Question

Sa se determine x ∈ R astfel incat numerele sa fie in progresie aritmetica:
a) -2x + 17, $x^{2}$ + 3x + 5, $x^{2}$ + 12 x + 14;
b) $x^{2}$ + 3x - 1, $x^{2}$ - 2x + 5, x - 4.

Answer 1

$\displaystyle a).-2x+17,~x^2+3x+5,~x^2+12x+14 \\ x^2+3x+5= \frac{-2x+17+x^2+12x+14}{2} \\ \\ x^2+3x+5= \frac{x^2+10x+31}{2} \\ \\ 2(x^2+3x+5)=x^2+10x+31 \\ 2x^2+6x+10=x^2+10x+31 \\ 2x^2+6x-x^2-10x=31-10 \\ x^2-4x=21 \\ x^2-4x-21=0 \\ a=1,b=-4,c=-21 \\ \Delta=b^2-4ac=(-4)^2-4 \cdot 1 \cdot (21)=16+84=100\ \textgreater \ 0 \\ x_1= \frac{4+ \sqrt{100} }{2 \cdot 1} = \frac{4+10}{2} = \frac{14}{2} =7 \\ \\ x_2= \frac{4- \sqrt{100} }{2 \cdot 1} = \frac{4-10}{2} = \frac{-6}{2} =-3 \\ \\ x \in \{-3;7\}$

$\displaystyle b).x^2+3x-1,~x^2-2x+5,~x-4 \\ x^2-2x+5= \frac{x^2+3x-1+x-4}{2} \\ \\ x^2-2x+5= \frac{x^2+4x-5}{2} \\ \\ 2(x^2-2x+5)=x^2+4x-5 \\ 2x^2-4x+10=x^2+4x-5 \\ 2x^2-4x-x^2-4x=-5-10 \\ x^2-8x=-15 \\ x^2-8x+15=0 \\ a=1,b=-8,c=15 \\ \Delta=b^2-4ac=(-8)^2-4 \cdot 1 \cdot 15=64-60=4\ \textgreater \ 0 \\ x_1= \frac{8+ \sqrt{4} }{2 \cdot 1}= \frac{8+2}{2} = \frac{10}{2} =5 \\ \\ x_2= \frac{8- \sqrt{4} }{2 \cdot 1} = \frac{8-2}{2} = \frac{6}{2} =3 \\ \\ x \in \{3;5\}$