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Is sqrt ((x-3)^2) = 3-x?

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Re: Is sqrt ((x-3)^2) = 3-x?  [#permalink]

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22 Nov 2014, 06:20
2
hanschris5 wrote:
Bunuel wrote:
gautamsubrahmanyam wrote:
I understand that 1) is insuff

But for 2) -x|x| > 0 means x cant be +ve => |x| = -x so that -x (-x) = x^2> 0

If x is -ve => (x-3)^2 = X^2+9-6x = (-ve)^2+9-6(-ve) = +ve+9-(-ve) = +ve +9 + (+ve) = +ve

=> sqrt ((x-3)^2) = +X-3

=> sqrt ( (x-3) ^2 ) is not equal to 3-x

=> Option B

Yes, the answer for this question is B.

Is $$\sqrt{(x-3)^2}=3-x$$?

Remember: $$\sqrt{x^2}=|x|$$. Why?

Couple of things:

The point here is that square root function can not give negative result: wich means that $$\sqrt{some \ expression}\geq{0}$$.

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! That is why $$\sqrt{x^2}=|x|$$

Back to the original question:

So $$\sqrt{(x-3)^2}=|x-3|$$ and the question becomes is: $$|x-3|=3-x$$?

When $$x>3$$, then RHS (right hand side) is negative, but LHS (absolute value) is never negative, hence in this case equations doesn't hold true.

When $$x\leq{3}$$, then $$LHS=|x-3|=-x+3=3-x=RHS$$, hence in this case equation holds true.

Basically question asks is $$x\leq{3}$$?

(1) $$x\neq{3}$$. Clearly insufficient.

(2) $$-x|x| >0$$, basically this inequality implies that $$x<0$$, hence $$x<3$$. Sufficient.

Hope it helps.

can u pls help in decoding option B where , you say x <0
i mean -x|x|> 0 , could you elaborate on this .

thanks.

Sure. We have that $$-x|x| >0$$, so the product of two multiples -x and |x| is positive. Now, we know that |x| must be positive, hence -x must also be positive for the product to be positive, thus -x > 0, which is the same as x < 0.

Hope it's clear.
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Re: Is sqrt ((x-3)^2) = 3-x?  [#permalink]

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22 Nov 2014, 07:26
okay gr8 ...
I have a doubt about this we have sqrt of x is always greater than equal to 0 .

Now same concept is what we applied on the above problem.

But there is one problem that u explained on a GMAT PREP question .

Let's consider following examples:

I am concern about the second one when x=-5 . So x should not be negative anyhow rite ??? but how is it negative and we say its 5=-x
I tried posting this on the

BUT was not accepting the expression ... havent yet learnt to use expression ...

Bunuel wrote:
gautamsubrahmanyam wrote:
I understand that 1) is insuff

But for 2) -x|x| > 0 means x cant be +ve => |x| = -x so that -x (-x) = x^2> 0

If x is -ve => (x-3)^2 = X^2+9-6x = (-ve)^2+9-6(-ve) = +ve+9-(-ve) = +ve +9 + (+ve) = +ve

=> sqrt ((x-3)^2) = +X-3

=> sqrt ( (x-3) ^2 ) is not equal to 3-x

=> Option B

Yes, the answer for this question is B.

Is $$\sqrt{(x-3)^2}=3-x$$?

Remember: $$\sqrt{x^2}=|x|$$. Why?

Couple of things:

The point here is that square root function can not give negative result: wich means that $$\sqrt{some \ expression}\geq{0}$$.

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! That is why $$\sqrt{x^2}=|x|$$

Back to the original question:

So $$\sqrt{(x-3)^2}=|x-3|$$ and the question becomes is: $$|x-3|=3-x$$?

When $$x>3$$, then RHS (right hand side) is negative, but LHS (absolute value) is never negative, hence in this case equations doesn't hold true.

When $$x\leq{3}$$, then $$LHS=|x-3|=-x+3=3-x=RHS$$, hence in this case equation holds true.

Basically question asks is $$x\leq{3}$$?

(1) $$x\neq{3}$$. Clearly insufficient.

(2) $$-x|x| >0$$, basically this inequality implies that $$x<0$$, hence $$x<3$$. Sufficient.

Hope it helps.
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Joined: 02 Sep 2009
Posts: 52438
Re: Is sqrt ((x-3)^2) = 3-x?  [#permalink]

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22 Nov 2014, 08:09
hanschris5 wrote:
okay gr8 ...
I have a doubt about this we have sqrt of x is always greater than equal to 0 .

Now same concept is what we applied on the above problem.

But there is one problem that u explained on a GMAT PREP question .

Let's consider following examples:

I am concern about the second one when x=-5 . So x should not be negative anyhow rite ??? but how is it negative and we say its 5=-x
I tried posting this on the

BUT was not accepting the expression ... havent yet learnt to use expression ...

