Question

Fie x1 si x2 radacinile ecuatiei x^2-mx-1=0 Det.nr real m, stiind ca x1^2+x2^2=x1+x2+2.

Answer 1

x²-mx-1=0
Conform formei reduse a ecuatiei de gradul ll:
p=-m
q=-1

Relatiile lui Viete:
x1x2=q
x1+x2=-p

x1²+x2²=x1+x2+2
(x1+x2)²=x1²+2·x1·x2+x2²

→(x1+x2)²-2·x1·x2=x1+x2+2

x1+x2=-p=-(-m)=m
x1·x2=-1
→ m²-2·(-1)=m+2
m²+2-m-2=0
m²-m=0
m(m-1)=0
m-1=0 sau m=0
m=1 sau m=0

1) m=1
x²-x-1=0
-p=-1 → p=1
q=-1
2)m=0
x²-1=0
p = 0
q=-1