x=√3+1/√3-1 and y=√3-1/√3+1 find the value of x²-xy+y²
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we know,
(x+y)^2-3xy=x^2-xy+y^2
x=1. as (√3+1)(√3-1 )/(√3-1)(√3+1 )=1
y=1. as (√3-1)(√3+1 )/(√3+1)(√3-1 )=1
so,
(x+y)^2=4,
3xy=3,
x²-xy+y²=4-3=1
(x+y)^2-3xy=x^2-xy+y^2
x=1. as (√3+1)(√3-1 )/(√3-1)(√3+1 )=1
y=1. as (√3-1)(√3+1 )/(√3+1)(√3-1 )=1
so,
(x+y)^2=4,
3xy=3,
x²-xy+y²=4-3=1
jayashree65:
but answer is 8√3+1
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