Problems using Vieta's formulas: Difficult Problems with Solutions

Solution:
First, we rewrite [tex]x_1^2+x_2^2[/tex] in terms of elementary symmetric polynomials:

[tex]x_1^2+x_2^2=x_1^2+2x_1x_2+x_2^2-2x_1x_2=(x_1+x_2)^2-2x_1x_2[/tex].

By Vieta's formulas we have [tex]x_1+x_2=-5[/tex] and [tex]x_1x_2=-3[/tex].

We substitute in the expression in order to get [tex]x_1^2+x_2^2=(-5)^2-2\cdot(-3)=25+6=31[/tex].

Solution:
First, we rewrite [tex]x_1^2+x_2^2[/tex] in terms of elementary symmetric polynomials:

[tex]x_1^2+x_2^2=x_1^2+2x_1x_2+x_2^2-2x_1x_2=(x_1+x_2)^2-2x_1x_2[/tex].

By Vieta's formulas we have [tex]x_1+x_2=-11[/tex] and [tex]x_1x_2=12[/tex].

We substitute in the expression in order to get [tex]x_1^2+x_2^2=(-11)^2-2\times 12=121-24=97[/tex].

Solution:
[tex]\frac{1}{x_1}+\frac{1}{x_2}=\frac{x_2}{x_1x_2}+\frac{x_1}{x_1x_2}=\frac{x_1+x_2}{x_1x_2}[/tex].

From Vieta's formulas, we have [tex]x_1+x_2=-9[/tex], [tex]x_1x_2=33[/tex].

We substitute and get: [tex]\frac{x_1+x_2}{x_1x_2}=\frac{-9}{33}=-\frac{3}{11}[/tex]

Solution:
Since [tex]n[/tex] is a root, [tex]n^2-n-2018=0[/tex], so [tex]n^2=n+2018[/tex].
Then [tex]m+n^2=m+n+2018[/tex].
By Vieta's formulas [tex]m+n=1[/tex], so [tex]m+n^2=1+2018=2019[/tex].

Solution:
We rewrite the expression as [tex]x_1-x_1^2+x_2-x_2^2=x_1+x_2-(x_1^2+x_2^2)=x_1+x_2-(x_1^2+2x_1x_2+x_2^2-2x_1x_2)=x_1+x_2-((x_1+x_2)^2-2x_1x_2)=(x_1+x_2)-(x_1+x_2)^2+2x_1x_2=(x_1+x_2)(1-(x_1+x_2))+2x_1x_2[/tex].

From Vieta's formulas we know that [tex]x_1+x_2=12[/tex] and [tex]x_1x_2=19[/tex].

We substitute and get [tex]12(1-12)+2\times 19=-12\times 11+38=-132+38=-94[/tex]

Solution:
First, we will use the following identity: [tex]|a|=\sqrt{a^2}[/tex].

This leads us to [tex]|x_1-x_2|=\sqrt{(x_1-x_2)^2}=\sqrt{x_1^2-2x_1x_2+x_2^2}=\sqrt{x_1^2+2x_1x_2+x_2^2-4x_1x_2}=\sqrt{(x_1+x_2)^2-4x_1x_2}[/tex].

Then we use Vieta's formulas to get [tex]x_1+x_2=15[/tex], [tex]x_1x_2=36[/tex].

The result is [tex]\sqrt{15^2-4 \times 36}=\sqrt{225-144}=\sqrt{81}=9[/tex]

Solution:
Let [tex]x_1=x_2+3[/tex]. By Vieta's formulas, [tex]x_1+x_2=15[/tex], so [tex]2x_2+3=15[/tex], [tex]x_2=6[/tex], [tex]x_1=9[/tex].
Then [tex]x_1x_2=54=k^2-10[/tex], so [tex]k^2=64[/tex] and [tex]k=\pm 8[/tex].
Check: for [tex]k=\pm 8[/tex] the equation is [tex]x^2-15x+54=0[/tex] with roots [tex]6[/tex] and [tex]9[/tex].

Solution:
[tex]D=9-4=5>0[/tex], [tex]x_1+x_2=3>0[/tex], [tex]x_1x_2=1>0[/tex], so both roots are positive and the square roots exist.
Let [tex]A=\sqrt{x_1}+\sqrt{x_2}>0[/tex]. Square it:
[tex]A^2=x_1+x_2+2\sqrt{x_1x_2}=3+2\sqrt{1}=5[/tex].
Since [tex]A>0[/tex], [tex]A=\sqrt{5}[/tex].

