Question

Encontre o produto das raízes da equação: raiz quadrada de (1/x+3 - 1/x-3) = 8

Answer 1

$\mathrm{\sqrt{\dfrac{1}{x+3}-\dfrac{1}{x-3}}=8\ \to\ \sqrt{\dfrac{(x-3)-(x+3)}{(x+3)(x-3)}}=8}}\\\\ \mathrm{\sqrt{\dfrac{x-3-x-3}{x^2-3^2}}=8\ \to\ \dfrac{-6}{x^2-9}=64}\\\\ \mathrm{-6=64(x^2-9)\ \to\ -3=32x^2-288}\\\\ \mathrm{32x^2=-3+288\ \to\ x^2=\dfrac{285}{32}\ \to\ x=\pm\sqrt{\dfrac{285}{32}}}\\\\\ \mathrm{x=\pm\dfrac{\sqrt{285}}{4\sqrt{2}}=\pm\dfrac{\sqrt{285}.\sqrt{2}}{4\sqrt{2}.\sqrt{2}}=\pm\dfrac{\sqrt{570}}{8}}$

$\mathbf{Produto\ das\ ra\'izes:}}\\\\ \mathrm{\dfrac{\sqrt{570}}{8}\times\bigg(\dfrac{-\sqrt{570}}{8}\bigg)=\dfrac{-570}{64}=\boxed{\mathbf{-\dfrac{285}{32}}}}$

Answer 2

Resposta:

\begin{gathered}\mathrm{\sqrt{\dfrac{1}{x+3}-\dfrac{1}{x-3}}=8\ \to\ \sqrt{\dfrac{(x-3)-(x+3)}{(x+3)(x-3)}}=8}}\\\\ \mathrm{\sqrt{\dfrac{x-3-x-3}{x^2-3^2}}=8\ \to\ \dfrac{-6}{x^2-9}=64}\\\\ \mathrm{-6=64(x^2-9)\ \to\ -3=32x^2-288}\\\\ \mathrm{32x^2=-3+288\ \to\ x^2=\dfrac{285}{32}\ \to\ x=\pm\sqrt{\dfrac{285}{32}}}\\\\\ \mathrm{x=\pm\dfrac{\sqrt{285}}{4\sqrt{2}}=\pm\dfrac{\sqrt{285}.\sqrt{2}}{4\sqrt{2}.\sqrt{2}}=\pm\dfrac{\sqrt{570}}{8}}\end{gathered}

\begin{gathered}\mathbf{Produto\ das\ ra\'izes:}}\\\\ \mathrm{\dfrac{\sqrt{570}}{8}\times\bigg(\dfrac{-\sqrt{570}}{8}\bigg)=\dfrac{-570}{64}=\boxed{\mathbf{-\dfrac{285}{32}}}}\end{gathered}