Question

como resolvo essa equação do 2º grau x²-1 sobre 3 = 1 sobre 6 x²

Answer 1

$\dfrac{x^2-1}{3}=\dfrac{1}{6x^2}~~~~~~(\text{com }x\ne 0)$


Multiplicando em cruz, temos

$(x^2-1)\cdot 6x^2=3\\\\ 6x^4-6x^2=3\\\\ 6x^4-6x^2-3=0\\\\ 3\cdot (2x^4-2x^2-1)=0\\\\ 2x^4-2x^2-1=0\\\\ 2(x^2)^2-2x^2-1=0$


Mudando de variável, fazendo 

$x^2=t~~~~~(x\ne 0~\Rightarrow~t>0)$


a equação fica

$2t^2-2t-1=0~~~\Rightarrow\left\{\!\begin{array}{l} a=2\\b=-2\\c=-1 \end{array} \right.\\\\\\ \Delta=b^2-4ac\\\\ \Delta=(-2)^2-4\cdot 2\cdot (-1)\\\\ \Delta=4+8\\\\ \Delta=12\\\\ \Delta=2^2\cdot 3$

$t=\dfrac{-(-2)\pm \sqrt{2^2\cdot 3}}{2\cdot 2}\\\\\\ t=\dfrac{2\pm 2\sqrt{3}}{2\cdot 2}\\\\\\ t=\dfrac{\diagup\!\!\!\! 2\cdot (1\pm \sqrt{3})}{\diagup\!\!\!\! 2\cdot 2}\\\\\\ t=\dfrac{1\pm \sqrt{3}}{2}~~~~~(\text{mas }\sqrt{3}>1)\\\\\\ \begin{array}{rcl} t=\dfrac{1+\sqrt{3}}{2}&~\text{ ou }~&t=\dfrac{1-\sqrt{3}}{2}~~\text{(n\~ao serve)} \end{array}\\\\\\ \therefore~~t=t=\dfrac{1+\sqrt{3}}{2}$

Voltando para a variável $x:$

$x^2=\dfrac{1+\sqrt{3}}{2}\\\\ x=\pm \sqrt{\dfrac{1+\sqrt{3}}{2}}\\\\\\ \boxed{\begin{array}{rcl} x=\sqrt{\dfrac{1+\sqrt{3}}{2}}&~\text{ ou }~&x=-\sqrt{\dfrac{1+\sqrt{3}}{2}} \end{array}}$