Solution:
Opposite roots have sum [tex]0[/tex]: [tex]x_1+x_2=\frac{m^2-m-12}{m}=0[/tex], so [tex]m^2-m-12=0[/tex], giving [tex]m=4[/tex] or [tex]m=-3[/tex].
For real opposite roots [tex]x[/tex] and [tex]-x[/tex], the product is [tex]-x^2\le 0[/tex]; here it cannot be [tex]0[/tex], so [tex]x_1x_2=\frac{37}{m}<0[/tex], i.e. [tex]m<0[/tex].
[tex]m=4[/tex] gives [tex]4x^2+37=0[/tex], which has no real roots.
[tex]m=-3[/tex] gives [tex]-3x^2+37=0[/tex], with roots [tex]\pm\sqrt{\frac{37}{3}}[/tex].
Answer: [tex]m=-3[/tex].

Solution:
We remove the parentheses and express in terms of the elementary symmetric polynomials:

[tex]x_1^2-2x_1\frac{1}{x_1}+\frac{1}{x_1^2}+x_2^2-2x_2\frac{1}{x_2}+\frac{1}{x_2^2}=x_1^2+x_2^2+\frac{1}{x_1^2}+\frac{1}{x_2^2}-4=x_1^2+x_2^2+\frac{x_1^2+x_2^2}{(x_1x_2)^2}-4=(x_1^2+x_2^2)(1+\frac{1}{(x_1x_2)^2})-4=((x_1+x_2)^2-2x_1x_2)(1+\frac{1}{(x_1x_2)^2})-4[/tex].

By Vieta's formulas we have [tex]x_1+x_2=4[/tex] and [tex]x_1x_2=1[/tex].

We substitute and get: [tex](4^2-2)(1+1)-4=(16-2)\times 2-4=28-4=24[/tex]

Solution:
We first rewrite the desired expression in terms of the elementary symmetric polynomials.

[tex](x_1^3+x_2^3)+(x_1^2+x_2^2)+(x_1+x_2)=(x_1+x_2)(x_1^2+x_2^2-x_1x_2)+(x_1^2+2x_1x_2+x_2^2-2x_1x_2)+(x_1+x_2) = (x_1+x_2)(x_1^2+2x_1x_2+x_2^2-3x_1x_2)+(x_1+x_2)^2-2x_1x_2+(x_1+x_2)=(x_1+x_2)((x_1+x_2)^2-3x_1x_2)+(x_1+x_2)^2+(x_1+x_2)-2x_1x_2[/tex].

From Vieta's formulas, we know that [tex]x_1+x_2=8[/tex] and [tex]x_2x_2=11[/tex].

We substitute: [tex]8(8^2-3\times 11)+8^2+8-2\times 11=8(64-33)+64+8-22=8\times 31+50=8\times 31+8 \times 6 +2=8\times 37 + 2=298[/tex]

Solution:
[tex]\frac{D}{4}=36-28=8>0[/tex]. By Vieta's formulas, [tex]x_1+x_2=\frac{12}{7}[/tex], [tex]x_1x_2=\frac{4}{7}[/tex].
Sum of the new roots: [tex]x_1^2+x_2^2=(x_1+x_2)^2-2x_1x_2=\frac{144}{49}-\frac{56}{49}=\frac{88}{49}[/tex].
Product of the new roots: [tex]x_1^2x_2^2=(x_1x_2)^2=\frac{16}{49}[/tex].
So [tex]x^2-\frac{88}{49}x+\frac{16}{49}=0[/tex], i.e. [tex]49x^2-88x+16=0[/tex].

Solution:
From Vieta's formulas we have [tex]x_1x_2=a^2-2a+1=(a-1)^2[/tex]. Since it is a perfect square, [tex](a-1)^2 \ge 0[/tex], with equality for [tex]a=1[/tex]. So the answer is [tex]a=1[/tex] and the minimal value is 0.

Solution:
Two distinct positive roots require [tex]D>0[/tex], [tex]x_1+x_2>0[/tex] and [tex]x_1x_2>0[/tex].
[tex]D=(2m-1)^2+4(m+3)=4m^2+13>0[/tex] for every [tex]m[/tex].
[tex]x_1+x_2=1-2m>0[/tex] gives [tex]m\lt \frac{1}{2}[/tex].
[tex]x_1x_2=-m-3>0[/tex] gives [tex]m\lt -3[/tex].
All three hold for [tex]m\lt -3[/tex].

Solution:
By Vieta's formulas, [tex]x_1^2+x_2^2=(x_1+x_2)^2-2x_1x_2=k^2-2[/tex].
The condition [tex]k^2-2\le 2[/tex] gives [tex]k^2\le 4[/tex].
But the roots must be real: [tex]D=k^2-4\ge 0[/tex], i.e. [tex]k^2\ge 4[/tex].
Both hold only when [tex]k^2=4[/tex], so [tex]k=\pm 2[/tex] (then the equation has the double root [tex]\mp 1[/tex]).

