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