Question

Essa equação está certa?
Como deu 3x-3x = 0 eu acho que ta errado, confere?

8^x^2-x = 4

2^3(x2-x)=2^2
3(x2-x)=2
3x+6-3x=2
3x-3x=2-6
0x=-4.(-1)
x- 4

Answer 1

Primeiramente vamos desenvolver a equação a fim de encontrar uma igualdade entre potências de mesma base:

$\large\begin{array}{l}\mathsf{8^{x^2-x}=4~\Leftrightarrow~(2^3)^{x^2-x}=2^2~\Leftrightarrow~2^{3(x^2-x)}=2^2}\end{array}$

Igualando os expoentes:

$\begin{array}{l}\mathsf{3(x^2-x)=2~\Leftrightarrow~3x^2-3x=2~\Leftrightarrow~3x^2-3x-2=0}\end{array}$

Aplique Bhaskara para encontrar as raízes dessa equação quadrática:

$\begin{array}{l}\mathsf{\dfrac{-b\pm\sqrt{\Delta}}{2a}~~~~onde~\Delta=b^2-4ac}\end{array}$

Primeiro vamos calcular o Δ

$\begin{array}{l}\mathsf{\Delta=(-3)^2-4\cdot3\cdot(-2)}\\\\\mathsf{\Delta=9-(-24)}\\\\\mathsf{\Delta=9+24}\\\\\mathsf{\Delta=33}\end{array}$

Substituindo Δ na fórmula de Bhaskara

$\begin{array}{l}\mathsf{\dfrac{-(-3)\pm \sqrt{33}}{2\cdot 3}~\Leftrightarrow~ \dfrac{3\pm\sqrt{33}}{6}}\\\\\\\left\{\begin{matrix}\mathsf{\dfrac{3+\sqrt{33}}{6} ~\Leftrightarrow~\dfrac{3}{6}+\dfrac{\sqrt{33}}{6}~\Leftrightarrow~\dfrac{1}{2}+\dfrac{\sqrt{33}}{6}}\\\\\\\mathsf{\dfrac{3-\sqrt{33}}{6}~\Leftrightarrow~ \dfrac{3}{6}-\dfrac{\sqrt{33}}{6}~\Leftrightarrow~\dfrac{1}{2}- \dfrac{ \sqrt{33} }{6} }\end{matrix}\right.\end{array}$

$\begin{array}{l}\mathsf{Solu\c{c}\~ao\left\{\begin{matrix}\mathsf{x= \dfrac{1}{2}+ \dfrac{\sqrt{33}}{6}}\\\\\mathsf{ou}\\\\\mathsf{x=\dfrac{1}{2}-\dfrac{ \sqrt{33} }{6} }\end{matrix}\right.}\end{array}$