Question

If x =√3+√2/√3 -√5 & y=√3-√2/√3+√5 ​then find: x^2 + y^2 + xy

Answer 1

x=√3+√2/√3-√5
∴, x²=(√3+√2)²/(√3-√5)²
=(3+2√6+2)/(3-2√15+5)
=(5+2√6)/(8-2√15)
y=(√3-√2)/(√3+√5)
∴, y²=(√3-√2)²/(√3+√5)²
=(3-2√6+2)/(3+2√15+5)
=(5-2√6)/(8+2√15)
xy=(√3+√2)/(√3-√5)×(√3-√2)/(√3+√5)
={(√3)²-(√2)²}/{(√3)²-(√5)²}
=(3-2)/(3-5)
=1/-2
=-1/2
∴, x²+y²+xy
=(5+2√6)/(8-2√15)+(5-2√6)/(8+2√15)+(-1/2)
={(5+2√6)(8+2√15)+(5-2√6)(8-2√15)}/{(8)²-(2√15)²}-1/2
=(40+16√6+10√15+4√90+40-16√6-10√15+4√90)/(64-60)-1/2
=(80+8√90)/4-1/2
=4(20+2√90)/4-1/2
=20+2√90-1/2
=(40+4√90-1)/2
=(39+4√90)/2