Explicación paso a paso:
Resolución:
[tex]3x^2+x-1=0[/tex]
Dividimos por "3" ambos lados para dejar a "x" sola:
[tex]\frac{3x^2+x-1}{3}=\frac{0}{3}[/tex]
[tex]\frac{3x^2}{3} +\frac{x}{3 }-\frac{1}{3} =\frac{0}{3}[/tex]
[tex]x^2+\frac{x}{3} -\frac{1}{3} =0[/tex]
Encontramos "x" completando cuadrado:
[tex]x^2+\frac{x}{3} =\frac{1}{3}[/tex]
[tex]x^2+\frac{x}{3} +(\frac{1}{6})^2-(\frac{1}{6} )^2=\frac{1}{3}[/tex]
[tex]x^2+\frac{x}{3} +\frac{1}{36} -\frac{1}{36} =\frac{1}{3}[/tex]
[tex](x+\frac{1}{6} )^2=\frac{1}{3} +\frac{1}{36}[/tex]
[tex](x+\frac{1}{6} )^2=\frac{13}{36}[/tex]
[tex]\sqrt{(x+\frac{1}{6})^2} =\sqrt{\frac{13}{36} }[/tex]
[tex]|x+\frac{1}{6}|=\frac{\sqrt{13} }{6}[/tex]
Sacamos raíces:
[tex]x_1=\frac{\sqrt{13} }{6} -\frac{1}{6}[/tex] [tex]x_2=\frac{-\sqrt{13} }{6} -\frac{1}{6}[/tex]
[tex]x_1=\frac{\sqrt{13}-1 }{6}[/tex] [tex]x_2=\frac{-\sqrt{13}-1 }{6}[/tex]
Solución:
[tex]x_1=\frac{\sqrt{13}-1 }{6}[/tex] [tex]x_2=\frac{-\sqrt{13}-1 }{6}[/tex]
