The binomial expansion of (1-ax)' is:
(1-ax)' = 1 - ax + a²x² - a³x³ + ...
To find the coefficient of x², we need to look at the term with x², which is a²x². Therefore, the coefficient of x² in the expansion is a².
Given that the coefficient of x² is 60, we can solve for a:
a² = 60
a = ±√60
Since a > 0, we take the positive square root:
a = √60 = √(2²×3×5) = 2√15
Therefore, the value of a is 2√15.