Math, asked by summaiyashabre, 1 year ago

If 2+root3/2-root3=a+broot3,then find the value of 'a' and 'b'

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Answered by MaheswariS
15

\underline{\textbf{Given:}}

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

\underline{\textbf{To find:}}

\textsf{The values of a and b}

\underline{\textbf{Solution:}}

\mathsf{Consider,}

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

\textsf{Multiply both numerator and denominator of}

\mathsf{L.H.S\;by\;2+\sqrt{3}}

\mathsf{\dfrac{2+\sqrt{3}}{2-\sqrt{3}}{\times}\dfrac{2+\sqrt{3}}{2+\sqrt{3}}=a+b\sqrt{3}}

\mathsf{Using\;identity,}\;\mathsf{(a-b)(a+b)=a^2-b^2}

\mathsf{\dfrac{(2+\sqrt{3})^2}{2^2-(\sqrt{3})^2}=a+b\sqrt{3}}

\mathsf{Using\;identity,}

\mathsf{(a+b)^2=a^2+b^2+2ab}

\mathsf{\dfrac{2^2+(\sqrt{3})^2+4\sqrt{3}}{4-3}=a+b\sqrt{3}}

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

\mathsf{7+4\sqrt{3}=a+b\sqrt{3}}

\textsf{Comparing on bothsides we get}

\textbf{a=7 and b=4}

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