if x2+y2 = 29 and xy = 2, find the value of
(i) x+y
(ii) x-y
(iii) x4+y4
(i) x+y
Given that,
x2 + y2 = 29
x2 + y2 + 2xy – 2xy = 29
(x + y)2 – 2 (2) = 29
(x + y)2 = 29 + 4
x + y = �
(ii) x-y
x2 + y2 = 29
x2 + y2 + 2xy – 2xy = 29
(x – y)2 + 2 (2) = 29
(x – y)2 + 4 = 29
(x – y)2 = 25
(x – y) = � 5
(iii) x4+y4
x2 + y2 = 29
Squaring both sides, we get
(x2 + y2)2 = (29)2
x4 + y4 + 2x2y2 = 841
x4 + y4 + 2 (2)2 = 841
x4 + y4 = 841 – 8
x4 + y4 = 833