Question

1. A equação modular |x ^ 2 - x - 1| = 1 tem como solução: a) S=\ 0,1,2,3,\ b) S = (-2,12,3,) c S=\ -2,-1,0,1\ d IS=\ -1,0,1,2\ c) S = (1, 2, 3, 4)​

Answer 1

$\boxed{\begin{array}{l}\boldsymbol{Resposta:}~\sf S=\{-1,0,1,2\}\\\boldsymbol{Explicac_{\!\!,}\tilde ao~passo~a~passo:}\\\sf |x^2-x-1|=1\\\sf x^2-x-1=1\\\sf x^2-x-2=0\\\sf\Delta=b^2-4ac\\\sf\Delta=(-1)^2-4\cdot1\cdot(-2)\\\sf\Delta=1+8\\\sf\Delta=9\\\sf x=\dfrac{-b\pm\sqrt{\Delta}}{2a}\\\\\sf x=\dfrac{-(-1)\pm\sqrt{9}}{2\cdot1}\\\\\sf x=\dfrac{1\pm3}{2}\begin{cases}\sf x_1=\dfrac{1+3}{2}=\dfrac{4}{2}=2\\\sf x_2=\dfrac{1-3}{2}=-\dfrac{2}{2}=-1\end{cases}\end{array}}$

$\large\boxed{\begin{array}{l}\sf x^2-x-\diagup\!\!\!1=-\diagup\!\!\!1\\\sf x^2-x=0\\\sf x\cdot(x-1)=0\\\sf x=0\\\sf x-1=0\\\sf x=1\\\sf S=\{0,1\}\end{array}}$

$\large\boxed{\begin{array}{l}\underline{\rm Soluc_{\!\!,}\tilde ao~geral:}\\\sf S=\{-1,0,1,2\}\\\huge\boxed{\boxed{\boxed{\boxed{\sf\maltese~alternativa~d}}}}\end{array}}$