Bunuel wrote:
gautamsubrahmanyam wrote:
I understand that 1) is insuff

But for 2) -x|x| > 0 means x cant be +ve => |x| = -x so that -x (-x) = x^2> 0

If x is -ve => (x-3)^2 = X^2+9-6x = (-ve)^2+9-6(-ve) = +ve+9-(-ve) = +ve +9 + (+ve) = +ve

=> sqrt ((x-3)^2) = +X-3

=> sqrt ( (x-3) ^2 ) is not equal to 3-x

=> Option B

Yes, the answer for this question is B.

Is $$\sqrt{(x-3)^2}=3-x$$?

Remember: $$\sqrt{x^2}=|x|$$. Why?

Couple of things:

The point here is that square root function can not give negative result: wich means that $$\sqrt{some \ expression}\geq{0}$$.

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! That is why $$\sqrt{x^2}=|x|$$

Back to the original question:

So $$\sqrt{(x-3)^2}=|x-3|$$ and the question becomes is: $$|x-3|=3-x$$?

When $$x>3$$, then RHS (right hand side) is negative, but LHS (absolute value) is never negative, hence in this case equations doesn't hold true.

When $$x\leq{3}$$, then $$LHS=|x-3|=-x+3=3-x=RHS$$, hence in this case equation holds true.

Basically question asks is $$x\leq{3}$$?

(1) $$x\neq{3}$$. Clearly insufficient.

(2) $$-x|x| >0$$, basically this inequality implies that $$x<0$$, hence $$x<3$$. Sufficient.

Hope it helps.

$$\sqrt{x^2}=\sqrt{25}=5$$ --> x = 5 or x = -5. The square root gives positive result but x itself can be 5 or -5. What is confusing there?
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Re: Is sqrt ((x-3)^2) = 3-x?  [#permalink]

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22 Nov 2014, 09:26
okay so its only do with the result ie the x that comes out as result after the operation ... whatever it may have been inside the expression...

thanks. doubt cleared.
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Re: Is sqrt ((x-3)^2) = 3-x?  [#permalink]

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27 Nov 2015, 08:11
Given to us $$\sqrt{(x-3)^2} = 3-x$$

As a general principle, whenever we are given square root of a square of a number, we can have 2 solutions for such a number. It is illustrated below:

$$\sqrt{x^2} = x (if x>=0)$$
$$\sqrt{x^2} = -x (if x<=0)$$
This square root of a square of a number is essentially the modulus of number and thus always gives a positive answer.

So in questions like the one illustrating the above principle we have to consider both the possibilities.

Now, for the given equation:

The two possibilities are

1st possibility : $$\sqrt{(x-3)^2} = x-3$$ (if x>=3)
2nd possibility : $$\sqrt{(x-3)^2} = 3-x$$ (if x<=3)

Now considering the given statements:

Using Statement 1

if $$x\neq{3}$$

This statement satisfies both the possibilities
Hence we cannot reach to a single possibility, so statement 1 is not sufficient.

Using Statement 2

$$-x|x| > 0$$

For this condition to be true, x needs to be necessarily less than 0.

Now if x<0, then it definitely is less than 3, hence will satisy the 2nd possibility, thus, giving a distcict answer.

Hence statement 2 is sufficent to answer $$\sqrt{(x-3)^2} = 3-x$$

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Re: Is sqrt ((x-3)^2) = 3-x?  [#permalink]

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26 Jan 2017, 18:55
1
Be definition:
√(x²) = |x|.
|x-y| is the DISTANCE between x and y.
The DISTANCE between two numbers must be greater than or equal to 0.

The question stem above, rephrased: Is |x-3| = 3-x?
In words:
Is the DISTANCE between x and 3 equal to the DIFFERENCE of 3 and x?
The answer will be YES if the DIFFERENCE of 3 and x is greater than or equal to 0:
3-x≥0
x≤3.

The question stem rephrased: Is x≤3?

Statement 1: x is not equal to 3.
It is possible that x<3 or that x>3.
INSUFFICIENT.

Statement 2: -x*|x| > 0 .
Thus, the left-hand side must be positive*positive or negative*negative.
Since |x| cannot be negative, both factors on the left-hand side must be positive.
Thus:
-x>0
x<0.
Since x<0, we know that x≤3.
SUFFICIENT.

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Re: Is sqrt ((x-3)^2) = 3-x?  [#permalink]

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02 Feb 2017, 01:51
hanschris5 wrote:
Bunuel wrote:
gautamsubrahmanyam wrote:
I understand that 1) is insuff

But for 2) -x|x| > 0 means x cant be +ve => |x| = -x so that -x (-x) = x^2> 0

If x is -ve => (x-3)^2 = X^2+9-6x = (-ve)^2+9-6(-ve) = +ve+9-(-ve) = +ve +9 + (+ve) = +ve

=> sqrt ((x-3)^2) = +X-3

=> sqrt ( (x-3) ^2 ) is not equal to 3-x

=> Option B

Yes, the answer for this question is B.

