},\ldots,x_{6}[/tex] of [tex]p(x)[/tex] (counted with multiplicity) are real. Since the coefficients of [tex]x^{5}[/tex] and [tex]x^{4}[/tex] are zero, Vieta's formulas give
[tex]x_{1}+x_{2}+\ldots+x_{6}=0,\qquad \sum_{i\lt j}x_{i}x_{j}=0[/tex]
Then
[tex]x_{1}^{2}+x_{2}^{2}+\ldots+x_{6}^{2}=\left(x_{1}+\ldots+x_{6}\right)^{2}-2\sum_{i\lt j}x_{i}x_{j}=0-0=0[/tex]
A sum of squares of real numbers is zero only if all of them are zero, so [tex]x_{1}=x_{2}=\ldots=x_{6}=0[/tex] and [tex]p(x)=x^{6}[/tex]. This means [tex]a=b=c=d=0[/tex], which contradicts the condition. Hence [tex]p(x)[/tex] has at least one non-real root.

Solution:
For [tex]x\gt 0[/tex] we can write
[tex]P(x)=x^n\left(1-g(x)\right),\qquad g(x)=\frac{a_1}{x}+\frac{a_2}{x^2}+\ldots+\frac{a_n}{x^n}[/tex]
and since [tex]x^n\gt 0[/tex], for positive [tex]x[/tex] the equation [tex]P(x)=0[/tex] is equivalent to [tex]g(x)=1[/tex].
Uniqueness. On [tex]\left(0, +\infty\right)[/tex] every term [tex]\frac{a_k}{x^k}[/tex] is positive, continuous and strictly decreasing (because [tex]a_k\gt 0[/tex]: if [tex]0\lt u\lt v[/tex], then [tex]\frac{a_k}{u^k}\gt\frac{a_k}{v^k}[/tex]). Hence [tex]g[/tex] is continuous and strictly decreasing on [tex]\left(0, +\infty\right)[/tex], so the equation [tex]g(x)=1[/tex] has at most one positive solution.
Existence. Let [tex]S=a_1+a_2+\ldots+a_n[/tex] and [tex]x_0=S+1\gt 1[/tex]. Then [tex]x_0^k\ge x_0[/tex] for [tex]k\ge 1[/tex], so
[tex]g(x_0)\le\frac{a_1+a_2+\ldots+a_n}{x_0}=\frac{S}{S+1}\lt 1[/tex]
For [tex]x_1=\frac{a_1}{2}[/tex] (note that [tex]x_1\lt x_0[/tex]) all terms of [tex]g(x_1)[/tex] are positive, so
[tex]g(x_1)\ge\frac{a_1}{x_1}=2\gt 1[/tex]
The continuous function [tex]g(x)-1[/tex] changes sign on [tex]\left[x_1, x_0\right][/tex], so by Bolzano's theorem there is [tex]c\in\left(x_1, x_0\right)[/tex] with [tex]g(c)=1[/tex], i.e. [tex]P(c)=c^n\left(1-g(c)\right)=0[/tex]. Therefore the equation [tex]P(x)=0[/tex] has exactly one positive root.

Solution:
The equation has real roots: its left side is continuous and equals [tex]7[/tex] for [tex]x=-2[/tex], [tex]-1[/tex] for [tex]x=0[/tex] and [tex]1[/tex] for [tex]x=1[/tex], so there is a root in [tex]\left(-2, 0\right)[/tex] and a root in [tex]\left(0, 1\right)[/tex]. Let [tex]a\ne b[/tex] be two real roots and let [tex]c[/tex] and [tex]d[/tex] be the other two roots (real or complex). By Vieta's formulas
[tex]a+b+c+d=-1,\quad ab+ac+ad+bc+bd+cd=0,\quad abc+abd+acd+bcd=0,\quad abcd=-1[/tex]
Put [tex]s=a+b[/tex], [tex]p=ab[/tex], [tex]t=c+d[/tex], [tex]q=cd[/tex]. The formulas become
[tex]s+t=-1,\quad p+q+st=0,\quad pt+qs=0,\quad pq=-1[/tex]
Here [tex]p[/tex] is a real number different from [tex]0[/tex] (because [tex]0[/tex] is not a root), so [tex]q=-\frac{1}{p}[/tex] and [tex]t=-1-s[/tex]. The third relation gives [tex]p(-1-s)-\frac{s}{p}=0[/tex]; multiplying by [tex]p[/tex]:
[tex]-p^{2}-p^{2}s-s=0\Rightarrow s\left(p^{2}+1\right)=-p^{2}\Rightarrow s=-\frac{p^{2}}{p^{2}+1}[/tex]
The second relation gives [tex]p-\frac{1}{p}-s(1+s)=0[/tex], i.e. [tex]p-\frac{1}{p}-s-s^{2}=0[/tex]. Multiply it by [tex]p\left(p^{2}+1\right)^{2}\ne 0[/tex] and substitute [tex]s[/tex]:
[tex]p^{2}\left(p^{2}+1\right)^{2}-\left(p^{2}+1\right)^{2}+p^{3}\left(p^{2}+1\right)-p^{5}=0[/tex]
Since [tex]\left(p^{2}-1\right)\left(p^{2}+1\right)^{2}=p^{6}+p^{4}-p^{2}-1[/tex] and [tex]p^{3}\left(p^{2}+1\right)-p^{5}=p^{3}[/tex], this is
[tex]p^{6}+p^{4}+p^{3}-p^{2}-1=0[/tex]
So the product [tex]p=ab[/tex] is a root of the equation [tex]x^{6}+x^{4}+x^{3}-x^{2}-1=0[/tex].