Question

X (elevado a x^2+2x-8)=1. Qual é a soma das soluções reais?

Answer 1

$\mathrm{x^{(x^2+2x-8)}=1\ \to\ 1\ pode\ ser\ escrito\ como\ x^0,\ logo:}\\ \mathrm{x^{(x^2+2x-8)}=x^0\ \to\ Comparando\ os\ expoentes:}\\\\ \mathrm{x^2+2x-8=0\ \to\ a=1\ \ \| \ \ b=2\ \ \| \ \ c=-8}\\\\ \textrm{Aplicando a f\'ormula quadr\'atica, teremos que:}\\\\ \mathrm{x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}=\dfrac{-2\pm\sqrt{2^2-4.1.(-8)}}{2.1}=}\\\\ \mathrm{=\dfrac{-2\pm\sqrt{4+32}}{2}=\dfrac{-2\pm\sqrt{36}}{2}=\dfrac{-2\pm6}{2}=-1\pm3}\\\\ \mathrm{x_1=-1+3=2\ \ \| \ \ x_2=-1-3=-4}\\\\ \mathbf{x=\{x_1,x_2\}=\{2,-4\}}$

$\textrm{Soma das solu\c{c}\~oes:}\\\\ \mathrm{S=x_1+x_2=2+(-4)=2-4=\mathbf{-2}}$