Răspuns :
daca numerele a , b si c sunt in prog. aritm daca :
2b = a + c
a. 2 · ( x² + 3x + 5) = - 2x + 17 + x² + 12x + 14
2x² + 6x + 10 = x² + 10x + 31
x² - 4x - 21 = 0 Δ = 16 - 4 ·( -21) = 16 + 84 = 100 ;
√Δ=√100 =10
x₁ = ( 4 -10 ) /2 = - 6 /2 = - 3
-2· ( -3) + 17 ; (-3)² + 3· ( -3) + 5 ; ( -3)² + 12 ·( -3) + 14
prog: 23 ; 5 ; - 13
x₂ = ( 4 + 10 ) /2 = 7
prog: 3 ; 75 ; 147
b . 2 · ( x² - 2x + 5) = x² + 3x - 1 + x - 4
2x² - 4x + 10 = x² + 4x - 5
x² - 8x + 15 = 0 ; Δ =64 - 60 = 4 ; √Δ = 2
x₁ = ( 8 -2) /2 = 6 /2 = 3
prog. 17 ; 8 ; - 1
x₂ = ( 8 + 2) / 2 = 5
prog. 39 ; 20 ; 1
2b = a + c
a. 2 · ( x² + 3x + 5) = - 2x + 17 + x² + 12x + 14
2x² + 6x + 10 = x² + 10x + 31
x² - 4x - 21 = 0 Δ = 16 - 4 ·( -21) = 16 + 84 = 100 ;
√Δ=√100 =10
x₁ = ( 4 -10 ) /2 = - 6 /2 = - 3
-2· ( -3) + 17 ; (-3)² + 3· ( -3) + 5 ; ( -3)² + 12 ·( -3) + 14
prog: 23 ; 5 ; - 13
x₂ = ( 4 + 10 ) /2 = 7
prog: 3 ; 75 ; 147
b . 2 · ( x² - 2x + 5) = x² + 3x - 1 + x - 4
2x² - 4x + 10 = x² + 4x - 5
x² - 8x + 15 = 0 ; Δ =64 - 60 = 4 ; √Δ = 2
x₁ = ( 8 -2) /2 = 6 /2 = 3
prog. 17 ; 8 ; - 1
x₂ = ( 8 + 2) / 2 = 5
prog. 39 ; 20 ; 1
[tex]\displaystyle a).-2x+17,~x^2+3x+5,~x^2+12x+14 \\ x^2+3x+5= \frac{-2x+17+x^2+12x+14}{2} \\ \\ x^2+3x+5= \frac{x^2+10x+31}{2} \\ \\ 2(x^2+3x+5)=x^2+10x+31 \\ 2x^2+6x+10=x^2+10x+31 \\ 2x^2+6x-x^2-10x=31-10 \\ x^2-4x=21 \\ x^2-4x-21=0 \\ a=1,b=-4,c=-21 \\ \Delta=b^2-4ac=(-4)^2-4 \cdot 1 \cdot (21)=16+84=100\ \textgreater \ 0 \\ x_1= \frac{4+ \sqrt{100} }{2 \cdot 1} = \frac{4+10}{2} = \frac{14}{2} =7 \\ \\ x_2= \frac{4- \sqrt{100} }{2 \cdot 1} = \frac{4-10}{2} = \frac{-6}{2} =-3 \\ \\ x \in \{-3;7\}[/tex]
[tex]\displaystyle b).x^2+3x-1,~x^2-2x+5,~x-4 \\ x^2-2x+5= \frac{x^2+3x-1+x-4}{2} \\ \\ x^2-2x+5= \frac{x^2+4x-5}{2} \\ \\ 2(x^2-2x+5)=x^2+4x-5 \\ 2x^2-4x+10=x^2+4x-5 \\ 2x^2-4x-x^2-4x=-5-10 \\ x^2-8x=-15 \\ x^2-8x+15=0 \\ a=1,b=-8,c=15 \\ \Delta=b^2-4ac=(-8)^2-4 \cdot 1 \cdot 15=64-60=4\ \textgreater \ 0 \\ x_1= \frac{8+ \sqrt{4} }{2 \cdot 1}= \frac{8+2}{2} = \frac{10}{2} =5 \\ \\ x_2= \frac{8- \sqrt{4} }{2 \cdot 1} = \frac{8-2}{2} = \frac{6}{2} =3 \\ \\ x \in \{3;5\}[/tex]
[tex]\displaystyle b).x^2+3x-1,~x^2-2x+5,~x-4 \\ x^2-2x+5= \frac{x^2+3x-1+x-4}{2} \\ \\ x^2-2x+5= \frac{x^2+4x-5}{2} \\ \\ 2(x^2-2x+5)=x^2+4x-5 \\ 2x^2-4x+10=x^2+4x-5 \\ 2x^2-4x-x^2-4x=-5-10 \\ x^2-8x=-15 \\ x^2-8x+15=0 \\ a=1,b=-8,c=15 \\ \Delta=b^2-4ac=(-8)^2-4 \cdot 1 \cdot 15=64-60=4\ \textgreater \ 0 \\ x_1= \frac{8+ \sqrt{4} }{2 \cdot 1}= \frac{8+2}{2} = \frac{10}{2} =5 \\ \\ x_2= \frac{8- \sqrt{4} }{2 \cdot 1} = \frac{8-2}{2} = \frac{6}{2} =3 \\ \\ x \in \{3;5\}[/tex]
