Question

Exercitiile 12 și 13 ...

Answer 1

Ex:13.
(m-1)x²+2x+m=0
x₁+x₂=-2/m-1
x₁*x₂=m/m-1
Deci:
x₁*x₂+x₁+x₂=5⇒m/m-1-2/m-1=5⇒m/m-1-2/m-1-5=0⇒
⇒m-2-5(m-1)/m-1=0⇒m-2-5m+5/m-1⇒-4m+3/m-1=0⇒
⇒-4m+3=0⇒-4m=-3⇒m=3/4;
Deci am folosit relatiile lui Viète:
ax²+bx+c=0
x₁+x₂=-b/a;
x₁*x₂=c/a;

Ex:12.
2x²-5x-1=0
Δ=(-5)²-4*2*(-1)=25+8=33>0
x₁=5-√33/4;
x₂=5+√33/4;

Deci:
y₁=x₁+1⇒y₁=5-√33/4+1=9-√33/4;
y₂=x₂+1⇒y₂=5+√33/4+1=9+√33/4;

y₁+y₂=-b
9-√33/4+9+√33/4=-b
18/4=-b
9/2=-b

y₁*y₂=c
9-√33/4*9+√33/4=c
(9-√33)(9+√33)/4*4=c
48/16=c
c=3

Deci vom avea:
a²-9/2b+3;
In cazul nostru avem:
y²-9/2y+3;

Bafta!