Pls i’m really confused

find ab round to the nearest tenth if necessary

Comment
By similar triangles it can be shown that AD^2 = AB*AC
If you want the proof, Google tangents and secants of a circle.

Find
So we want
AB

Givens
AD = 16
BC = 9

AB = ??
CA = CB + AB
CA = 9 + AB

Formula
AB * (AB + BC) = AD^2 

Sub and Solve
AB*(AB + 9) = 16^2
AB*(AB + 9) = 256 Remove the brackets.
AB^2 + 9AB = 256 Subtract 256 from both sides.
AB^2 + 9AB - 256 = 0

You can only do this either with a graph or the quadratic formula. I'll get the graph for you. You can made these yourself at Desmos.

x = [-b +/- sqrt(b^2 - 4ac)] / (2a) 
a = 1 
b = 9
c = -256

Answer
When you substitute these into the quadratic formula, you get
x1 = 12.12 and 
x2 = -21.12
x2 is meaningless. The solution is
x = 12.12 

Comment
But that's not your question. Your question is what is this rounded to the nearest 1/10th? That's a fancy way of saying round to the first decimal place. Since the hundredth place (or second place) is 2, 12.12 rounds to 12.1

The answer is
x = 12.1 <<<<< answer.