Pentru ca numerele date sa fie in progresie geometrica,punem conditia ca:
[tex]b^2=a*c[/tex]
Folosim conditia in problema noastra:
[tex](x+7)^2=(5-x)(3x+11) \\\\ x^2+14x+49=15x +55-3x^2-11x \\\\ 4x^2+10x-6=0 \ \ \ |:2 \\\\ 2x^2+5x-3=0 \\\\\\ \Delta= 25-4*2*(-3)=25+24\to 49 \\\\\\ x_1=\frac{-5+7}{4}= \frac{2}{4} \to \frac{1}{2} \\\\\\ x_2= \frac{-5-7}{4}=\frac{-12}{4} \to -3[/tex]
Verificare:
[tex]Pt \ x \to \frac{1}{2} \\\\\\ (\frac{1}{2}+7)^2=(5-\frac{1}{2})*(3*\frac{1}{2}+11) \\\\ (\frac{15}{2})^2=\frac{9}{2}* \frac{25}{2} \\\\ \frac{225}{4}=\frac{225}{4} \ \ 'A' \\\\\\ Pt \ x \to -3 \\\\ (-3+7)^2=(5+3)(-9+11) \\\\ 16=8*2 \ \ \ 'A' [/tex]
Solutiile sunt:
[tex]S \in \{ -3;\frac{1}{2} \}[/tex]
