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:
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:
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{15^2-4.36}=\sqrt{225-144}=\sqrt{81}=9[/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:
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:
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:
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.