Question

Salut la toti . Cum se determina x>0 astfel incit numerele x+1, x+5,3x+7 sa fie in progresie geometrica ?
Imi trebuie urgent :(((
Mersi anticipat . :*

Answer 1

x+1
x+3
3x+7
pg⇒b=√axc⇒
x+1=a
x+3=b
3x+7=c
x+3=√(x+1)(3x+7)
fiind un radical de ordin ,pt a putea elimina radicalul se ridica la puerea a2
(x+3)²=(x+1)(3x+7)⇒
x²+6x+9=3x²+7x+3x+7⇒x²-3x²+6x-7x+9-7=0⇒
-2x²-x+2=0⇒
a=-2
b=-1
c=2
Δ=b²-4ac⇒Δ=(-1)²-4(-2)×2⇒Δ=1+8⇒Δ=9⇒√Δ=√9⇒Δ=3
x1x2=-b+-√Δ|2a
x1=1+3|2x(-2)⇒4|-4⇒x1=-1
x2=1-3|-4⇒-2|-4⇒x2=-1|-2⇒x2=-1|2
succes