if x4+1_x4=47 then value of x+1_x is:
Answers
Answer:
x
3
+
x
3
1
=18
Given:
x ^ { 4 } + \left( \frac { 1 } { x ^ { 4 } } \right) = 47x
4
+(
x
4
1
)=47
To find:
x ^ { 3 } + \frac { 1 } { x ^ { 3 } }x
3
+
x
3
1
Solution:
x ^ { 4 } + \left( \frac { 1 } { x ^ { 4 } } \right) = 47x
4
+(
x
4
1
)=47
Adding and subtracting 2 on LHS,
x ^ { 4 } + \left( \frac { 1 } { x ^ { 4 } } \right) + 2 - 2 = 47x
4
+(
x
4
1
)+2−2=47
\left[ \left( x ^ { 2 } \right) ^ { 2 } + \left( \frac { 1 } { x ^ { 2 } } \right) ^ { 2 } + 2 \times \left( \frac { 1 } { x ^ { 2 } } \right) x ^ { 2 } \right] = 47 + 2[(x
2
)
2
+(
x
2
1
)
2
+2×(
x
2
1
)x
2
]=47+2
\left( x ^ { 2 } + \frac { 1 } { x ^ { 2 } } \right) ^ { 2 } = 49(x
2
+
x
2
1
)
2
=49
\left( x ^ { 2 } + \frac { 1 } { x ^ { 2 } } \right) ^ { 2 } = 7 ^ { 2 }(x
2
+
x
2
1
)
2
=7
2
x ^ { 2 } + \frac { 1 } { x ^ { 2 } } = 7 [neglecting the negative sign]x
2
+
x
2
1
=7[neglectingthenegativesign]
x ^ { 2 } + \frac { 1 } { x ^ { 2 } } + 2 - 2 = 7x
2
+
x
2
1
+2−2=7
x ^ { 2 } + 2 \left( x ^ { 2 } \right) \frac { 1 } { x ^ { 2 } } + \frac { 1 } { x ^ { 2 } } = 7 + 2x
2
+2(x
2
)
x
2
1
+
x
2
1
=7+2
\left( x + \frac { 1 } { x } \right) ^ { 2 } = 9(x+
x
1
)
2
=9
\left( x + \frac { 1 } { x } \right) ^ { 2 } = 3 ^ { 2 }(x+
x
1
)
2
=3
2
x + \frac { 1 } { x } = 3 [neglecting the negative sign]x+
x
1
=3[neglectingthenegativesign]
x ^ { 3 } + \frac { 1 } { x ^ { 3 } } = \left( x + \frac { 1 } { x } \right) ^ { 3 } - 3 ( x ) \frac { 1 } { x } \left( x + \frac { 1 } { x } \right)x
3
+
x
3
1
=(x+
x
1
)
3
−3(x)
x
1
(x+
x
1
)
= ( 3 ) ^ { 3 } - 3 ( 3 )=(3)
3
−3(3)
= 27-9= 18
x ^ { 3 } + \frac { 1 } { x ^ { 3 } }= 18x
3
+
x
3
1
=18 "