Interesting question. First, before looking at the statements, let's look at the question itself. Actually, let's start with a simpler question:
What is the sqrooot(y^2) equal to? By sqroot, I mean the 'non-negative sqrooot' (I can't write the root symbol here unfortunately!).
If y is positive, sqrooot(y^2) is just y.
But, if y is negative, sqrooot(y^2) is not y; the sqroot can't be negative. Instead, it's equal to |y|, which is equal to -y.
Now let's look at the question:
Is sqroot((x-5)^2) = 5 - x ?
Well, just as with y above:
sqroot((x-5)^2) will be equal to x-5 if x-5 is positive.
sqroot((x-5)^2) will be equal to -(x-5) = 5-x if x-5 is negative.
So the question is just asking: is x-5 negative? That is, is x<5? Statement 2) tells us that exactly, so it's sufficient. Statement 1) requires a bit more work:
-x|x| > 0
We can divide both sides by |x| here since we know |x| is positive (no need to worry about whether you'd need to reverse the inequality), so
-x > 0
x < 0
So again, x is certainly less than 5, and Statement 1 is also sufficient on its own.
D.
(technically, I should have also included the possibility that x-5 = 0 in the above, but it doesn't affect the solution in this case, and I've left it out for clarity).
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Square root problem!
Source: Beat The GMAT — Data Sufficiency |
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Ian Stewart
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The answer is correct according to the GMAT solution but approach is wrong as per me.
I am still stuck at one thing. Please find a problem with the following.
Sqroot of a^2 is normally written as +/- a. As both +a and -a on squaring will become a^2.
So roots of the given problem are (x-5) and -(x-5).
So considering the negative value of sqroot of (x-5)^2,
which is -(x-5), it is actually (5-x). Proves the question by itself.
Now positive, that is (x-5). (The actual value may or may not be positive).
i) As per first condition, we find that x is negative. Lets say -2 or -10.
That is, (x-5) will surely be a negative value. As in, (-2-5=-7) or (-10-5=-15)
Whereas, (5-x) will surely have a positive value. As in, 5-(-2)=7 or 5-(-10)=15
How can these 2 values ever be equal!!?? So its definitely NOT EQUAL.
ii) 5-x>0, that is 5>x. x can be positive or negative. Eg. 2 or -10.
For either, (x-5) will surely be negative. As in, -3 or -15.
Again both values are NOT EQUAL.
If we just look at the positive root of the initial question, both given conditions are SUFFICIENT as we can say for sure that it is not equal. Answer will then be D.
BUT, what about the negative root, -(x-5) which is equal to (5-x)??
I am still stuck at one thing. Please find a problem with the following.
Sqroot of a^2 is normally written as +/- a. As both +a and -a on squaring will become a^2.
So roots of the given problem are (x-5) and -(x-5).
So considering the negative value of sqroot of (x-5)^2,
which is -(x-5), it is actually (5-x). Proves the question by itself.
Now positive, that is (x-5). (The actual value may or may not be positive).
i) As per first condition, we find that x is negative. Lets say -2 or -10.
That is, (x-5) will surely be a negative value. As in, (-2-5=-7) or (-10-5=-15)
Whereas, (5-x) will surely have a positive value. As in, 5-(-2)=7 or 5-(-10)=15
How can these 2 values ever be equal!!?? So its definitely NOT EQUAL.
ii) 5-x>0, that is 5>x. x can be positive or negative. Eg. 2 or -10.
For either, (x-5) will surely be negative. As in, -3 or -15.
Again both values are NOT EQUAL.
If we just look at the positive root of the initial question, both given conditions are SUFFICIENT as we can say for sure that it is not equal. Answer will then be D.
BUT, what about the negative root, -(x-5) which is equal to (5-x)??
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Ian Stewart
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There is a technicality here. It is definitely true that a^2 has two square roots, a and -a, as you point out (unless, of course, a = 0). But the square root symbol, which I'll try to draw:anshul265 wrote: I am still stuck at one thing. Please find a problem with the following.
Sqroot of a^2 is normally written as +/- a. As both +a and -a on squaring will become a^2.
_____
\/
*means* the non-negative square root *only*. If you see this:
______
\/ a^2
this has only one solution, either a or -a, whichever of these is positive (or zero). It does not have two solutions, and you cannot assume it to be equal to a unless you know whether a is positive. So it is incorrect to say that
________
\/ (x-5)^2 = x-5
This is only true if x - 5 >= 0. And that's where your answer differs from the GMAT's (and mine). Hope that clears things up.
I am getting what you are trying to explain, but in GMAT I have gotten in trouble a lot of times for not considering the negative values. Anyway, thanks a lot.
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Ian Stewart
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Yes, and you will often get in trouble for not considering negatives. If you see the equation x^2 = 1, for example, there are two solutions for x, 1 and -1. The difference in the question you posted above is that it contains the square root symbol. That symbol means 'the non-negative square root', so can never produce a negative result. If you see the number 1 underneath the square root symbol, the value is 1, not -1.anshul265 wrote:I am getting what you are trying to explain, but in GMAT I have gotten in trouble a lot of times for not considering the negative values. Anyway, thanks a lot.
