You want 9x^2 + bx + 9a to be a perfect square trinomial. Note that 9 x 2 is incorrect and should be written as 9x^2, where "^" represents "exponentiation."
What about a? Are we supposed to find a also?
One way in which to do this problem is to factor 9 out of the trinomial:
9 (x^2 + (b/9)x + a )
Concentrate now on making x^2 + (b/9)x + a into a perfect square trinomial.
x^2 + (b/9)x + a
Take half of the coefficient (b/9) and square the result: [(b/9)/2]^2 = b^2/81.
Then, x^2 + (b/9)x + b^2/81 - b^2/81 + a.
The above quadratic expression can be re-written as
(x + b/9)^2 - b^2/81 + a. This is a perfect square trinomial if
-b^2/81 + a = 0. Solve for b: b^2/81 = a,
b/9 = sqrt(a)
b = 9 sqrt a
