Why|x(sqaure)-4| must be greater than 0?

[JK]

Someone posted the answer within all of these messages. If you scroll up you can read it.

[StaceyKoprince]

Answer is C and JK's explanation is right. Sumit, you're also absolutely right that we could interpret statement 1 in the same way regardless of what was between the absolute value signs - whatever you find inside absolute value signs, once you apply the absolute value, you cannot have a negative. You've either got zero or a positive number.

So, it seems that we do have consensus that neither statement is sufficient alone. We also have consensus that statement 2 solves to y = -8 or y = 14.

We don't have consensus on statement 1. So let's examine that one a bit further. When we try the two statements together, the only valid solutions are those that work for both statements, right? We know that -8 and 14 both work for statement 2. Let's see whether they work for statement 1. (Note: I far prefer JK's method - using theory to be able to avoid all of the nasty work below. But let's first make sure that you understand the math. Then, maybe, you'll be able to believe the theory and use it instead.)

3*lx^2 - 4l = y-2
try 14 first:
3lx^2-4l = 14-2
3lx^2-4l = 12
lx^2-4l = 4
x^2 = 8 or x^2 = 0
x = SQRT8 or x = 0
Plug either one back in to see whether it works: 3* l(SQRT8)^2 - 4l = 14-2
That works, so 14 is one possible solution. (You could also try zero - and I'd actually recommend that because it's easier to try - but as long as one works, then 14 is a possible solution.)

Try -8 next:
3lx^2-4l = -8-2
3lx^2-4l = -10
lx^2-4l = -10/3
x^2 = -10/3 + 4 OR x^2 = 10/3 + 4
x^2 = -10/3 + 12/3 OR x^2 = 10/2 + 12/3
x^2 = 2/3 OR x^2 = 22/3
x = SQRT(2/3) or x = SQRT (22/3)
Plug it back in: 3* l(SQRT2/3)^2 - 4l =-8-2
3*l2/3-4l = -10 (right around here, it should be obvious that you can't get a negative answer from the left-hand side)
3*l-10/3l = -10
3*10/3 = -10
10=-10
that's not true. Now we have to plug in the other solution, though - that one might work.
3* l(SQRT22/3)^2 - 4l =-8-2
3*l22/43- 4l = -10 (again, look at that. are you going to get a negative number on the left? no.)
3*l10/3l = -10
3*10/3 = -10
10=-10
again, not true. -8 is not a valid solution for BOTH statements 1 and 2, so we ignore it. We're left with a single valid solution (14), so the info is sufficient.

Back to the theory. In the original equation, you've got two parts on the left-hand side:
3 multiplied by some stuff between absolute value signs
3 is positive. The stuff between the absolute value signs has to be either zero or positive - it can never be negative.
If that stuff equals zero, then you have 3*0 on the left-hand side. If that stuff is positive, then you have 3*something positive on the left-hand side.

The stuff on the left-hand side equals whatever's on the right-hand side (by definition, right? it's an equation). So y-2 is either zero or positive. If y-2 is zero, then y must be 2. If y-2 is positive, then y must be greater than 2. So the valid solution(s) for y must be greater than or equal to 2. If we can figure that out theoretically, we get to skip the nasty math above and realize, once we've solved statement 2, that only one of the two solutions from statement 2 is valid when we also look at statement 1 in combination with statement 2.