Question

Determinati numarul real x pentru care x-1, x+1, x+5 sunt, in aceasta ordine, numere in progresie geometrica.

Answer 1

$\hbox{Conditia ca 3 numere sa fie intr-o progresie geometrica este:} \\\\ \underline{b^2=a*c} \\\\\\ (x+1)^2=(x-1)(x+5) \\\\ x^2+2x+1=x^2+5x-x-5 \\\\ 2x+1=4x-5 \\\\ 2x-4x=-5-1 \\\\ -2x=-6 \ \ \ |:(-2) \\\\ \boxed{x=3} \\\\\\ \hbox{Verificam daca rezultatul obtinut este corect pentru ecuatia data} \\\\ (3+1)^2=(3-1)(3+5)\\\\ 9+6+1= 2*8 \\\\ 16=16 \ \ \ 'A'$