Question

Resolva 2 log²x + log x-1 = 0

Answer 1

$2log^{2}x+log~x - 1=0\\2(log~x)^{2}+log~x-1=0\\2(log~x)^{2}+1(log~x)-1=0$

Temos uma equação do segundo grau com variável 'log x'

$\Delta=b^{2}-4ac\\\Delta=1^{2}-4\cdot2\cdot(-1)\\\Delta=1+8\\\Delta=9$

Logo:

$log~x=\dfrac{-b\pm\sqrt{\Delta}}{2a}\\\\\\log~x=\dfrac{-1\pm\sqrt{9}}{2\cdot2}\\\\\\log~x=\dfrac{-1\pm3}{4}$

Achando as raízes:

$log~x=\dfrac{-1+3}{4}=\dfrac{1}{2}~~~~\therefore~~~10^{1/2}=x~~~~\therefore~~~~\boxed{\boxed{x=\sqrt{10}}}\\\\\\log~x=\dfrac{-1-3}{4}=-1~~~~\therefore~~~10^{-1}=x~~~~\therefore~~~~\boxed{\boxed{x=\dfrac{1}{10}}}$