Indian Institute of Foreign Trade 2010 Solved Paper

If the quadratic equation is of type $\mathit{A}\mathit{x}+\mathit{B}\mathit{x}+\mathit{C}=0$ then the roots of the quadratic equation are given by
$\mathit{x}=\frac{-\mathit{B}\pm \sqrt{{\mathit{B}}^2-4\mathit{A}\mathit{C}}}{2\mathit{A}}$
Comparing $\mathit{A}\mathit{x}+\mathit{B}\mathit{x}+\mathit{C}=0$ with $\mathit{b}{\mathit{x}}^2-2\mathit{a}\mathit{x}+\mathit{a}=0$,$\mathit{A}=\mathit{b},\mathit{B}=-2\mathit{a},\mathit{C}=\mathit{a}$
Hence, the roots = $\frac{2\mathit{a}\pm \sqrt{4{\mathit{a}}^2-4\mathit{b}\mathit{a}}}{2\mathit{b}}$
$\mathit{x}=\frac{\mathit{a}\pm \sqrt{{\mathit{a}}^2-\mathit{b}\mathit{a}}}{\mathit{b}}$
Let ${\mathit{x}}_1=\frac{\mathit{a}-\sqrt{{\mathit{a}}^2-\mathit{b}\mathit{a}}}{\mathit{b}},{\mathit{x}}_2=\frac{\mathit{a}+\sqrt{{\mathit{a}}^2-\mathit{b}\mathit{a}}}{\mathit{b}}$
Rationalizing ${\mathit{x}}_1=\frac{\mathit{a}-\sqrt{{\mathit{a}}^2-\mathit{b}\mathit{a}}}{\mathit{b}}$
$\Rightarrow {\mathit{x}}_1=\frac{\mathit{a}-\sqrt{{\mathit{a}}^2-\mathit{b}\mathit{a}}}{\mathit{b}}\times \frac{\mathit{a}+\sqrt{{\mathit{a}}^2-\mathit{b}\mathit{a}}}{\mathit{a}+\sqrt{{\mathit{a}}^2-\mathit{b}\mathit{a}}}$
$\Rightarrow {\mathit{x}}_1=\frac{{\mathit{a}}^2-\left({\mathit{a}}^2-\mathit{b}\mathit{a}\right)}{\mathit{b}\times \left(\mathit{a}+\sqrt{{\mathit{a}}^2-\mathit{b}\mathit{a}}\right)}$
$\Rightarrow {\mathit{x}}_1=\mathit{b}\times \frac{\mathit{a}\mathit{b}}{\left(\mathit{a}+\sqrt{{\mathit{a}}^2-\mathit{b}\mathit{a}}\right)}$
$\Rightarrow {\mathit{x}}_1=\mathit{a}+\frac{\mathit{a}}{\sqrt{{\mathit{a}}^2-\mathit{b}\mathit{a}}}$
$\Rightarrow {\mathit{x}}_1=\mathit{a}+\frac{\mathit{a}}{\sqrt{\mathit{a}}\times \sqrt{\mathit{a}-\mathit{b}}}$
$\Rightarrow {\mathit{x}}_1=\mathit{a}+\frac{\mathit{a}}{\sqrt{\mathit{a}}+\sqrt{\mathit{a}-\mathit{b}}}$
Similarly,${\mathit{x}}_2=\frac{\sqrt{\mathit{a}}}{\sqrt{\mathit{a}}-\sqrt{\mathit{a}-\mathit{b}}}$
Therefore,$\mathit{x}=\frac{\sqrt{\mathit{a}}}{\sqrt{\mathit{a}}\pm \sqrt{\mathit{a}-\mathit{b}}}$
Hence, option C is the correct answer.