Is $$\sqrt{(x-3)^2}=3-x$$?

Remember: $$\sqrt{x^2}=|x|$$. Why?

Couple of things:

The point here is that square root function can not give negative result: wich means that $$\sqrt{some \ expression}\geq{0}$$.

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! That is why $$\sqrt{x^2}=|x|$$

Back to the original question:

So $$\sqrt{(x-3)^2}=|x-3|$$ and the question becomes is: $$|x-3|=3-x$$?

When $$x>3$$, then RHS (right hand side) is negative, but LHS (absolute value) is never negative, hence in this case equations doesn't hold true.

When $$x\leq{3}$$, then $$LHS=|x-3|=-x+3=3-x=RHS$$, hence in this case equation holds true.

Basically question asks is $$x\leq{3}$$?

(1) $$x\neq{3}$$. Clearly insufficient.

(2) $$-x|x| >0$$, basically this inequality implies that $$x<0$$, hence $$x<3$$. Sufficient.

Hope it helps.

can u pls help in decoding option B where , you say x <0
i mean -x|x|> 0 , could you elaborate on this .

thanks.

Whatever you put into the |x|, it always gives you positive value. If you multiply 2 positive or 2 negative values, it always gives you positive result, which is ''some expression''>0. Here, |x| already gives you positive value. So, the green part has to be negative to legit the statement. Here, is test case:
-x*positive value>0
--> -(-10)*positive value>0
-->positive value*positive value>0
So, to legit the statement 2, x must be negative. The statement is sufficient.
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Re: Is sqrt ((x-3)^2) = 3-x?  [#permalink]

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23 Apr 2017, 19:01
Bunuel wrote:
gautamsubrahmanyam wrote:
I understand that 1) is insuff

But for 2) -x|x| > 0 means x cant be +ve => |x| = -x so that -x (-x) = x^2> 0

If x is -ve => (x-3)^2 = X^2+9-6x = (-ve)^2+9-6(-ve) = +ve+9-(-ve) = +ve +9 + (+ve) = +ve

=> sqrt ((x-3)^2) = +X-3

=> sqrt ( (x-3) ^2 ) is not equal to 3-x

=> Option B

Yes, the answer for this question is B.

Is $$\sqrt{(x-3)^2}=3-x$$?

Remember: $$\sqrt{x^2}=|x|$$. Why?

Couple of things:

The point here is that square root function can not give negative result: wich means that $$\sqrt{some \ expression}\geq{0}$$.

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! That is why $$\sqrt{x^2}=|x|$$

Back to the original question:

So $$\sqrt{(x-3)^2}=|x-3|$$ and the question becomes is: $$|x-3|=3-x$$?

When $$x>3$$, then RHS (right hand side) is negative, but LHS (absolute value) is never negative, hence in this case equations doesn't hold true.

When $$x\leq{3}$$, then $$LHS=|x-3|=-x+3=3-x=RHS$$, hence in this case equation holds true.

Basically question asks is $$x\leq{3}$$?

(1) $$x\neq{3}$$. Clearly insufficient.

(2) $$-x|x| >0$$, basically this inequality implies that $$x<0$$, hence $$x<3$$. Sufficient.

Hope it helps.

Hi Bunuel,

From $$|x-3|=3-x$$, since we know |x-3| is positive, can we not go one step further and simplify the equation to:

x-3=3-x
=>x=3?

What am I doing wrong?
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Posts: 52438
Re: Is sqrt ((x-3)^2) = 3-x?  [#permalink]

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23 Apr 2017, 21:44
puto wrote:
Bunuel wrote:
gautamsubrahmanyam wrote:
I understand that 1) is insuff

But for 2) -x|x| > 0 means x cant be +ve => |x| = -x so that -x (-x) = x^2> 0

If x is -ve => (x-3)^2 = X^2+9-6x = (-ve)^2+9-6(-ve) = +ve+9-(-ve) = +ve +9 + (+ve) = +ve

=> sqrt ((x-3)^2) = +X-3

=> sqrt ( (x-3) ^2 ) is not equal to 3-x

=> Option B

Yes, the answer for this question is B.

Is $$\sqrt{(x-3)^2}=3-x$$?

Remember: $$\sqrt{x^2}=|x|$$. Why?

Couple of things:

The point here is that square root function can not give negative result: wich means that $$\sqrt{some \ expression}\geq{0}$$.

