问题 解答题
已知实数a,b分别满足3a4+2a2-4=0和b4+b2-3=0,求
4
a4
+b4
的值.
答案

由3a4+2a2-4=0,得a2=

-2±
4-4×3×(-4)
2×3
,即a2=
-1±
13
3

∵a2≥0,

∴a2=

-1+
13
3

由b4+b2-3=0,得b2=

-1±
1-4×1×(-3)
2×1
,即b2=
-1±
13
2

又∵b2≥0,

∴b2=

-1+
13
2

4
a4
+b4

=

4
(
-1+
13
3
)2
+(
-1+
13
2
)2

=

18
7-
13
+
7-
13
2

=

18×(7+
13
)
(7-
13
)(7+
13
)
+
7-
13
2

=

7+
13
2
+
7-
13
2

=7,

4
a4
+b4=7.

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