[tex]\left(x^2+x+1\right)\left(x^2+x+2\right)-12=0[/tex]
[tex]x^4+2x^3+4x^2+3x-10=0[/tex]
[tex]\left(x-1\right)\left(x+2\right)\left(x^2+x+5\right)=0[/tex]
[tex]x-1=0:\quad x=1[/tex]
[tex]x+2=0:\quad x=-2[/tex]
[tex]x^2+x+5=0:\quad x=-\frac{1}{2}+i\frac{\sqrt{19}}{2},\:x=-\frac{1}{2}-i\frac{\sqrt{19}}{2}[/tex]
Răspuns:
S={-2; 1}
Explicație pas cu pas:
(x la a 2+x+1)(x la a 2+x+2)-12=0 (1)
Notăm x²+x+1=y. Din (1), ⇒y·(y+1)-12=0, ⇒y²+y-12=0, Δ=1²-4·1·(-12)=1+48=49.
y1=(-1-7)/2=-4, y2=(-1+7)/2=3
Deci x²+x+1=-4 (2) sau x²+x+1=3 (3)
Din (2), ⇒x²+x+5=0, Δ=1²-4·1·5=1-20=-19<0, deci (2) n-are soluții reale.
Din (3), ⇒x²+x-2=0, Δ=1²-4·1·(-2)=1+8=9, x1=(-1-3)/2=-2, x2=(-1+3)/2=1.
Deci S={-2; 1}.
