Question

Verify that the conjugate of $\sqrt{3}+\sqrt{2}$ is $\sqrt{3}-\sqrt{2}$

Answer 1

Consider $(\sqrt{3}+\sqrt{2})(\sqrt{3}-\sqrt{2})=(\sqrt{3}{)}^2-(\sqrt{2}{)}^2=3-2=1$ which is a rational number. [using the identity $(a+b)(a-b)=a^2-b^2]$
$\therefore$ The conjugate of $(\sqrt{3}+\sqrt{2})$ is $(\sqrt{3}-\sqrt{2})$.