Question

Resolva a equação irracionais V3x+6=x

Answer 1

$\large\boxed{\begin{array}{l}\sf\sqrt{3x+6}=x\\\sf(\sqrt{3x+6})^2=x^2\\\sf 3x+6=x^2\\\sf x^2-3x-6=0\\\sf\Delta=b^2-4ac\\\sf\Delta=(-3)^2-4\cdot1\cdot(-6)\\\sf\Delta=9+24\\\sf\Delta=33\\\sf x=\dfrac{3\pm\sqrt{33}}{2}\end{array}}$    

$\large\boxed{\begin{array}{l}\underline{\sf Verificac_{\!\!,}\tilde ao}\\\sf para~x=\dfrac{3+\sqrt{33}}{2}\\\\\sf \sqrt{3\cdot\bigg(\dfrac{3+\sqrt{33}}{2}\bigg)+6}\\\\\sf\sqrt{\dfrac{9+3\sqrt{33}+12}{2}}\\\\\sf\sqrt{\dfrac{21+3\sqrt{33}}{2}}=\sqrt{\dfrac{21}{2}+\dfrac{3\sqrt{33}}{2}}\\\\\sf=\sqrt{\dfrac{21}{2}+\sqrt{\dfrac{297}{4}}}\end{array}}$

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

$\large\boxed{\begin{array}{l}\sf\sqrt{\bigg(\dfrac{21}{2}\bigg)^2-\dfrac{297}{4}}=\sqrt{\dfrac{441-297}{4}}\\\\\sf\sqrt{\dfrac{144}{4}}=\sqrt{36}=6\\\\\sf\sqrt{\dfrac{21}{2}+\sqrt{\dfrac{297}{4}}}=\sqrt{\dfrac{\frac{21}{2}+6}{2}}+\sqrt{\dfrac{\frac{21}{2}-6}{2}}\\\\\sf\sqrt{\dfrac{21}{2}+\sqrt{\dfrac{297}{4}}}=\sqrt{\dfrac{\frac{33}{2}}{2}}+\sqrt{\dfrac{\frac{9}{2}}{2}}\\\\\sf\sqrt{\dfrac{21}{2}+\sqrt{\dfrac{297}{4}}}=\sqrt{\dfrac{33}{2}\cdot\dfrac{1}{2}}+\sqrt{\dfrac{9}{2}\cdot\dfrac{1}{2}}\end{array}}$

$\large\boxed{\begin{array}{l}\sf\sqrt{\dfrac{21}{2}+\sqrt{\dfrac{297}{4}}}=\dfrac{\sqrt{33}+3}{2}\end{array}}$

$\Large\boxed{\begin{array}{l}\sf portanto\\\sf S=\bigg\{\dfrac{3+\sqrt{33}}{2}\bigg\}\end{array}}$