Question

determine o conjunto solução da equação. ​

Answer 1

$\boxed{\begin{array}{l}\sf\sqrt{2x^2+3x-1}=x\\\sf(\sqrt{2x^2+3x-1})^2=x^2\\\sf 2x^2+3x-1=x^2\\\sf 2x^2-x^2+3x-1=0\\\sf x^2+3x-1=0\\\sf\Delta=b^2-4ac\\\sf\Delta=3^2-4\cdot1\cdot(-1)\\\sf\Delta=9+4\\\sf\Delta=13\\\sf x=\dfrac{-b\pm\sqrt{\Delta}}{2a}\\\\\sf x=\dfrac{-3\pm\sqrt{13}}{2\cdot1}\\\\\sf x=\dfrac{-3\pm\sqrt{13}}{2}\begin{cases}\sf x_1=\dfrac{-3+\sqrt{13}}{2}\\\\\sf x_2=\dfrac{-3-\sqrt{13}}{2}\end{cases}\end{array}}$

$\boxed{\begin{array}{l}\underline{\rm Verificac_{\!\!,}\tilde ao\!:}\\\sf\bigodot para~x=\dfrac{-3-\sqrt{13}}{2}:\\\sf\sqrt{2\cdot\bigg(\dfrac{-3-\sqrt{13}}{2}\bigg)^2+3\cdot\bigg(\dfrac{-3-\sqrt{13}}{2}\bigg)-1}\\\\\sf=\sqrt{\backslash\!\!\!2\cdot\bigg(\dfrac{9+6\sqrt{13}+13}{\backslash\!\!\!4_2}\bigg)-\dfrac{9+3\sqrt{13}}{2}-1}\\\\\sf\sqrt{\dfrac{22+6\sqrt{13}}{2}-\dfrac{9+3\sqrt{13}}{2}-1}\\\\\sf\sqrt{\dfrac{22+6\sqrt{13}-9-3\sqrt{13}-2}{2}}\end{array}}$

$\boxed{\begin{array}{l}\sf\sqrt{\dfrac{11+3\sqrt{3}}{2}}\ne\dfrac{-3-\sqrt{13}}{2}\\\\\sf portanto~x=\dfrac{-3-\sqrt{13}}{2}~n\tilde ao~\acute e~soluc_{\!\!,}\tilde ao\end{array}}$

$\boxed{\begin{array}{l}\sf para~x=\dfrac{-3+\sqrt{13}}{2}\\\\\sf\sqrt{2\cdot\bigg(\dfrac{-3+\sqrt{13}}{2}\bigg)^2+3\cdot\bigg(\dfrac{-3+\sqrt{13}}{2}-\bigg)-1}\\\\\sf\sqrt{\backslash\!\!\!2\cdot\bigg(\dfrac{9-6\sqrt{13}+13}{\backslash\!\!\!2\cdot2}\bigg)+\dfrac{-9+3\sqrt{13}}{2}-\dfrac{2}{2}}\\\\\sf\sqrt{\dfrac{22-6\sqrt{13}-9+3\sqrt{13}-2}{2}}\\\\\sf\sqrt{\dfrac{11-3\sqrt{13}}{2}}=\sqrt{\dfrac{11}{2}-\dfrac{3}{2}\sqrt{13}}\end{array}}$

$\boxed{\begin{array}{l}\underline{\rm Radical~duplo}\\\sf\sqrt{A\pm\sqrt{B}}=\sqrt{\dfrac{A+C}{2}}\pm\sqrt{\dfrac{A-C}{2}}\\\sf onde~C=\sqrt{A^2-B}\end{array}}$

$\boxed{\begin{array}{l}\underline{\rm note~que}\\\sf\dfrac{3}{2}\sqrt{13}=\sqrt{\bigg(-\dfrac{3}{2}\bigg)^2\cdot13}=\sqrt{\dfrac{9}{4}\cdot13}=\sqrt{\dfrac{117}{4}}\\\sf DA\acute I:\\\sf\sqrt{\dfrac{11}{2}-\dfrac{3}{2}\sqrt{13}}=\sqrt{\dfrac{11}{2}-\sqrt{\dfrac{117}{4}}}\\\sf C=\sqrt{A^2-B}\\\sf C=\sqrt{\bigg(\dfrac{11}{2}\bigg)^2-\dfrac{117}{4}}=\sqrt{\dfrac{121-117}{4}}\\\\\sf C=\sqrt{\dfrac{4}{4}}=\sqrt{1}=1\\\underline{\rm ent\tilde ao}\end{array}}$

$\boxed{\begin{array}{l}\sf\sqrt{\dfrac{11}{2}-\sqrt{\dfrac{117}{4}}}=\sqrt{\dfrac{\frac{11}{2}+1}{2}}-\sqrt{\dfrac{\frac{11}{2}-1}{2}}\\\\\sf\sqrt{\dfrac{11}{2}-\sqrt{\dfrac{117}{4}}}=\sqrt{\dfrac{1}{2}\cdot\dfrac{13}{2}}-\sqrt{\dfrac{1}{2}\cdot\dfrac{9}{2}}\\\\\sf\sqrt{\dfrac{11}{2}-\sqrt{\dfrac{117}{4}}}=\dfrac{\sqrt{13}}{2}-\dfrac{3}{2}=-\dfrac{3}{2}+\dfrac{\sqrt{13}}{2}\\\\\sf\sqrt{\dfrac{11}{2}-\sqrt{\dfrac{117}{4}}}=\dfrac{-3+\sqrt{13}}{2}\end{array}}$

$\large\boxed{\begin{array}{l}\underline{\rm portanto}\\\sf x=\dfrac{-3+\sqrt{13}}{2}~\acute e~soluc_{\!\!,}\tilde ao\\\\\sf S=\bigg\{\dfrac{-3+\sqrt{13}}{2}\bigg\} \end{array}}$