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# What is (x^2 - 6x + 9)^(1/2)/(x - 3) + (-x)^(1/2) if each term in this

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What is (x^2 - 6x + 9)^(1/2)/(x - 3) + (-x)^(1/2) if each term in this [#permalink]

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02 Jun 2015, 23:43
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What is $$\frac{\sqrt{x^2 -6x+9}}{x-3}+\sqrt{-x}$$ if each term in this expression is correctly defined?

A. $$\sqrt{-x}$$
B. $$1 + \sqrt{x}$$
C. $$1 + \sqrt{-x}$$
D. $$-1 + \sqrt{x}$$
E. $$-1 + \sqrt{-x}$$

Source: OptimusPrep
[Reveal] Spoiler: OA

Last edited by Bunuel on 03 Jun 2015, 02:48, edited 1 time in total.
Renamed the topic and edited the question.

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What is (x^2 - 6x + 9)^(1/2)/(x - 3) + (-x)^(1/2) if each term in this [#permalink]

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03 Jun 2015, 02:55
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naeln wrote:
What is $$\frac{\sqrt{x^2 -6x+9}}{x-3}+\sqrt{-x}$$ if each term in this expression is correctly defined?

A. $$\sqrt{-x}$$
B. $$1 + \sqrt{x}$$
C. $$1 + \sqrt{-x}$$
D. $$-1 + \sqrt{x}$$
E. $$-1 + \sqrt{-x}$$

Source: OptimusPrep

STEP 1:
Even roots from a negative numbers are not defined, thus $$\sqrt{-x}$$ to be defined -x must be more than or equal to 0: $$-x\geq{0}$$. Therefore $$x\leq{0}$$.

STEP 2:
Recall that $$\sqrt{x^2}=|x|$$.

$$\sqrt{x^2 -6x+9}=\sqrt{(x-3)^2}=|x-3|$$. Since from above $$x\leq{0}$$, then x-3 is negative and thus |x - 3| = -(x - 3).

STEP 3:
$$\frac{\sqrt{x^2 -6x+9}}{x-3}+\sqrt{-x}=\frac{-(x-3)}{x-3} +\sqrt{-x}=-1+\sqrt{-x}$$.

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Re: What is (x^2 - 6x + 9)^(1/2)/(x - 3) + (-x)^(1/2) if each term in this [#permalink]

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03 Jun 2015, 03:15
Bunuel wrote:
naeln wrote:
What is $$\frac{\sqrt{x^2 -6x+9}}{x-3}+\sqrt{-x}$$ if each term in this expression is correctly defined?

A. $$\sqrt{-x}$$
B. $$1 + \sqrt{x}$$
C. $$1 + \sqrt{-x}$$
D. $$-1 + \sqrt{x}$$
E. $$-1 + \sqrt{-x}$$

Source: OptimusPrep

STEP 1:
Even roots from a negative numbers are not defined, thus $$\sqrt{-x}$$ to be defined -x must be more than or equal to 0: $$-x\geq{0}$$. Therefore $$x\leq{0}$$.

STEP 2:
Recall that $$\sqrt{x^2}=|x|$$.

$$\sqrt{x^2 -6x+9}=\sqrt{(x-3)^2}=|x-3|$$. Since from above $$x\leq{0}$$, then x-3 is negative and |x - 3| = -(x - 3).

STEP 3:
$$\frac{\sqrt{x^2 -6x+9}}{x-3}+\sqrt{-x}=\frac{-(x-3)}{x-3} +\sqrt{-x}=-1+\sqrt{-x}$$.

Thank you Bunuel for the clear explanation. Love it

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Re: What is (x^2 - 6x + 9)^(1/2)/(x - 3) + (-x)^(1/2) if each term in this [#permalink]

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03 Jun 2015, 03:55
√-x is defined means -x is positive hence x is negative.
For the first term if we simplify top term it becomes √(x-3)^2 which is equal to x-3 or 3-x. Since only positive quantity comes out of sq root 3-x, which is positive comes out and cancels with denominator to give -1.

