Question:
If \[ x+y=2 \] and \[ xy=1, \] then \[ x^4+y^4= \]
(a) 6
(b) 4
(c) 8
(d) 2
Solution:
\[ x^2+y^2=(x+y)^2-2xy \]
\[ =2^2-2(1) \]
\[ =4-2 \]
\[ =2 \]
\[ x^4+y^4=(x^2+y^2)^2-2x^2y^2 \]
\[ =2^2-2(1)^2 \]
\[ =4-2 \]
\[ =2 \]
\[ \boxed{2} \]
If \[ x+y=2 \] and \[ xy=1, \] then \[ x^4+y^4= \]
(a) 6
(b) 4
(c) 8
(d) 2
\[ x^2+y^2=(x+y)^2-2xy \]
\[ =2^2-2(1) \]
\[ =4-2 \]
\[ =2 \]
\[ x^4+y^4=(x^2+y^2)^2-2x^2y^2 \]
\[ =2^2-2(1)^2 \]
\[ =4-2 \]
\[ =2 \]
\[ \boxed{2} \]