Problems using Vieta's formulas: Very Difficult Problems with Solutions

Solution:
Case [tex]m=2[/tex]: the equation is linear, [tex]-4x-1=0[/tex], [tex]x=-\frac{1}{4}\lt 0[/tex]. So [tex]m=2[/tex] works.
Case [tex]m\ne 2[/tex]: we need real roots that are both negative:
[tex]\frac{D}{4}=m^2+(m-2)=(m+2)(m-1)\ge 0[/tex], so [tex]m\le -2[/tex] or [tex]m\ge 1[/tex];
[tex]x_1x_2=-\frac{1}{m-2}>0[/tex], so [tex]m\lt 2[/tex];
[tex]x_1+x_2=\frac{2m}{m-2}\lt 0[/tex], so [tex]0\lt m\lt 2[/tex].
Together: [tex]1\le m\lt 2[/tex] (for [tex]m=1[/tex] there is a double root [tex]x=-1[/tex]).
Answer: [tex]m\in[1, 2][/tex].

Solution:
By Vieta's formulas, [tex]x_1+x_2=1-a[/tex], [tex]x_1x_2=\frac{2a^2-5a}{6}[/tex].
[tex]x_1^3+x_2^3=(x_1+x_2)\left((x_1+x_2)^2-3x_1x_2\right)=(1-a)\left((1-a)^2-\frac{2a^2-5a}{2}\right)=\frac{(1-a)(a+2)}{2}=\frac{-a^2-a+2}{2}[/tex].
The roots must be real: [tex]\frac{D}{4}=9(a-1)^2-6(2a^2-5a)=-3a^2+12a+9\ge 0[/tex], i.e. [tex]a^2-4a-3\le 0[/tex], [tex]2-\sqrt{7}\le a\le 2+\sqrt{7}[/tex].
The parabola [tex]\frac{-a^2-a+2}{2}[/tex] opens downward and has its vertex at [tex]a=-\frac{1}{2}[/tex], which lies in this interval because [tex]2-\sqrt{7}<-\frac{1}{2}[/tex].
So the sum of cubes is greatest for [tex]a=-\frac{1}{2}[/tex], and the greatest value is [tex]\frac{9}{8}[/tex].

Solution:
Let [tex]f(t)=t^2-2kt+4[/tex]. A number lies strictly between the roots of [tex]f[/tex] exactly when [tex]f[/tex] is negative there. So we need two distinct numbers [tex]y_1, y_2[/tex] with [tex]f(y_1)\lt 0[/tex] and [tex]f(y_2)\lt 0[/tex].
If [tex]y[/tex] is a root of the second equation, then [tex]y^2+4=2(k-1)y[/tex], hence
[tex]f(y)=y^2+4-2ky=2(k-1)y-2ky=-2y[/tex].
So [tex]f(y_i)\lt 0[/tex] exactly when [tex]y_i>0[/tex]: the second equation must have two distinct positive roots.
[tex]\frac{D}{4}=(k-1)^2-4>0[/tex] gives [tex]k\lt -1[/tex] or [tex]k>3[/tex]; [tex]y_1+y_2=2(k-1)>0[/tex] gives [tex]k>1[/tex]; [tex]y_1y_2=4>0[/tex] always.
Answer: [tex]k>3[/tex].

Solution:
From the first equation: [tex]x_1+x_2=-p[/tex], [tex]x_1x_2=q[/tex].
From the second equation:
[tex](x_1+1)+(x_2+1)=p^2[/tex], i.e. [tex]-p+2=p^2[/tex], so [tex]p^2+p-2=0[/tex] and [tex]p=1[/tex] or [tex]p=-2[/tex];
[tex](x_1+1)(x_2+1)=x_1x_2+x_1+x_2+1=pq[/tex], i.e. [tex]q-p+1=pq[/tex].
[tex]p=-2[/tex]: [tex]q+3=-2q[/tex], [tex]q=-1[/tex]; the equation [tex]x^2-2x-1=0[/tex] has real roots.
[tex]p=1[/tex]: [tex]q=q[/tex] holds for every [tex]q[/tex], but [tex]x^2+x+q=0[/tex] must have real roots: [tex]D=1-4q\ge 0[/tex], i.e. [tex]q\le\frac{1}{4}[/tex].
Answer: [tex]p=-2, q=-1[/tex], or [tex]p=1[/tex] with any [tex]q\le\frac{1}{4}[/tex].

