sa notam coeficientii cu literele lor obisnuite: a=2,b=-4,c=-5 si atunci ai ecuatia standard: [tex]ax^{2}+bx+c[/tex] Folosind relatiile lui Viete stim ca:
[tex]S=x1+x2=\frac{-b}{a}=\frac{4}{2}=2[/tex]
[tex]P=x1x2=\frac{c}{a}=\frac{-5}{2}[/tex]
Acum sa ne apucam sa rescriem acele ecuatii
a) [tex](x1+x2)^{3}=x1^{3}+x2^{3}+3x1^{2}x2+3x1x2^{2}\Rightarrow
x1^{3}+x2^{3}=(x1+x2)^{3}-3x1x2(x1+x2)=2^{3}+3\frac{5}{2}2=8+15=23 [/tex]
b) [tex](x1^{3}+x2^{3})*(x1+x2)=x1^{4}+x2^{4}+x1x2^{3}+x1^{3}x2\Rightarrow x1^{4}+x2^{4}=(x1^{3}+x2^{3})*(x1+x2)-x1x2(x1^{2}+x2^{2})=23*2+\frac{5}{2}*(x1^{2}+x2^{2})[/tex]
Hai sa vedem cat face si ultimul termen
[tex](x1+x2)^{2}=x1^{2}+x2^{2}+2x1x2\Rightarrow x1^{2}+x2^{2}=(x1+x2)^{2}-2x1x2=2^{2}+\frac{5}{2}2=4+5=9[/tex]
Atunci:[tex]x1^{4}+x2^{4}=46+\frac{5}{2}9=46+\frac{45}{2}=68.5[/tex]
c) [tex](x1^{3}+x2^{3})^{2}=x1^{6}+x2^{6}+2x1^{3}x2^{3}\Rightarrow x1^{6}+x2^{6}=(x1^{3}+x2^{3})^{2}-2(x1x2)^{3}=23^{2}+2\frac{125}{8}=529+\frac{125}{4}=529+31.25=560/25 [/tex]
d) [tex]\frac{1}{x1-2}+\frac{1}{x2-2}=\frac{x2-2+x1-2}{(x1-2)(x2-2)}=\frac{x1+x2-4}{x1x2-2(x1+x2)+4}=\frac{2-4}{-\frac{5}{2}-2*2+4}=\frac{4}{5}[/tex]