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What is (x^2 - 6x + 9)^(1/2)/(x - 3) + (-x)^(1/2) if each term in this [#permalink]

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03 Jun 2015, 06:06
naeln wrote:
What is $$\frac{\sqrt{x^2 -6x+9}}{x-3}+\sqrt{-x}$$ if each term in this expression is correctly defined?

A. $$\sqrt{-x}$$
B. $$1 + \sqrt{x}$$
C. $$1 + \sqrt{-x}$$
D. $$-1 + \sqrt{x}$$
E. $$-1 + \sqrt{-x}$$

Source: OptimusPrep

At first, There has to be an observation to be made

Since each term is correctly defined in the expression
$$\frac{\sqrt{x^2 -6x+9}}{x-3}+\sqrt{-x}$$

therefore we must realize that $$\sqrt{-x}$$ should be a real number which means that x MUST be a Negative Number

Therefore, (x-3) MUST be Negative

therefore, (x-3) = $$-\sqrt{(x-3)^2}$$ = $$-\sqrt{x^2 -6x+9}$$

Now, Expression, $$\frac{\sqrt{x^2 -6x+9}}{x-3}$$ can be written here as

$$\frac{\sqrt{x^2 -6x+9}}{-\sqrt{x^2 -6x+9}}$$ i.e. equal to (-1)

Hence the result becomes

$$-1+\sqrt{-x}$$

[Reveal] Spoiler:
E

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Re: What is (x^2 - 6x + 9)^(1/2)/(x - 3) + (-x)^(1/2) if each term in this [#permalink]

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03 Jun 2015, 06:18
naeln wrote:
What is $$\frac{\sqrt{x^2 -6x+9}}{x-3}+\sqrt{-x}$$ if each term in this expression is correctly defined?

A. $$\sqrt{-x}$$
B. $$1 + \sqrt{x}$$
C. $$1 + \sqrt{-x}$$
D. $$-1 + \sqrt{x}$$
E. $$-1 + \sqrt{-x}$$

Source: OptimusPrep

ALTERNATE METHOD:

SUBSTITUTION of a correct numbers works very well in such cases but there has to be an observation to be made

Since each term is correctly defined in the expression
$$\frac{\sqrt{x^2 -6x+9}}{x-3}+\sqrt{-x}$$

therefore we must realize that $$\sqrt{-x}$$ should be a real number which means that x MUST be a Negative Number

So let's take X = -4 [I am not taking X = -1 because that may confuse us among options]

Now, $$\frac{\sqrt{x^2 -6x+9}}{x-3}+\sqrt{-x}$$ = $$\frac{\sqrt{(-4)^2 -6(-4)+9}}{(-4)-3}+\sqrt{-(-4)}$$

= $$\frac{\sqrt{16 + 24+9}}{-4-3}+\sqrt{(4)}$$
= $$\frac{\sqrt{49}}{-7}+2$$
= $$\frac{7}{-7}+2$$
= $$-1+2$$
= $$1$$

Check Options By Substituting $$X=-4$$

A. $$\sqrt{-x}$$ = 2 INCORRECT
B. $$1 + \sqrt{x}$$ = NOT REAL NUMBER therefore INCORRECT
C. $$1 + \sqrt{-x}$$ = $$1+2$$ = $$3$$ INCORRECT
D. $$-1 + \sqrt{x}$$ = NOT REAL NUMBER therefore INCORRECT
E. $$-1 + \sqrt{-x}$$ = $$-1 + 2$$ = $$+1$$ CORRECT

[Reveal] Spoiler:
E

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Re: What is (x^2 - 6x + 9)^(1/2)/(x - 3) + (-x)^(1/2) if each term in this [#permalink]

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18 Dec 2015, 18:34
i thought it is way to easy to be 1+sqrt(-x)
if we have sqrt(-x), it must be that x is either 0 or negative.
if x is negative, then x-3 is negative, thus |x-3| = -(x-3) or -1(x-3).
now, -1+sqrt(-x) is E, which should be the correct answer.