Solution:
By the converse of Vieta's theorem, [tex]x[/tex] and [tex]y[/tex] are the roots of [tex]t^2-2at+a^2-1=0[/tex].
The number [tex]1[/tex] lies strictly between them exactly when [tex](x-1)(y-1)\lt 0[/tex]:
[tex](x-1)(y-1)=xy-(x+y)+1=a^2-1-2a+1=a^2-2a\lt 0[/tex].
So [tex]a(a-2)\lt 0[/tex], i.e. [tex]0\lt a\lt 2[/tex].
Check: the roots are [tex]a-1[/tex] and [tex]a+1[/tex], and [tex]a-1\lt 1\lt a+1[/tex] means [tex]0\lt a\lt 2[/tex].

Solution:
From the condition, [tex]x_1=2x_2+3[/tex]. By Vieta's formulas, [tex]x_1x_2=5[/tex]:
[tex](2x_2+3)x_2=5[/tex], so [tex]2x_2^2+3x_2-5=0[/tex], giving [tex]x_2=1[/tex] or [tex]x_2=-\frac{5}{2}[/tex].
[tex]x_2=1[/tex]: [tex]x_1=5[/tex] and [tex]a=-(x_1+x_2)=-6[/tex].
[tex]x_2=-\frac{5}{2}[/tex]: [tex]x_1=-2[/tex] and [tex]a=-(x_1+x_2)=\frac{9}{2}[/tex].
Answer: [tex]a=-6[/tex] or [tex]a=\frac{9}{2}[/tex].

Solution:
Let [tex]x_1=x_2^2[/tex]. By Vieta's formulas:
[tex]x_2^2+x_2=\frac{15}{4}[/tex], so [tex]4x_2^2+4x_2-15=0[/tex] and [tex]x_2=\frac{-4\pm 16}{8}[/tex], i.e. [tex]x_2=\frac{3}{2}[/tex] or [tex]x_2=-\frac{5}{2}[/tex].
[tex]x_1x_2=x_2^3=a^3[/tex], so [tex]a=x_2[/tex].
Check [tex]a=\frac{3}{2}[/tex]: roots [tex]\frac{3}{2}[/tex] and [tex]\frac{9}{4}[/tex]; check [tex]a=-\frac{5}{2}[/tex]: roots [tex]-\frac{5}{2}[/tex] and [tex]\frac{25}{4}[/tex]. Both work.
Answer: [tex]a=\frac{3}{2}[/tex] or [tex]a=-\frac{5}{2}[/tex].

Solution:
[tex]D=(p-3)^2+4>0[/tex], so the roots are always real; their product is [tex]-1\ne 0[/tex].
By Vieta's formulas, [tex]x_1+x_2=3-p[/tex], [tex]x_1x_2=-1[/tex]:
[tex]\frac{x_1}{x_2}+\frac{x_2}{x_1}=\frac{(x_1+x_2)^2-2x_1x_2}{x_1x_2}=\frac{(p-3)^2+2}{-1}[/tex].
The condition becomes [tex]-(p-3)^2-2+3=0[/tex], i.e. [tex](p-3)^2=1[/tex], so [tex]p=2[/tex] or [tex]p=4[/tex].
For [tex]p=2[/tex]: [tex]x^2-x-1=0[/tex], [tex]x_{1,2}=\frac{1\pm\sqrt{5}}{2}[/tex]; for [tex]p=4[/tex]: [tex]x^2+x-1=0[/tex], [tex]x_{1,2}=\frac{-1\pm\sqrt{5}}{2}[/tex].

Solution:
By Vieta's formulas, [tex]S=(x_1+x_2)^2-2x_1x_2=4-2a^2[/tex].
The roots are real only when [tex]\frac{D}{4}=1-a^2\ge 0[/tex], i.e. [tex]0\le a^2\le 1[/tex].
Then [tex]2\le 4-2a^2\le 4[/tex].
Greatest value [tex]S=4[/tex] at [tex]a=0[/tex] (roots [tex]0[/tex] and [tex]2[/tex]); least value [tex]S=2[/tex] at [tex]a=\pm 1[/tex] (double root [tex]1[/tex]).
Note: without the condition [tex]D\ge 0[/tex] one would wrongly conclude that [tex]S[/tex] has no least value.

Solution:
Real roots: [tex]D=a^2-4a-12=(a-6)(a+2)\ge 0[/tex], so [tex]a\le -2[/tex] or [tex]a\ge 6[/tex].
Both sides of [tex]|x_1-x_2|\le 3[/tex] are non-negative, so we may square:
[tex](x_1-x_2)^2=(x_1+x_2)^2-4x_1x_2=a^2-4(a+3)=a^2-4a-12\le 9[/tex].
[tex]a^2-4a-21\le 0[/tex], i.e. [tex](a-7)(a+3)\le 0[/tex], so [tex]-3\le a\le 7[/tex].
Intersecting with the first condition: [tex]a\in[-3, -2]\cup[6, 7][/tex].

