Odpowiedź:
[tex]36x^{4}-43x^{3}+12=0[/tex]
Podstawiam [tex]x^{2}=t[/tex]
[tex]36t^{2}-43t+12=0\\ \\\Delta=b^{2}-4ac\\\Delta=(-43)^{2}-4*36*12\\\Delta=1849-1728\\\Delta=121\\\sqrt{\Delta}=11\\\\t_{1}=\frac{-b-\sqrt\Delta}{2a}=\frac{43-11}{2*36}=\frac{32}{2*36}=\frac{4}{9}\\\\t_{2}=\frac{-b+\sqrt\Delta}{2a}=\frac{43+11}{2*36}=\frac{54}{2*36}=\frac{3}{4}\\[/tex]
[tex]x^{2}=t,\ \ to\ x^{2}=\frac{4}{9}\ i\ x^{2}=\frac{3}{4}\\\\x^{2}-\frac{4}{9}=0\\\\(x-\frac{2}{3})(x+\frac{2}{3})=0\\\\x_{1}=\frac{2}{3}\ \ \vee\ \ x_{2}=-\frac{2}{3}\\\\\\x^{2}-\frac{3}{4}=0\\\\(x-\frac{\sqrt{3}}{2})(x+\frac{\sqrt{3}}{2})=0\\\\x_{3}=\frac{\sqrt{3}}{2}\ \ \vee\ \ x_{4}=-\frac{\sqrt{3}}{2}[/tex]
Suma pierwiastków dodatnich:
[tex]x_{1}+x_{3}=\frac{2}{3}+\frac{\sqrt{3}}{2}=\frac{4+3\sqrt{3}}{6}[/tex]
Szczegółowe wyjaśnienie:
