Math, asked by rashmimatwaya315, 14 days ago

if a+1/a = 1/√3-1 then value of a²+1/a² is?​

Answers

Answered by MaheswariS
2

If a+1/a = 1/√3-1 then value of a²+1/a² is?​

\underline{\textbf{Given:}}

\mathsf{a+\dfrac{1}{a}=\dfrac{1}{\sqrt{3}-1}}

\underline{\textbf{To find:}}

\mathsf{The\;value\;of\;\;a^2+\dfrac{1}{a^2}}

\underline{\textbf{Solution:}}

\mathsf{Consider,}

\mathsf{a+\dfrac{1}{a}=\dfrac{1}{\sqrt{3}-1}{\times}\dfrac{\sqrt{3}+1}{\sqrt{3}+1}}

\mathsf{a+\dfrac{1}{a}=\dfrac{\sqrt{3}+1}{3-1}}

\mathsf{a+\dfrac{1}{a}=\dfrac{\sqrt{3}+1}{2}}

\textsf{Squaring on bothsides, we get}

\mathsf{\left(a+\dfrac{1}{a}\right)^2=\dfrac{(\sqrt{3}+1)^2}{4}}

\mathsf{a^2+\dfrac{1}{a^2}+2{\times}a{\times}\dfrac{1}{a}=\dfrac{3+1+2\sqrt{3}}{4}}

\mathsf{a^2+\dfrac{1}{a^2}+2=\dfrac{4+2\sqrt{3}}{4}}

\mathsf{a^2+\dfrac{1}{a^2}=\dfrac{4+2\sqrt{3}}{4}-2}

\mathsf{a^2+\dfrac{1}{a^2}=\dfrac{2+\sqrt{3}}{2}-2}

\mathsf{a^2+\dfrac{1}{a^2}=\dfrac{2+\sqrt{3}-4}{2}}

\implies\boxed{\mathsf{a^2+\dfrac{1}{a^2}=\dfrac{\sqrt{3}-2}{2}}}

Answered by akrout5
0

Answer:

3

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