Find two values of b that will make 9 x 2 + b x + 9 a perfect square trinomial.

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

Using the rise/run rule. you would get 2/6.simp. 1/3