Question

Sa se determine x€R stiind ca numerele 4-x,x+8,3x+14 sunt termeni consecutivi ai unei progresii geometrice.

Answer 1

(x+8)=√(4-x)*(3x+14)

Ridic totul la pătrat

(x+8)^2=(4-x)*(3x+14)

x^2+16x+64=4*3x+4*14-x*3x-x*14

x^2+16x+64=12x+56-3x^2-14x

x^2+16x+64=-3x^2+56-2x

x^2+3x^2+16x+2x-56+64=0

4x^2+18x+8=0/:2

2x^2+9x+4=0

a=2, b=9, c=4

Δ=b^2-4ac

Δ=9^2-4*2*4

Δ=81-32

Δ=49

x1=(-b+√Δ)/2a

x1=(-9+√49)/2*2

x1=(-9+7)/4

x1=-2/4 simplificat prin 2

x1=-1/2

x2=(-b-√Δ)/2a

x2=(-9-√49)/2*2

x2=(-9-7)/4

x2=-16/4

x2=-4

S={-4, -1/2}

Answer 2

Răspuns:

Explicație pas cu pas:

daca numerele a,b,c sunt termeni consecutivi ai unei progresii geometrice, atunci b²=a·c. Aplicam aceasta proprietate,

⇒(x+8)²=(4-x)·(3x+14), ⇒x²+16x+64=12x+56-3x²-14x, ⇒4x²+18x+8=0 |:2, ⇒2x²+9x+4=0, Δ=9²-4·2·4=81-32=49>0, √49=7

Deci x1=(-9-7)/4=-4 si x2=(-9+7)/4=-1/2.

x∈{-1/2; -4}