Question

If xy = 1 x-y = 2root over 3, find x+y

Answer 1

Answer:

( x + y ) = 4

Step-by-step explanation:

Given that -

xy = 1

x - y = $2\sqrt{3}$

Squaring on both sides

We have, $(x-y)^{2} = [2\sqrt{x}]^{2}$

Formula $(a+b)^{2} = a^{2} + b^{2} + 2ab \\x^{2} + y^{2} - 2xy = 2^{2} \sqrt{3} ^{2} \\x^{2} + y^{2} -2.1 = 4 \times 3 \\x^{2} + y^{2} - 2 = 12 \\x^{2} + y^{2} = 12 + 2 \\x^{2} + y^{2} = 14 \\Adding \ 2xy \ on \ both \  sides, \\We \  get, x^{2} + y^{2} + 2xy = 14 + 2xy \\We \  know  : a^{2} + b^{2} + 2ab = (a+b)^{2}$

$(x+y)^{2}$ = 14 + 2

$(x+y)^{2}$  = 16

( x + y ) = $\sqrt{16}$

( x + y ) = 4