Question

Resolva a equação a seguir:
sqrt(x^3+ -7x^2+2x+3)=sqrt(x-2)^3

Answer 1

Caso esteja pelo app, e tenha problemas para visualizar esta resposta, experimente abrir pelo navegador https://brainly.com.br/tarefa/39105677

                                                     

$\boxed{\begin{array}{l}\sf\sqrt{x^3-7x^2+2x+3}=\sqrt{(x-2)^3}\\\sf(\sqrt{x^3-7x^2+2x+3})^2=(\sqrt{(x-2)^3})^2\\\sf x^3-7x^2+2x+3=(x-2)^3\\\sf \diagup\!\!\!\!\!x^3-7x^2+2x+3=\diagup\!\!\!\1\!\1x^3-6x^2+12x-8\\\sf -7x^2+6x^2+2x-12x+3+8=0\\\sf-x^2-10x+11=0\cdot(-1)\\\sf x^2+10x-11=0\\\sf\Delta=b^2-4ac\\\sf\Delta=10^2-4\cdot1\cdot(-11)\\\sf\Delta=100+44\\\sf\Delta=144\end{array}}$

$\boxed{\begin{array}{l}\sf x=\dfrac{-b\pm\sqrt{\Delta}}{2a}\\\sf x=\dfrac{-10\pm\sqrt{144}}{2\cdot1}\\\sf x=\dfrac{-10\pm12}{2}\begin{cases}\sf x_1=\dfrac{-10+12}{2}=\dfrac{2}{2}=1\\\sf x_2=\dfrac{-10-12}{2}=-\dfrac{22}{2}=-11\end{cases}\end{array}}$

$\boxed{\begin{array}{l}\underline{\rm verificac_{\!\!,}\tilde ao}\\\sf para~x=1:\\\sf\sqrt{1^3-7\cdot1^2+2\cdot1+3}=\sqrt{1-7+2+3}=\sqrt{-1}\not\in\mathbb{R}\\\sf\sqrt{(-1-2)^3}=\sqrt{(-3)^3}=\sqrt{-27}\not\in\mathbb{R}\\\sf logo~x=1~n\tilde ao~\acute e~soluc_{\!\!,}\tilde ao.\end{array}}$

$\boxed{\begin{array}{l}\underline{\rm verificac_{\!\!,}\tilde ao}\\\sf para~x=-11:\\\sf\sqrt{(-11)^3-7\cdot(-11)^2+2\cdot(-11)+3}=\sqrt{-2197}\not\in\mathbb{R}\\\sf\sqrt{(-11-2)^3}=\sqrt{(-13)^3}=\sqrt{-2197}\not\in\mathbb{R}.\\\sf portanto~ x=-11~n\tilde ao~\acute e~soluc_{\!\!,}\tilde ao~da~equac_{\!\!,}\tilde ao.\\\sf S=\bigg\{\bigg\}\end{array}}$