Solution:
[tex](x_1-b)(x_2-b)=x_1x_2-b(x_1+x_2)+b^2[/tex]. By Vieta's formulas:
[tex](x_1-b)(x_2-b)=\frac{2a-1}{a-1}-b\times\frac{a+1}{a-1}+b^2=\frac{(2-b)a-(1+b)}{a-1}+b^2[/tex].
The fraction does not depend on [tex]a[/tex] exactly when its numerator is a constant multiple of [tex]a-1[/tex]: [tex](2-b)a-(1+b)=\lambda(a-1)[/tex], i.e. [tex]\lambda=2-b[/tex] and [tex]\lambda=1+b[/tex].
So [tex]2-b=1+b[/tex] and [tex]b=\frac{1}{2}[/tex]. Then the fraction equals [tex]\frac{3}{2}[/tex] and [tex](x_1-b)(x_2-b)=\frac{3}{2}+\frac{1}{4}=\frac{7}{4}[/tex] for every admissible [tex]a[/tex].
(Admissible [tex]a[/tex] exist: for example, for [tex]a=\frac{3}{2}[/tex] we get [tex]D=\frac{9}{4}>0[/tex].)
Answer: [tex]b=\frac{1}{2}[/tex].

Solution:
Since [tex]a, b, c[/tex] form an arithmetic progression, [tex]2b=a+c[/tex].
[tex]\frac{D}{4}=b^2-ac=\frac{(a+c)^2}{4}-ac=\frac{(a-c)^2}{4}\ge 0[/tex], so the roots are real.
Let [tex]x_1=3x_2[/tex]. By Vieta's formulas: [tex]4x_2=\frac{2b}{a}[/tex], so [tex]x_2=\frac{b}{2a}[/tex]; and [tex]3x_2^2=\frac{c}{a}[/tex], so [tex]\frac{3b^2}{4a^2}=\frac{c}{a}[/tex], i.e. [tex]3b^2=4ac[/tex].
Substitute [tex]c=2b-a[/tex]: [tex]3b^2=8ab-4a^2[/tex], i.e. [tex]3b^2-8ab+4a^2=0[/tex], [tex](3b-2a)(b-2a)=0[/tex].
[tex]b=2a[/tex]: [tex]c=3a[/tex], so [tex]a:b:c=1:2:3[/tex].
[tex]b=\frac{2a}{3}[/tex]: [tex]c=\frac{a}{3}[/tex], so [tex]a:b:c=3:2:1[/tex].
Both ratios give positive numbers, so both are answers.

Solution:
By Vieta's formulas, [tex]x_1x_2=2012[/tex] and [tex]m=-(x_1+x_2)[/tex].
So the integer roots form a pair of integers with product [tex]2012=2^2\times 503[/tex], where [tex]503[/tex] is prime. Both roots have the same sign.
Positive pairs: [tex](1, 2012)[/tex], [tex](2, 1006)[/tex], [tex](4, 503)[/tex] with sums [tex]2013[/tex], [tex]1008[/tex], [tex]507[/tex]; the negative pairs give the opposite sums.
Hence [tex]m\in\{\pm 2013, \pm 1008, \pm 507\}[/tex], and conversely each of these [tex]m[/tex] gives integer roots.
Answer: [tex]6[/tex] values.

