Question

Resolva a equacao; log√2 (3x²+7x+3) =0 

Answer 1

$log_{ \sqrt{2} } (3x^2 + 7x + 3) = 0 \\ \\ (\sqrt{2}) ^0 =3x^2+7x+3 \\ \\ 1=3x^2+7x+3\\ \\ 3x^2+7x+3-1=0\\ \\ 3x^2+7x+2=0\\ \\ \\ \Delta = b^2-4ac\\ \Delta = 7^2 - 4.3.2\\ \Delta = 49 - 24\\ \Delta = 25\\ \\ x= \dfrac{-b\pm \sqrt{\Delta} }{2a} \\ \\ x= \dfrac{-7\pm \sqrt{25} }{2.3} \\ \\ x= \dfrac{-7\pm 5}{6} \\ \\ \\ x'= \dfrac{-7- 5}{6} = \dfrac{-12}{6}=-2 \\ \\ x"= \dfrac{-7+ 5}{6} =\dfrac{-2}{6} = -\dfrac{1}{3}$

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