Question

$\frac{ \sqrt{3} + 1}{ \sqrt{3} - 1 } = a + b \sqrt{3}$
Find the value of a and b
please help with explanation​

Answer 1

$\green{\large\underline{\bf{ Question-}}}$

$\sf{ \dfrac{ \sqrt{3} + 1}{ \sqrt{3} - 1 } = a + b \sqrt{3}}$

$\green{\large\underline{\bf{Solution-}}}$

$\sf{ \dfrac{ \sqrt{3} + 1}{ \sqrt{3} - 1 } = a + b \sqrt{3}}$

$\sf{By\:Rationalising\: it,\: we\:get}$

$\sf\longmapsto{\dfrac{ (\sqrt{3} + 1)(\sqrt{3} + 1 )}{ (\sqrt{3} - 1 )(\sqrt{3} + 1) }}$

$\sf{Using\:identity \:\mapsto {a}^{2}- {b}^{2} =(a +b)(a-b)}$

Now, the equation becomes $\sf\longmapsto{\dfrac{{ (\sqrt{3} + 1)}^{2}}{{{(\sqrt {3})}^{2}}-{(1)}^{2}}= a + b \sqrt{3}}$

$\sf{Using\:identity \:\mapsto {(a+b)}^{2}={a}^{2} + {b}^{2}+2ab}$

$\sf\longmapsto{\dfrac{3+1-2{\sqrt{3}}}{2}=a + b \sqrt{3}}$

$\sf\longmapsto{2+{\sqrt{3}}=a + b \sqrt{3}}$

$\sf{On\:comparing}$

$\sf\longmapsto{a\:=\:2\:and\:b\:=\:1}$

$\green{\large\underline{\bf{Answer-}}}$

• Value of a = 2
• Value of b = 1

@BrainlicaLDoll