Presupun ca toate ecuatiile sunt egale cu 0.
1) x²+5x-6=0
Calculam Δ=b²-4ac => Δ=25+24=49 => √Δ=T=7
x1=[tex]\frac{-b+T}{2}[/tex]=[tex]\frac{-5+7}{2}[/tex]=1
x2=[tex]\frac{-b-T}{2}[/tex]=[tex]\frac{-5-7}{2}[/tex]=-6
2) x²-x-6=0
Calculam Δ=b²-4ac => Δ=1+24=25 => √Δ=T=5
x1=[tex]\frac{-b+T}{2}[/tex]=[tex]\frac{1+5}{2}[/tex]=3
x2=[tex]\frac{-b+T}{2}[/tex]=[tex]\frac{1-5}{2}[/tex]=-2
3) x²+11x-12=0
Ecuatia mai poate fi scrisa si asa: x²+12x-x-12=0, grupand cate 2 termenii, ne rezulta: x(x+12)-(x+12)=0=> (x-1)(x+12)=0 => x1=1, x2=-12.
4) x²-6x-27=0
Calculam Δ=b²-4ac => Δ=36+108=144 => √Δ=T=12
x1=[tex]\frac{-b+T}{2}[/tex]=[tex]\frac{6+12}{2}[/tex]=9
x2=[tex]\frac{-b-T}{2}[/tex]=[tex]\frac{6-12}{2}[/tex]=-3
5) (x²+8)(x²+x-8)+12=0
Cred ca ai gresit aici, ecuatia nu are solutii reale sau complexe.
6) x(x+4)(x²+4x+6)+8=0
=> [tex]x^{4}+8x^{3}+22x^{2} +24x+8=0[/tex]
Scriem ecuatia sub aceasta forma:
[tex]x^{4} +2x^{3}+6x^{3}+12x^{2}+10x^{2}+20x+4x+8=0[/tex]
=> [tex]x^{3}(x+2)+6x^{2}(x+2)+10x(x+2)+4(x+2)=0[/tex]
=> [tex](x+2)(x^{3}+6x^{2}+10x+4)=0[/tex]
=>[tex](x+2)(x^{3}+2x^{2}+4x^{2}+8x+2x+4)=0[/tex]
=> [tex](x+2)(x^{2}(x+2)+4x(x+2)+2(x+2))[/tex]
=> [tex](x+2)^{2} (x^{2}+4x+2)=0[/tex]
De aici ne rezulta ca x1=x2=-2 si ca x²+4x+2=0.
Rezolvam ultima ecuatie:
Calculam Δ=b²-4ac => Δ=16-8=8 => √Δ=T=2√2
x3=[tex]\frac{-b+T}{2}[/tex]=[tex]\frac{-4+T}{2}[/tex]=-2+√2
x4=[tex]\frac{-b-T}{2}[/tex]=[tex]\frac{-4-T}{2}[/tex]=-2-√2
