Question

If root3-1/ root3+1 = a+b root2 , find the values a and b

Answer 1

$\large{\underline{\underline{\red{\sf{ Given:}}}}}$

• $\tt{\dfrac{\sqrt{3}-1}{\sqrt{3}+1}=a+b\sqrt{3}(correct)}$

$\large{\underline{\underline{\red{\sf{To\:Find: }}}}}$

• $\tt{Values\:of\:a\:and\:b.}$

$\large{\underline{\underline{\red{\sf{ Answer:}}}}}$

$\tt{We\:would\:firsrt\:rationalise\:the\:denominator.}$$\tt{Then\:we\:will\:compare\:them\:to\:find\:a\:and\:b.}$

$\purple{\bf{\leadsto Taking\:the\:given\:equ^n:-}}$

$\green{\tt{\longmapsto \dfrac{\sqrt{3}-1}{\sqrt{3}+1}=a+b\sqrt{3}.}}$

$\tt{\implies \dfrac{(\sqrt{3}-1)(\sqrt{3}-1)}{(\sqrt{3}+1)(\sqrt{3}-1)}=a+b\sqrt{3}.}$

$\tt{\implies \dfrac{(\sqrt{3}-1)^2}{(\sqrt{3})^2-(1)^2}=a+b\sqrt{3}.}$

$\tt{\implies \dfrac{(\sqrt{3}^2+(1)^2-2\times\sqrt{3}\times1}{3-1}=a+b\sqrt{3}.}$

$\tt{\implies\dfrac{3+1-2\sqrt{3}}{2}=a+b\sqrt{3} .}$

$\tt{\implies\dfrac{4-2\sqrt{3}}{2}=a+b\sqrt{3} .}$

$\tt{\implies 2-\sqrt{3}=a+b\sqrt{3}.}$

$\underline{\boxed{\purple{\tt{\implies 2+(-\sqrt{3})=a+b\sqrt{3}.}}}}$

$\blue{\tt{On\:comparing\:L.H.S.\:and\:R.H.S.\:we\:get,}}$

• $\pink{\tt{a=2}}$
• $\pink{\tt{b=(-1)}}$