A takes 9 days more than B to do a certain piece of work. Together they can do the work in 6 days. How many days will A alone take to do the work ?

Let A alone takes x days and B alone takes y days.

Given,

A takes 9 days more than B to do a certain piece of work

∴ x = y + 9 ………(1)

As, A takes x days to do a work

So, one day work of A = $\dfrac{1}{x}$

As, B takes y days to do a work

So, one day work of B = $\dfrac{1}{y}$

As, together they can do work in 6 days.

∴ In one day both of them do $\dfrac{1}{6}$ th part of work.

$\therefore \dfrac{1}{x} + \dfrac{1}{y} = \dfrac{1}{6} \\[1em] \Rightarrow \dfrac{1}{y + 9} + \dfrac{1}{y} = \dfrac{1}{6} \space [\text{From (1)}] \\[1em] \Rightarrow \dfrac{y + y + 9}{y(y + 9)} = \dfrac{1}{6} \\[1em] \Rightarrow \dfrac{2y + 9}{y^2 + 9y} = \dfrac{1}{6} \\[1em] \Rightarrow 6(2y + 9) = y^2 + 9y \\[1em] \Rightarrow 12y + 54 = y^2 + 9y \\[1em] \Rightarrow y^2 + 9y - 12y - 54 = 0 \\[1em] \Rightarrow y^2 - 3y - 54 = 0\\[1em] \Rightarrow y^2 - 9y + 6y - 54 = 0 \\[1em] \Rightarrow y(y - 9) + 6(y - 9) = 0 \\[1em] \Rightarrow (y + 6)(y - 9) = 0 \\[1em] \Rightarrow y + 6 = 0 \text{ or } y - 9 = 0 \\[1em] \Rightarrow y = -6 \text{ or } y = 9.$

Since, days cannot be negative so, y ≠ -6.

x = y + 9 = 18.

Hence, A alone will take 18 days to complete the work.