Question

se considera ecuatia x^2-mx+10=0. sa se afle m apartine lui R, stiind ca solutiile x1 si x2 ale ecuatiei verifica relatia x2-x1=3

Answer 1

x^2-mx+10=0, m∈R

x2-x1=3 => x2=3+x1;

x1,x2 solutile ecuatiei =>

(3+x1)^2-m(3+x1)+10=0;

9+6x1+x1^2-3m-mx1+10=0;

x1^2-x1*(m-6)+19-3m=0;

Δ=(m-6)^2+12m-46;

Δ=m^2-12m+12m+36-76

Stind ca ecuatia are 2 solutii reale diferite, Δ>0=>

Δ=m^2-40>0 => |m|>2√10=> m∈(-∞,2√10)∪(2√10,+∞);