Răspuns:
[tex]x^{2}+px+q=0\\Dupa~relatiile~lui~Viette,~\left \{ {{x_{1}+x{2}=-p} \atop {x_{1}*x_{2}=q}} \right. ~deci~\left \{ {{-p=\frac{1-\sqrt{5} }{2} +\frac{1+\sqrt{5} }{2} } \atop {q=\frac{1-\sqrt{5} }{2} *\frac{1+\sqrt{5} }{2} }} \right. ~deci~\left \{ {{-p=1} \atop {q=\frac{1^{2}-(\sqrt{5})^{2} }{4} }} \right.~deci~\left \{ {{p=-1} \atop {q=-1}} \right.[/tex]
Explicație pas cu pas:
Atunci ecuatia este: x²-1·x-1=0, deci x²-x-1=0.