Is ((y - 4)^2)^(1/2) = 4 - y ? (1) |y - 3| less than or equal to 1

Absolute value properties:
When \(x\leq{0}\) then \(|x|=-x\), or more generally when \(some \ expression\leq{0}\) then \(|some \ expression|={-(some \ expression)}\). For example: \(|-5|=5=-(-5)\);

When \(x\geq{0}\) then \(|x|=x\), or more generally when \(some \ expression\geq{0}\) then \(|some \ expression|={some \ expression}\). For example: \(|5|=5\);

Next:
Must know for the GMAT: \(\sqrt{x^2}=|x|\).

The point here is that as square root function can not give negative result then \(\sqrt{some \ expression}\geq{0}\).

So \(\sqrt{x^2}\geq{0}\). But what does \(\sqrt{x^2}\) equal to?

Let's consider following examples:
If \(x=5\) --> \(\sqrt{x^2}=\sqrt{25}=5=x=positive\);
If \(x=-5\) --> \(\sqrt{x^2}=\sqrt{25}=5=-x=positive\).

So we got that:
\(\sqrt{x^2}=x\), if \(x\geq{0}\);
\(\sqrt{x^2}=-x\), if \(x<0\).

What function does exactly the same thing? The absolute value function: \(|x|=x\), if \(x\geq{0}\) and \(|x|=-x\), if \(x<0\). That is why \(\sqrt{x^2}=|x|\).

BACK TO THE ORIGINAL QUESTION:

Question asks: is \(\sqrt{(y-4)^2}=4-y\)?

Now, according to the above \(\sqrt{(y-4)^2}=|y-4|\)? So the question becomes: is \(|y-4|=4-y\)? --> or is \(|y-4|=-(y-4)\)? Which, again according to the properties of absolute value, is true when \(y-4\leq{0}\) or when \(y\leq{4}\).

Hope it's clear.
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Is \(\sqrt{(y-4)^2} = 4-y\)?

Is \(\sqrt{(y-4)^2}=4-y\)? --> is \(|y-4|=4-y\)? is \(y\leq{4}\)?

(1) |y-3| less than or equal to 1 --> \(|y-3|\) is just the distance between 3 and \(y\) on the number line. We are told that this distance is less than or equal to 1: --2----3----4-- so, \(2\leq{y}\leq{4}\). Sufficient.

(2) y*|y|>0 --> just says that \(y>0\). Not sufficient.

Answer: A.



Just a little correction, which didn't play a part for this question though may be crucial for others:
\(|y-4|=4-y\) if \(y-4\leq{0}\), you missed "=" sign --> if \(y=4\) the equation still holds true.
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sqrt{(y-4)^2} = |y-4|

And |y-4| = 4-y if y-4 < 0

or |y-4| = 4-y if y < 4

1) |y-3| less than or equal to 1

|y-3| <= 1

Square both sides
y^2 - 6y + 9 <= 1

y^2 -4y - 2y + 8 <=0
(y-2)(y-4) <=0

=> 2 <= y <= 3

So y < 4

Sufficient

2) y*|y| > 0

This is insufficient becuase

y = 5 and y = 1, both satisfy the condition

Answer - A
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I just cannot understand why do you use the inequality to restate the question? does not is simply states that |y-4| = 4-y -> y-4=4-y -> y=4 ? Otherwise, it is 0=0.

Because \(|y-4|=4-y\) holds true for \(y\leq{4}\) and not only for \(y=4\).
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Bunuel - I did the same thing as JubtaGubar and I'm still confused on how you and subhashghosh came to that conclusion so quickly (that \(y\leq{4}\)).

Thank you

I think I figured it out --- this \(\sqrt{(y-4)^2}\) must be positive (anything squared and/or taken to its square root) which is essentially like saying that the left side must be positive or absolute value of the left side of the equation and so the right side must also be positive (or > 0). The only way that can happen is when \(y\leq{4}\)

Am I thinking about that correctly?

Reposting---------

So we got that:
\(\sqrt{x^2}=x\), if \(x\geq{0}\);
\(\sqrt{x^2}=-x\), if \(x<0\).

What function does exactly the same thing? The absolute value function: \(|x|=x\), if \(x\geq{0}\) and \(|x|=-x\), if \(x<0\). That is why \(\sqrt{x^2}=|x|\).

BACK TO THE ORIGINAL QUESTION:

Question asks: is \(\sqrt{(y-4)^2}=4-y\)?

Now, according to the above \(\sqrt{(y-4)^2}=|y-4|\)? So the question becomes: is \(|y-4|=4-y\)? --> or is \(|y-4|=-(y-4)\)? Which, again according to the properties of absolute value, is true when \(y-4\leq{0}\) or when \(y\leq{4}\).

Hope it's clear.[/quote][/quote]

Not sure understand your question. Can you please elaborate?

P.S. Note that the absolute value is ALWAYS non-negative, also I did not multiply anything by -1...[/quote]

Hello Bunnel, I have one small doubt.
When you say |X| = +X, when X<eq0 and -X when <0 , then in this question how could you compare |y-4| to be 4-y when comparing y<eq4 ? Shouldnt it be y<4 only? On x=0 how does the mod function behave, its neither postive nor negative then dont you think at x= 0, we cant be sure wether F(X) is positive or negative ?

Check your approach with numbers: if \(y=4\) --> \(\sqrt{(4-4)^2}=0\) and \(4-4=0\) --> \(0=0\).

Hope it's clear.
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Bunuel chetan2u

I have a question here

I am clear with the concept of absolute function. |x| = +x when x>=0
-x when x<0

Here, you have considered the case as |y-4| = y-4, when y>4
4-y, when y<=4

My question is at y =4, the case can be either y-4 OR 4-y. Both yield 0 as the answer. Then why have you only considered y<=4 in (4-y) case

Thanking in anticipation

Cheers

Hi Keats,

It doesn't matter where you take EQUAL sign, with less than or greater than.. only point is it should be catered in any one.....
You are having TWO ranges..
1) y is LESS than or EQUAL to 4 and y is more than 4.... so this completes the range
2) y is less than 4 and y is GREATER than or EQUAL to 4... this again completes the range

But you cannot have EQUAL to in both as it is duplicated while you combine the two to find the different ranges..


The MOD in my signature below may be of some use here
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Bunuel But isn't -(y-4) < 0? How is it -(y-4)<=0? Then what about y-4>=0??? How can both expressions equal to 0? In the GMAT Math book, the expression is positive when (x-1)>=0 and negative when (x-1)<0.

Can you please explain?

Absolute value properties:
When \(x\leq{0}\) then \(|x|=-x\), or more generally when \(some \ expression\leq{0}\) then \(|some \ expression|={-(some \ expression)}\). For example: \(|-5|=5=-(-5)\);

When \(x\geq{0}\) then \(|x|=x\), or more generally when \(some \ expression\geq{0}\) then \(|some \ expression|={some \ expression}\). For example: \(|5|=5\);

The question asks whether \(|y-4|=-(y-4)\)? The answer is when \(y\leq{4}\).

When y < 4, then (y - 4) is negative, so in this case \(|y-4|=-(y-4)\).
When y = 0, then (y - 4) is zero, so in this case \(|y-4|=-(y-4)=0\).

So, the answer is when \(y\leq{4}\).
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