Confused as well, and get an answer of B like you.

Statement 1 would yield +/- 3.
If you plug both of those numbers in Statement 2, neither will work. Only an x value of 0 would work for statement 2, which led me to B.

Is there an official explanation?

Statement (1) sqrt(x^4) = 9
This will give you x = -3 or +3

Statement (2) sqrt(x^2) = -x
This is not sufficient at all since this can be any number -2, -3, -4 or any -ve value

Statement (1) and (2) together will result in x = -3

So (C) is the answer.

I am sorry its still not clear:
Stmt 2:
Root(x^2)=-x
(x^2)^1/2= -x
x= -x
2x= 0
x= 0 Hence B

Would this be logical? DS experts please comment:

(1) sqrt(x^4) = 9
x^2 = 9
x can be +/- 3
BCE
(2) sqrt(x^2) = -x
x = -x
CE
C => x = -3

Here you have to make an assumption that right hand side is not a variable but a value. For ex: stmt 2 could have been sqrt(x^2) = -5 (in the stmt it is just -x). So deducing it to be a negative value would make it an acceptable solution for combining the 2 stmts (1) & (2).

LHS (left hand side) is a variable, RHS is value. (just applying computer terminology)

HTH.

in stmt 2 whenyou are getting x=-x,we know that this is possible only when x=0..so why is this not sufficient??!

In Statement (2), you are not really getting x = -x.

May be this will clarify.

When you take a square root of a number, there are two possible values, one is the negative square root value and the other is the positive square root, i.e. for example square root of 4 is either +2 or -2.

So what Statement (2) is saying is that x is the negative value and not the positive value, i.e. in the above example, the value is -2 and not +2.

uptowngirl92 wrote:in stmt 2 whenyou are getting x=-x,we know that this is possible only when x=0..so why is this not sufficient??!

This rule will be helpful: root(x^2)is nothing but |x|

Now in statement (2) root(x^2)= -x which is nothing but |x| = -x

This means that x is negative. But x can be any number.

With the details in A & B, we can get the value of X.

HTH