\(a^2 = a + c\)
AND
\(ac > 0\)
Now, \(a^2 = a + c\)\(= a^2-a=c\)
\(a(a-1)=c\)
This means that a and a-1 are consecutive integers such as 3 and 2
We do know also that ac > 0 and a and c are both positive or negative numbers
BUt in the original stem we had a^2 and this is a clue to know that a is + and therefore c is + as well
Now we do have all the pieces of information
A. a> 0 this is true because if we know that c is + also a must be positive
B. a < c this could be true
3*2=6 and 3 < 6 and this is true. however if we have 2*1=2 the condition is satisfied but 2(a) = 2 (c) So b can be or not true and we need a MUST be true
C. 0<a<1 suppose a is 1/2 then we would have 1/2-1 which would be negative and negative * positive is negative. This is not possible
D. a>1 yes because if a would 1 then a-1 would be zero and this is not possible
So A and D are the correct choices
As for your question I am not sure I got what you mean but I think is not feasible
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