Question

1)Determinati x apartine R,astfel incat numerele x-3,x+3,2x+4 sa fie termeni consecutivi ai unei progreii geometrice.
2)Sa se determine x apartine R stiind ca numerele x-1,x+1 si 2x+5 sunt termeni consecutivi ai unei progresii geometrice.

Answer 1

1)
Daca a,b,c sunt in progresie geometrica, inseamna ca ac=b², deci:
(x-3)(2x+4)=(x+3)²
2x² - 2x - 12 = x² + 6x + 9
x² - 8x - 21 = 0
Δ=b²-4ac
Δ = 64 + 84 = 148
x₁=(8+2√37)/2=4+√37
x₂=(8-2√37)/2=4-√37

2) Acelasi lucru:
(x-1)(2x+5)=(x+1)²
2x² + 3x - 5 = x² + 2x + 1
x² + x - 6 = 0
(x+3)(x-2)=0
Cazul I : x+3 =0 ==> x=-3
Cazul II : x-2 =0 ==> x=2

Answer 2

$\displaystyle 1).x-3,~x+3,~2x+4 \\ (x+3)^2=(x-3)(2x+4) \\ x^2+2 \cdot x \cdot 3+3^2=2x^2+4x-6x-12 \\ x^2+6x+9=2x^2+4x-6x-12 \\ x^2+6x-2x^2-4x+6x=-12-9 \\ -x^2+8x=-21 \\ -x^2+8x+21=0| \cdot (-1) \\ x^2-8x-21=0 \\ a=1,~b=-8,~c=-21 \\ \Delta=b^2-4ac=(-8)^2-4 \cdot 1 \cdot 21=64+84=148\ \textgreater \ 0 \\ x_1= \frac{-(-8)+ \sqrt{148} }{2 \cdot 1} = \frac{8+2 \sqrt{37} }{2} = \frac{\not2(4+ \sqrt{37}) }{\not2} =4+ \sqrt{37}$
$\displaystyle x_2= \frac{-(-8)- \sqrt{148} }{2 \cdot 1} = \frac{8-2 \sqrt{37} }{2} = \frac{\not2(4- \sqrt{37}) }{\not2} =4- \sqrt{37}$
$\displaystyle 2).x-1,~x+1,~2x+5 \\ (x+1)^2=(x-1)(2x+5) \\ x^2+2 \cdot x \cdot 1+1^2=2x^2+5x-2x-5 \\ x^2+2x+1=2x^2+5x-2x-5 \\ x^2+2x-2x^2-5x+2x=-5-1 \\ -x^2-x=-6 \\ -x^2-x+6=0| \cdot (-1) \\ x^2+x-6=0 \\ a=1,~b=1,~c=-6 \\ \Delta=b^2-4ac=1^2-4 \cdot 1 \cdot (-6)=1+24=25\ \textgreater \ 0 \\ x_1= \frac{-1+ \sqrt{25} }{2 \cdot 1}= \frac{-1+5}{2} = \frac{4}{2} =2 \\ x_2= \frac{-1- \sqrt{25} }{2 \cdot 1} = \frac{-1-5}{2} = \frac{-6}{2} =-3$