Question

x+y=root 3 and x-y=1,then x^4-y^4 is

Answer 1

x+y=√3 ------------------(1)
x-y=1 --------------------(2)
Adding we get,
2x=√3+1
or, x=(√3+1)/2
Putting in (1) we get,
y=√3-(√3+1)/2
or, y=(2√3-√3-1)/2
or, y=(√3-1)/2
∴, x+y
=(√3+1)/2+(√3-1)/2
=(√3+1+√3+1)/2
=(2√3+2)/2
=√3+1
x-y
=(√3+1)/2-(√3-1)/2
=(√3+1-√3+1)/2
=2/2
=1
xy
=[(√3+1)/2][(√3-1)/2]
=[(√3)²-(1)²]/4
=(3-1)/4
=2/4
=1/2
∴, x⁴-y⁴
=(x²+y²)(x²-y²)
=[(x+y)²-2xy][(x+y)(x-y)]
=[(√3+1)²-2×1/2][(√3+1)(1)]
=(3+2√3+1-1)(√3+1)
=(3+2√3)(√3+1)
=3√3+6+3+2√3
=5√3+9