Question

log2 (x²+×-4)=3!me ajudeeeeeeeeeeeem​

Answer 1

Explicação passo-a-passo:

$\sf log_{2}~(x^2+x-4)=3$

$\sf x^2+x-4=2^3$

$\sf x^2+x-4=8$

$\sf x^2+x-4-8=0$

$\sf x^2+x-12=0$

$\sf \Delta=1^2-4\cdot1\cdot(-12)$

$\sf \Delta=1+48$

$\sf \Delta=49$

$\sf x=\dfrac{-1\pm\sqrt{49}}{2\cdot1}=\dfrac{-1\pm7}{2}$

$\sf x'=\dfrac{-1+7}{2}~\Rightarrow~x'=\dfrac{6}{2}~\Rightarrow~x'=3$

$\sf x"=\dfrac{-1-7}{2}~\Rightarrow~x"=\dfrac{-8}{2}~\Rightarrow~x"=-4$

Logo, x = 3 ou x = -4