Solution:
By Vieta's formulas: [tex]c+d=10a[/tex] and [tex]a+b=10c[/tex].
Subtracting: [tex](c+d)-(a+b)=10a-10c[/tex], so [tex]b-d=11(c-a)[/tex].
Since [tex]c[/tex] is a root of the first equation and [tex]a[/tex] is a root of the second:
[tex]c^2-10ac-11b=0[/tex] and [tex]a^2-10ac-11d=0[/tex].
Subtracting: [tex]c^2-a^2=11(b-d)=121(c-a)[/tex].
Since [tex]c\ne a[/tex], divide by [tex]c-a[/tex]: [tex]a+c=121[/tex].
Then [tex]a+b+c+d=10c+10a=10\times 121=1210[/tex].

Solution:
The product of the roots is [tex]-\frac{1}{2p^2}<0[/tex], so the roots are real. By Vieta's formulas, [tex]x_1+x_2=2p[/tex], [tex]x_1x_2=-\frac{1}{2p^2}[/tex].
[tex]x_1^2+x_2^2=(x_1+x_2)^2-2x_1x_2=4p^2+\frac{1}{p^2}[/tex]
[tex]x_1^4+x_2^4=(x_1^2+x_2^2)^2-2x_1^2x_2^2=16p^4+8+\frac{1}{p^4}-\frac{1}{2p^4}=16p^4+\frac{1}{2p^4}+8[/tex].
By the AM-GM inequality, [tex]16p^4+\frac{1}{2p^4}\ge 2\sqrt{16p^4\times\frac{1}{2p^4}}=2\sqrt{8}=4\sqrt{2}[/tex].
Therefore [tex]x_1^4+x_2^4\ge 4\sqrt{2}+8[/tex], with equality when [tex]16p^4=\frac{1}{2p^4}[/tex], i.e. [tex]p^8=\frac{1}{32}[/tex].

Solution:
Since [tex]1, x_1, x_2[/tex] is a geometric progression, [tex]x_1^2=1\times x_2[/tex], so [tex]x_2=t^2[/tex], where [tex]t=x_1[/tex].
By Vieta's formulas: [tex]q=x_1x_2=t^3[/tex] and [tex]t+t^2=-p[/tex], i.e. [tex]t^2=-p-t[/tex].
Then [tex]t^3=t\times t^2=-pt-t^2=-pt+p+t[/tex], so [tex]q=(1-p)t+p[/tex].
If [tex]p=1[/tex], then [tex]t^2+t+1=0[/tex], which has no real roots; impossible. So [tex]p\ne 1[/tex] and [tex]t=\frac{q-p}{1-p}[/tex] is rational.
Write [tex]t=\frac{r}{s}[/tex] in lowest terms, [tex]s>0[/tex]. From [tex]t^2+t+p=0[/tex]: [tex]r^2+rs+ps^2=0[/tex], so [tex]s[/tex] divides [tex]r^2[/tex]. Since [tex]r[/tex] and [tex]s[/tex] have no common factor, [tex]s=1[/tex], i.e. [tex]t[/tex] is an integer.
Therefore [tex]q=t^3[/tex] is the cube of the integer [tex]t[/tex].

Solution:
Let [tex]S_n=x_1^n+x_2^n[/tex]. By Vieta's formulas, [tex]S_1=x_1+x_2=2001[/tex] and [tex]x_1x_2=1[/tex], so [tex]S_2=S_1^2-2x_1x_2=2001^2-2[/tex]. Both are integers.
Since [tex]x_1^2=2001x_1-1[/tex], multiplying by [tex]x_1^{n-2}[/tex] gives [tex]x_1^n=2001x_1^{n-1}-x_1^{n-2}[/tex]; the same holds for [tex]x_2[/tex]. Adding:
[tex]S_n=2001S_{n-1}-S_{n-2}[/tex] for [tex]n\ge 3[/tex].
Induction: if [tex]S_{n-1}[/tex] and [tex]S_{n-2}[/tex] are integers, then so is [tex]S_n[/tex]. Starting from the integers [tex]S_1[/tex] and [tex]S_2[/tex], every [tex]S_n[/tex] is an integer.