Answer:
Step-by-step explanation:
You can look at it and see that 1/3 is added to each term. The common difference is 1/3. If you want to be be more formal,
d = d4 - d3
d = 0 - - 1/3
d = 0 + 1/3
d = 1/3

Expanding the left side gives

which gives two solutions,

and

. But if

, then

, but this number isn't real, so

is an extraneous solution. Meanwhile if

, you get

, so this solution is correct.
"Potential solutions" might refer to both possibilities, but there is only one actual (real) solution.
Write the conjugates of them then FOIL both the top and bottom, then factor
Answer:
12g
Step-by-step explanation:
Answer:
-14-x
Step-by-step explanation:
Distribute the negative.