Question

if x=√3+√2/√3-√2 and y=√3-√2/√3+√2 then find x^2+y^2+xy​ and explain how to multiple it

Answer 1

Answer:

Step-by-step explanation:

$x=(\sqrt{3}+\sqrt{2})/(\sqrt{3} -\sqrt{2})\\ y=(\sqrt{3}-\sqrt{2})/(\sqrt{3}+\sqrt{2})\\x^2=((\sqrt{3}+\sqrt{2})/(\sqrt{3} -\sqrt{2}))^2\\ =((3+2+2\sqrt{6})/(3+2-2\sqrt{6}) \\ =(5+2\sqrt{6})/(5-2\sqrt{6})\\ y^2=((\sqrt{3}-\sqrt{2})/(\sqrt{3}+\sqrt{2}))^2\\ =(3+2-2\sqrt{6})/(3+2+2\sqrt{6})\\ y^2=(5-2\sqrt{6})/(5+2\sqrt{6}) \\xy={{[(\sqrt{3}+\sqrt{2})/(\sqrt{3}-\sqrt{2})]/[(\sqrt{3}-\sqrt{2})/(\sqrt{3}+\sqrt{2})]}}\\xy=1$$x^2+y^2+xy=(5+2\sqrt{6})/(5-2\sqrt{6}) + (5-2\sqrt{6})/(5+2\sqrt{6})+1\\ = ((5+2\sqrt{6})^2+(5-2\sqrt{6})^2+(5+2\sqrt{6})(5-2\sqrt{6})) / (5+2\sqrt{6})(5-2\sqrt{6})\\ =( 25+24+20\sqrt{6} + 25+ 24 - 20\sqrt{6}+25-24)/(25-24)\\ =98/1\\x^2+y^2+xy=98$