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! That is why $$\sqrt{x^2}=|x|$$

Back to the original question:

So $$\sqrt{(x-3)^2}=|x-3|$$ and the question becomes is: $$|x-3|=3-x$$?

When $$x>3$$, then RHS (right hand side) is negative, but LHS (absolute value) is never negative, hence in this case equations doesn't hold true.

When $$x\leq{3}$$, then $$LHS=|x-3|=-x+3=3-x=RHS$$, hence in this case equation holds true.

Basically question asks is $$x\leq{3}$$?

(1) $$x\neq{3}$$. Clearly insufficient.

(2) $$-x|x| >0$$, basically this inequality implies that $$x<0$$, hence $$x<3$$. Sufficient.

Hope it helps.

Hi Bunuel,

From $$|x-3|=3-x$$, since we know |x-3| is positive, can we not go one step further and simplify the equation to:

x-3=3-x
=>x=3?

What am I doing wrong?

I tried to explain exactly this point in may solutions. From |x - 3| = 3 - x, when x - 3 <= 0. If it's still unclear you should go through theory about absolute values:

Theory on Absolute Values: http://gmatclub.com/forum/math-absolute ... 86462.html
Absolute value tips: http://gmatclub.com/forum/absolute-valu ... 75002.html

Hope it helps.
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Re: Is sqrt ((x-3)^2) = 3-x?  [#permalink]

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02 Oct 2018, 05:19
Bunuel wrote:
gautamsubrahmanyam wrote:

(2) $$-x|x| >0$$, basically this inequality implies that $$x<0$$, hence $$x<3$$. Sufficient.

Hope it helps.

Why does -x|X|>0 imply that x<0?
Math Expert
Joined: 02 Sep 2009
Posts: 52438
Re: Is sqrt ((x-3)^2) = 3-x?  [#permalink]

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02 Oct 2018, 05:26
lostaish wrote:
Bunuel wrote:
gautamsubrahmanyam wrote:

(2) $$-x|x| >0$$, basically this inequality implies that $$x<0$$, hence $$x<3$$. Sufficient.

Hope it helps.

Why does -x|X|>0 imply that x<0?

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Re: Is sqrt ((x-3)^2) = 3-x?  [#permalink]

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02 Oct 2018, 10:30
Bunuel, Thanks for the explanation, but i had one question.
You mentioned : "Couple of things:
The point here is that square root function can not give negative result: wich means that some expression−−−−−−−−−−−−−−√≥0some expression≥0."

Can you please let me know why? Because in general square root gives two values, one positive and one negative.
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Posts: 52438
Re: Is sqrt ((x-3)^2) = 3-x?  [#permalink]

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02 Oct 2018, 19:35
1
PriyankaGehlawat wrote:
Bunuel, Thanks for the explanation, but i had one question.
You mentioned : "Couple of things:
The point here is that square root function can not give negative result: wich means that some expression−−−−−−−−−−−−−−√≥0some expression≥0."

Can you please let me know why? Because in general square root gives two values, one positive and one negative.

$$\sqrt{...}$$ is the square root sign, a function (called the principal square root function), which cannot give negative result. So, this sign ($$\sqrt{...}$$) always means non-negative square root.

The graph of the function f(x) = √x

Notice that it's defined for non-negative numbers and is producing non-negative results.

TO SUMMARIZE:
When the GMAT provides the square root sign for an even root, such as a square root, fourth root, etc. then the only accepted answer is the non-negative root. That is:

$$\sqrt{9} = 3$$, NOT +3 or -3;
$$\sqrt[4]{16} = 2$$, NOT +2 or -2;

Notice that in contrast, the equation $$x^2 = 9$$ has TWO solutions, +3 and -3. Because $$x^2 = 9$$ means that $$x =-\sqrt{9}=-3$$ or $$x=\sqrt{9}=3$$.
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Re: Is sqrt ((x-3)^2) = 3-x?  [#permalink]

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03 Oct 2018, 02:25
gmatnub wrote:
Is $$\sqrt{(x-3)^2} = 3-x$$?

(1) $$x\neq{3}$$

(2) $$-x|x| > 0$$

Attachment:
fasdfasdfasdfasdf.JPG

This is an interesting question

1) states that x doesn’t equal 3

So it could be any number.

|x-3| = 3 - x

If x = -3 then it is equal
If x = 0 then it is equal

X = 5 then it is not equal

One side would be (5-3)^2 = 4^(1/2)= 2

The other side would be 3-5 = -2 these two aren’t equal

A is insufficient.

-x |x| > 0

Then x must be negative.

From the question stem we have |x-3| = 3 - x

This implies that any negative number will make the equation equal.

B is sufficient.

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Re: Is sqrt ((x-3)^2) = 3-x? &nbs [#permalink] 03 Oct 2018, 02:25

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