Question

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Answer 1

$\bf Given:\ \sf \dfrac{\sqrt{3}\ -\ 1}{\sqrt{3}\ +\ 1}\ =\ a\ -\ b\sqrt{3}.\\\\\\\bf To\ find:\ \sf The\ values\ of\ a\ and\ b.\\\\\\\bf Answer:\\\\\\\sf Let's\ first\ rationalize\ the\ denominator\ in\ \dfrac{\sqrt{3}\ -\ 1}{\sqrt{3}\ +\ 1}.\\\\\\\implies\ \dfrac{\bigg(\sqrt{3}\ -\ 1\bigg)\bigg(\sqrt{3}\ -\ 1\bigg)}{\bigg(\sqrt{3}\ +\ 1\bigg)\bigg(\sqrt{3}\ -\ 1\bigg)}\\\\\\=\ \ \ \ \dfrac{\bigg(\sqrt{3}\ -\ 1\bigg)^2}{(\sqrt{3})^2\ -\ 1}$

$\sf =\ \ \ \ \dfrac{3\ -\ 2\sqrt{3}\ +\ 1}{3\ -\ 1}\\\\\\\\=\ \ \ \ \ \dfrac{4\ -\ 2\sqrt{3}}{2}\\\\\\=\ \ \ \ \ \dfrac{2(2\ -\ \sqrt{3})}{2}\\\\\\=\ \ \ \ \ \ 2\ -\ \sqrt{3}\\\\\\\\\\\\\bf 2\ -\ \sqrt{3}\ \sf is\ of\ the\ form\ \bf a\ -\ b\sqrt{3},\ \sf where\ \bf a\ =\ 2\ \sf and\ \bf b\ =\ 1.$