If r = 2 + 13, then (x+1/x)equals to :
Answers
Answer:
Thanks for asking the question. Here is your answer:
Given in the question,
x+\frac{1}{x} = \sqrt{13}x+
x
1
=
13
and x^{5} + \frac{1}{x^{5}}=?x
5
+
x
5
1
=?
On squaring both sides we get,
(x+\frac{1}{x})^{2} = (\sqrt{13})^{2}(x+
x
1
)
2
=(
13
)
2
or, x^{2} + 2.x.\frac{1}{x} + (\frac{1}{x})^{2} = 13x
2
+2.x.
x
1
+(
x
1
)
2
=13
or, x^{2} + 2+\frac{1}{x^{2} } = 13x
2
+2+
x
2
1
=13
or, x^{2} +\frac{1}{x^{2}}= 13-2=11x
2
+
x
2
1
=13−2=11 ------------------(1)
On cubing both sides of the equation we get,
(x+\frac{1}{x})^{3} = (\sqrt{13} )^{3}(x+
x
1
)
3
=(
13
)
3
or, x^{3} + \frac{1}{x^{3} }+ 3.x.\frac{1}{x}(x+\frac{1}{x} )= 13\sqrt{13}x
3
+
x
3
1
+3.x.
x
1
(x+
x
1
)=13
13
or, x^{3}+\frac{1}{x^{3} } + 3\sqrt{13}= 13\sqrt{13}x
3
+
x
3
1
+3
13
=13
13
( given in the question, ∵x+\frac{1}{x} = \sqrt{13}x+
x
1
=
13
)
or, x^{3} +\frac{1}{x^{3} }=13\sqrt{13}-3\sqrt{13}x
3
+
x
3
1
=13
13
−3
13
or, x^{3} + \frac{1}{x^{3} }= 10\sqrt{13} ------------------(2)x
3
+
x
3
1
=10
13
−−−−−−−−−−−−−−−−−−(2)
Multiplication from (1) to (2) we get,
x^{2} + \frac{1}{x^{2} } (x^{3} +\frac{1}{x^{3} } )= 11 (10\sqrt{13})x
2
+
x
2
1
(x
3
+
x
3
1
)=11(10
13
)
or, x^{2} (x^{3}+\frac{1}{x}^{3})+\frac{1}{x^{2} }(x^{3}+\frac{1}{x}^{3})x
2
(x
3
+
x
1
3
)+
x
2
1
(x
3
+
x
1
3
) = 110\sqrt{13}110
13
or, x^{5}+\frac{1}{x}+\frac{x^{3} }{x^{2} } + \frac{1}{x^{5} } = 110\sqrt{13}x
5
+
x
1
+
x
2
x
3
+
x
5
1
=110
13
or, x^{5} + \frac{1}{x^{5} }= 110\sqrt{13} - (\frac{1}{x} + x)x
5
+
x
5
1
=110
13
−(
x
1
+x)
or, x^{5} + \frac{1}{x^{5} }= 109\sqrt{13}x
5
+
x
5
1
=109
13
( given in the question, ∵x+\frac{1}{x} = \sqrt{13}x+
x
1
=
13
)
Hence, the solution is x^{5} + \frac{1}{x^{5} }= 109\sqrt{13}x
5
+
x
5
1
=109
13