Question

Root 2+root 3/3 root2-2 root3

Answer 1

Answer:

$\dfrac{\sqrt{2}+\sqrt{3}}{3\sqrt{2}-2\sqrt{3}}=\dfrac{12+5\sqrt{6}}{6}\quad \left(\mathrm{Decimal:\quad }\:4.04124\dots \right)$

Step-by-step explanation:

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

$\mathrm{Multiply\:by\:the\:conjugate}\:\dfrac{3\sqrt{2}+2\sqrt{3}}{3\sqrt{2}+2\sqrt{3}}$

$=\dfrac{\left(\sqrt{2}+\sqrt{3}\right)\left(3\sqrt{2}+2\sqrt{3}\right)}{\left(3\sqrt{2}-2\sqrt{3}\right)\left(3\sqrt{2}+2\sqrt{3}\right)}$

$\left(\sqrt{2}+\sqrt{3}\right)\left(3\sqrt{2}+2\sqrt{3}\right)$

$\mathrm{Apply\:FOIL\:method}:\quad \left(a+b\right)\left(c+d\right)=ac+ad+bc+bd$

$a=\sqrt{2},\:b=\sqrt{3},\:c=3\sqrt{2},\:d=2\sqrt{3}$

$=\sqrt{2}\cdot \:3\sqrt{2}+\sqrt{2}\cdot \:2\sqrt{3}+\sqrt{3}\cdot \:3\sqrt{2}+\sqrt{3}\cdot \:2\sqrt{3}$

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

$=12+5\sqrt{6}$

$\left(3\sqrt{2}-2\sqrt{3}\right)\left(3\sqrt{2}+2\sqrt{3}\right)$

$\mathrm{Apply\:Difference\:of\:Two\:Squares\:Formula:\:}\left(a-b\right)\left(a+b\right)=a^2-b^2$

$a=3\sqrt{2},\:b=2\sqrt{3}$

$=\left(3\sqrt{2}\right)^2-\left(2\sqrt{3}\right)^2$

$=6$

$=\dfrac{12+5\sqrt{6}}{6}$