Find the roots of the quadratic equation $$bx^{2}-2ax+a=0$$

If the quadratic equation is of type $$Ax^+Bx+C=0$$ then the roots of the quadratic equation are given by

$$x = \dfrac{-B\pm\sqrt{B^2 - 4AC}}{2A}$$

Comparing $$Ax^+Bx+C=0$$ with $$bx^{2}-2ax+a=0$$, A = b,  B = -2a,  C = a

Hence, the roots = $$\dfrac{2a\pm\sqrt{4a^2 - 4ba}}{2b}$$

x = $$\dfrac{a\pm\sqrt{a^2 - ba}}{b}$$

Let $$x_{1}$$ = $$\dfrac{a-\sqrt{a^2 - ba}}{b}$$,  $$x_{2}$$ = $$\dfrac{a+\sqrt{a^2 - ba}}{b}$$

Rationalizing  $$x_{1}$$ = $$\dfrac{a-\sqrt{a^2 - ba}}{b}$$

$$\Rightarrow$$ $$x_{1}$$ = $$\dfrac{a-\sqrt{a^2 - ba}}{b}*\dfrac{a+\sqrt{a^2 - ba}}{a+\sqrt{a^2 - ba}}$$

$$\Rightarrow$$ $$x_{1}$$ = $$\dfrac{a^2-(a^2 - ba)}{b*(a+\sqrt{a^2 - ba})}$$

$$\Rightarrow$$ $$x_{1}$$ = $$\dfrac{ab}{b*(a+\sqrt{a^2 - ba})}$$

$$\Rightarrow$$ $$x_{1}$$ = $$\dfrac{a}{a+\sqrt{a^2 - ba}}$$

$$\Rightarrow$$ $$x_{1}$$ = $$\dfrac{a}{a+\sqrt{a}*\sqrt{a - b}}$$

$$\Rightarrow$$ $$x_{1}$$ = $$\dfrac{\sqrt{a}}{\sqrt{a}+\sqrt{a - b}}$$

Similarly, $$x_{2}$$ = $$\dfrac{\sqrt{a}}{\sqrt{a}-\sqrt{a - b}}$$

Therefore, x = $$\dfrac{\sqrt{a}}{\sqrt{a}\pm\sqrt{a - b}}$$. Hence, option C is the correct answer.