if x =√5+√3/√5-√3 find x²+1/x²
Answers
Now ,
Doing square both side -
Step-by-step explanation:
\large \underline \bold{Given :-}
Given:−
\: \: \: \: \: \: \sf{x =\dfrac{\sqrt{5} + \sqrt{3}}{\sqrt{5} - \sqrt{3}}}x=
5
−
3
5
+
3
\large \underline \bold{To \: Find :-}
ToFind:−
\: \: \: \: \: \: \sf{x^{2} + \dfrac{1}{x^{2}} = \: ?}x
2
+
x
2
1
=?
\large \underline \bold{Usable \: Identity -}
UsableIdentity−
\sf{1) \: (a + b)^{2} + (a - b)^{2} = 2(a^{2} + b^{2})}1)(a+b)
2
+(a−b)
2
=2(a
2
+b
2
)
\sf{2) \: (a + b)(a - b) = a^{2} - b^{2}}2)(a+b)(a−b)=a
2
−b
2
\sf{3) \: (a + b)^{2} = a^{2} + 2ab + b^{2}}3)(a+b)
2
=a
2
+2ab+b
2
\large \underline \bold{Solution :-}
Solution:−
\: \: \: \sf{x =\dfrac{\sqrt{5} + \sqrt{3}}{\sqrt{5} - \sqrt{3}}}x=
5
−
3
5
+
3
\: \: \: \sf{\dfrac{1}{x} =\dfrac{\sqrt{5} - \sqrt{3}}{\sqrt{5} + \sqrt{3}}}
x
1
=
5
+
3
5
−
3
Now ,
\: \: \: \sf{x + \dfrac{1}{x} =\dfrac{\sqrt{5} + \sqrt{3}}{\sqrt{5} - \sqrt{3}} + \dfrac{\sqrt{5} - \sqrt{3}}{\sqrt{5} + \sqrt{3}}}x+
x
1
=
5
−
3
5
+
3
+
5
+
3
5
−
3
\: \: \: \sf{x + \dfrac{1}{x} =\dfrac{(\sqrt{5} + \sqrt{3})^{2} + (\sqrt{5} - \sqrt{3})^{2}}{(\sqrt{5} - \sqrt{3})(\sqrt{5} + \sqrt{3})}}x+
x
1
=
(
5
−
3
)(
5
+
3
)
(
5
+
3
)
2
+(
5
−
3
)
2
\: \: \: \sf{x + \dfrac{1}{x} =\dfrac{2[(\sqrt{5})^{2} + (\sqrt{3})^{2}]}{(\sqrt{5})^{2} - (\sqrt{3})^{2}}}x+
x
1
=
(
5
)
2
−(
3
)
2
2[(
5
)
2
+(
3
)
2
]
\: \: \: \sf{x + \dfrac{1}{x} =\dfrac{2(5 + 3)}{5 - 3}}x+
x
1
=
5−3
2(5+3)
\: \: \: \sf{x + \dfrac{1}{x} = \dfrac{\cancel{2} (8)}{\cancel{2}}}x+
x
1
=
2
2
(8)
\: \: \: \sf{x + \dfrac{1}{x} = 8}x+
x
1
=8
\: Doing square both side -
\: \: \sf{(x + \dfrac{1}{x})^{2} = (8)^{2}}(x+
x
1
)
2
=(8)
2
\: \: \sf{x^{2} + \dfrac{1}{x^{2}} + 2(\cancel{x})(\dfrac{1}{\cancel{x}}) = 64}x
2
+
x
2
1
+2(
x
)(
x
1
)=64
\: \: \sf{x^{2} + \dfrac{1}{x^{2}} + 2 = 64}x
2
+
x
2
1
+2=64
\: \: \small \bold{x^{2} + \dfrac{1}{x^{2}} = 62}x
2
+
x
2
1
=62