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If a2 = a + c,where a.c > 0. Which of the following must be true ? a > [#permalink]
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\(a^2 = a + c\)

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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Re: If a2 = a + c,where a.c > 0. Which of the following must be true ? a > [#permalink]
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a2= a+c ac>0 which means a is not zero. Therefore a2= positive ; if a2 is positive a+c must be positive (1) . we know that ac >0 which means both a and c have same sign. i.e they both are positive or they Both are negative. since a+c is positive (from 1), both a and c cannot be negative(2) . which means a and c are greater than zero.
i.e a>0 and c>0
a2= a+c. -> a2-a=c ->. a(a-1) = c . we know that c is positive (from 2) and a>0 , for c to be positive "a" must be greater than 1 in the equation a(a-1)=c
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