The question is:
Is sqrt(x) a prime number?
1) Mod (3x - 7) = 2x + 2
2) x^2 = 9x

The answer is given as C but I really think it should be E.
When we solve the equations separately, we get X as 9 and 1 for the first part.
And for the second, 0 and 9 are the answers.

So the solution says, root of 9 is 3 and thus three is common in both the parts and hence the Answer is C.
But what about the negative root. Why isnt that considered? So You will get a common answer as 3 and -3. So doesnt narrow down to a single number. Thus the answer should be E)

Request some one to explain if I am wrong.

Prime numbers are positive, non-zero numbers that have exactly two factors.
I think sqrt(x) cannot be justified for negative values.

Hence C is sufficient

Substitution:-

sqr(x)= prime

BDACE,

x^2= 9x
x^2-9x=0
x(x-9)=0
x=0;9
2) is Insufficient

|3x-7|= 2x+2

3x-7=2x+2;
x=9

-3x+7=2x+2
-5x=-5
x=1

So, x= 9;1
1) is Insufficient

From 1) and 2), x=9

sqrt(9)=3 which is prime. -3 cannot be a prime number because it is negative.

Thanks

it seems as though you already understand the details of the solution; the only issue appears to be the definition of √x.

remember this: √x, like all other symbols, represents a FUNCTION. it therefore has only ONE VALUE.
there are no exceptions to the following: expressions (with symbols etc. but NO equals sign) can only have one value.

equations (which have equals signs), on the other hand, can have multiple solutions.
so, for instance, the equation x^2 = 9 has two solutions (3 and -3), while the expression √9 only has one value (= 3).

hth

Thank you Ron

You're welcome!