Question

Eu queria saber o resultado dessa conta: (x+3)²=1

Answer 1

Olá Bianca,

basta fazer:

$(x+3)^2=1\\\\ x+3= \pm\sqrt{1}\\ x+3=\pm1\\\\ \begin{cases}x+3=1\\ x=1-3\\ x'=-2\end{cases}\begin{cases} x+3=-1\\ x=-1-3\\ x''=-4\end{cases}\\\\\\ \boxed{S=\{-2,-4\}}$

Usando a fórmula geral (Báskara), teremos:

$(x+3)^2=1\\ x^2+2*x*3+3^2=1\\ x^2+6x+9=1\\ x^2+6x+9-1=0\\ x^2+6x+8=0$

$\begin{cases}a=1\\ b=6\\ c=8\end{cases}$

$\Delta=b^2-4ac\\ \Delta=6^2-4*1*8\\ \Delta=36-32\\ \Delta=4$

$x= \dfrac{-b\pm \sqrt{\Delta} }{2a}= \dfrac{-6\pm \sqrt{4} }{2}= \dfrac{-6\pm2}{2}\begin{cases}x'= \dfrac{-6+2}{2}= \dfrac{-4}{2}=-2\\\\ x''= \dfrac{-6-2}{2}= \dfrac{-8}{2} =-4 \end{cases} \\\\\\ \boxed{S=\{-2,-4\}}$

Tenha ótimos estudos =))