Question

Radiciação como faço?



$\sqrt{3+ 2\sqrt{2} }$

Answer 1

Utilize a seguinte identidade:

$\sqrt{A \pm B}=\sqrt{\dfrac{A+\sqrt{A^{2}-B^{2}}}{2}}\pm \sqrt{\dfrac{A-\sqrt{A^{2}-B^{2}}}{2}}$

onde $0 \leq B \leq A$.


Então, para

$A=3,\;\;B=2\sqrt{2}$

temos que

$\sqrt{3+2\sqrt{2}}=\sqrt{\dfrac{3+\sqrt{\left(3 \right )^{2}-\left(2\sqrt{2} \right )^{2}}}{2}}+ \sqrt{\dfrac{3-\sqrt{\left(3 \right )^{2}-\left(2\sqrt{2} \right )^{2}}}{2}}\\ \\ \sqrt{3+2\sqrt{2}}=\sqrt{\dfrac{3+\sqrt{9-8}}{2}}+ \sqrt{\dfrac{3-\sqrt{9-8}}{2}}\\ \\ \sqrt{3+2\sqrt{2}}=\sqrt{\dfrac{3+\sqrt{1}}{2}}+ \sqrt{\dfrac{3-\sqrt{1}}{2}}\\ \\ \sqrt{3+2\sqrt{2}}=\sqrt{\dfrac{3+1}{2}}+ \sqrt{\dfrac{3-1}{2}}\\ \\ \sqrt{3+2\sqrt{2}}=\sqrt{\dfrac{4}{2}}+ \sqrt{\dfrac{2}{2}}\\ \\ \sqrt{3+2\sqrt{2}}=\sqrt{2}+ \sqrt{1}\\ \\ \boxed{\sqrt{3+2\sqrt{2}}=\sqrt{2}+ 1}$