Solution:
[tex]x_1x_2=-(m^2+2m+2)=-\left((m+1)^2+1\right)<0[/tex] for every [tex]m[/tex], so the roots are real and of opposite signs.
Let [tex]x_1>x_2[/tex] and [tex]x_1-x_2=6[/tex]. Then [tex](x_1-x_2)^2=36[/tex]:
[tex](x_1-x_2)^2=(x_1+x_2)^2-4x_1x_2=4(m+1)^2+4\left((m+1)^2+1\right)=8(m+1)^2+4[/tex].
[tex]8(m+1)^2+4=36[/tex], so [tex](m+1)^2=4[/tex] and [tex]m+1=\pm 2[/tex].
Answer: [tex]m=1[/tex] or [tex]m=-3[/tex].

Solution:
Since [tex]\alpha[/tex] and [tex]\beta[/tex] are roots: [tex]\alpha^2=1-\alpha[/tex], [tex]\beta^2=1-\beta[/tex]. By Vieta's formulas, [tex]\alpha+\beta=-1[/tex], so [tex]\beta=-1-\alpha[/tex] and [tex]\beta^2=1-\beta=2+\alpha[/tex].
[tex]f(\alpha)=\beta^2[/tex]: [tex]1-\alpha+P\alpha+Q=2+\alpha[/tex], i.e. [tex](P-2)\alpha+(Q-1)=0[/tex].
In the same way [tex]f(\beta)=\alpha^2[/tex] gives [tex](P-2)\beta+(Q-1)=0[/tex].
Subtracting: [tex](P-2)(\alpha-\beta)=0[/tex]. Since [tex]D=5>0[/tex], [tex]\alpha\ne\beta[/tex], so [tex]P=2[/tex] and then [tex]Q=1[/tex].

Solution:
By Vieta's formulas: [tex](k+1)x_2=-\frac{b}{a}[/tex] and [tex]kx_2^2=\frac{c}{a}[/tex].
If [tex]k\ne -1[/tex], then [tex]x_2=-\frac{b}{a(k+1)}[/tex]; substituting into the second equation:
[tex]k\times\frac{b^2}{a^2(k+1)^2}=\frac{c}{a}[/tex], i.e. [tex]kb^2=(k+1)^2ac[/tex].
If [tex]k=-1[/tex] (opposite roots), then [tex]b=0[/tex], which also agrees with [tex]kb^2=(k+1)^2ac[/tex].
Answer: [tex]kb^2=(k+1)^2ac[/tex].
Example: for [tex]x^2-3x+2=0[/tex] (roots [tex]2[/tex] and [tex]1[/tex], [tex]k=2[/tex]): [tex]2\times 9=9\times 2[/tex].

Solution:
Using Vieta's formulas, $\alpha_1+\alpha_2+\alpha_3=2/3\Rightarrow\left(\alpha_1+\alpha_2+\alpha_3\right)^2=4/9$
On the other hand, $\left(\alpha_1+\alpha_2+\alpha_3\right)^2=\alpha_1^2+\alpha_2^2+\alpha_3^2+ 2\left(\alpha_1\alpha_2+\alpha_1\alpha_3+\alpha_2\alpha_3\right)$

Using again Vieta's formulas, $\alpha_1\alpha_2+\alpha_1\alpha_3+\alpha_2\alpha_3=5/3$ Then, $\alpha_1^2+\alpha_2^2+\alpha_3^2=4/9-10/3=-26/9$

Observation: the sum of three squares is negative. This is only possible if the values of the roots are complex.

Solution:
Since [tex]x_1[/tex] is a root, [tex]ax_1^2+bx_1+c=0[/tex]. Multiply by [tex]x_1^{n-2}[/tex]:
[tex]ax_1^n+bx_1^{n-1}+cx_1^{n-2}=0[/tex].
In the same way, [tex]ax_2^n+bx_2^{n-1}+cx_2^{n-2}=0[/tex].
Adding the two equalities: [tex]a(x_1^n+x_2^n)+b(x_1^{n-1}+x_2^{n-1})+c(x_1^{n-2}+x_2^{n-2})=0[/tex], i.e. [tex]aS_n+bS_{n-1}+cS_{n-2}=0[/tex].
This lets us compute power sums step by step. For example, for [tex]x^2-x-1=0[/tex]: [tex]S_0=2[/tex], [tex]S_1=1[/tex], [tex]S_n=S_{n-1}+S_{n-2}[/tex], so [tex]S_2=3[/tex], [tex]S_3=4[/tex], [tex]S_4=7[/tex], [tex]S_5=11[/tex].