Discussion

Explanation:

bx² – 2ax + a = 0

Using the quadratic formula,

x = $\frac{2a\pm \sqrt{4a^2-4ab}}{2b}$

= $\frac{2a\pm 2\sqrt{a^2-ab}}{2b}$

= $\frac{a\pm \sqrt{a(a-b)}}{b}$

= $\frac{a\pm \sqrt{a(a-b)}}{b}$

= $\frac{\left(a\pm \sqrt{a}\sqrt{a-b}\right)\left(a\mp \sqrt{a}\sqrt{a-b}\right)}{b\left(a\mp \sqrt{a}\sqrt{a-b}\right)}$

= $\frac{\left(a^2-a(a-b)\right)}{b\left(a\pm \sqrt{a}\sqrt{a-b}\right)}$

= $\frac{\left(a^2-a^2+ab\right)}{b\left(a\pm \sqrt{a}\sqrt{a-b}\right)}$

= $\frac{a}{\sqrt{a}\left(\sqrt{a}\pm \sqrt{a-b}\right)}$

= $\frac{\sqrt{a}}{\left(\sqrt{a}\pm \sqrt{a-b}\right)}$

Hence, option (c).

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