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Re: What is (x^2 - 6x + 9)^(1/2)/(x - 3) + (-x)^(1/2) if each term in this [#permalink]

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21 May 2016, 16:34
Bunuel wrote:
naeln wrote:
What is $$\frac{\sqrt{x^2 -6x+9}}{x-3}+\sqrt{-x}$$ if each term in this expression is correctly defined?

A. $$\sqrt{-x}$$
B. $$1 + \sqrt{x}$$
C. $$1 + \sqrt{-x}$$
D. $$-1 + \sqrt{x}$$
E. $$-1 + \sqrt{-x}$$

Source: OptimusPrep

STEP 1:
Even roots from a negative numbers are not defined, thus $$\sqrt{-x}$$ to be defined -x must be more than or equal to 0: $$-x\geq{0}$$. Therefore $$x\leq{0}$$.

STEP 2:
Recall that $$\sqrt{x^2}=|x|$$.

$$\sqrt{x^2 -6x+9}=\sqrt{(x-3)^2}=|x-3|$$. Since from above $$x\leq{0}$$, then x-3 is negative and thus |x - 3| = -(x - 3).

STEP 3:
$$\frac{\sqrt{x^2 -6x+9}}{x-3}+\sqrt{-x}=\frac{-(x-3)}{x-3} +\sqrt{-x}=-1+\sqrt{-x}$$.

Hi Brunel - why wouldn't the answer be D, since X <= 0, meaning why wouldn't sqrt(-x) become sqrt(x)?

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Re: What is (x^2 - 6x + 9)^(1/2)/(x - 3) + (-x)^(1/2) if each term in this [#permalink]

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21 May 2016, 20:33
1
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Expert's post
mdacosta wrote:
Bunuel wrote:
naeln wrote:
What is $$\frac{\sqrt{x^2 -6x+9}}{x-3}+\sqrt{-x}$$ if each term in this expression is correctly defined?

A. $$\sqrt{-x}$$
B. $$1 + \sqrt{x}$$
C. $$1 + \sqrt{-x}$$
D. $$-1 + \sqrt{x}$$
E. $$-1 + \sqrt{-x}$$

Source: OptimusPrep

STEP 1:
Even roots from a negative numbers are not defined, thus $$\sqrt{-x}$$ to be defined -x must be more than or equal to 0: $$-x\geq{0}$$. Therefore $$x\leq{0}$$.

STEP 2:
Recall that $$\sqrt{x^2}=|x|$$.

$$\sqrt{x^2 -6x+9}=\sqrt{(x-3)^2}=|x-3|$$. Since from above $$x\leq{0}$$, then x-3 is negative and thus |x - 3| = -(x - 3).

STEP 3:
$$\frac{\sqrt{x^2 -6x+9}}{x-3}+\sqrt{-x}=\frac{-(x-3)}{x-3} +\sqrt{-x}=-1+\sqrt{-x}$$.

Hi Brunel - why wouldn't the answer be D, since X <= 0, meaning why wouldn't sqrt(-x) become sqrt(x)?

Hi,

D is $$-1 + \sqrt{x}$$
so $$\sqrt{x}$$ means square root of some negative integers, and this is not defined in GMAT so this is out

what you are talking of is present in E
E. $$-1 + \sqrt{-x}$$
$$\sqrt{-x}$$.. since x<=0, the square root turns into that of +ive number..
so E is correct
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Absolute modulus :http://gmatclub.com/forum/absolute-modulus-a-better-understanding-210849.html#p1622372
Combination of similar and dissimilar things : http://gmatclub.com/forum/topic215915.html

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Re: What is (x^2 - 6x + 9)^(1/2)/(x - 3) + (-x)^(1/2) if each term in this [#permalink]

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20 Jun 2016, 06:59
Hi Bunuel,
I have completed the Manhattan GMAT Algebra Guide and it says that sqrt(y), where y>=0, is positive only; while for the expression x^2= y, where y>=0, is mod(y), i.e. y can be positive or negative.

Now you have solved the first term of this expression by considering both positive and negative values.
This is confusing me.

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Re: What is (x^2 - 6x + 9)^(1/2)/(x - 3) + (-x)^(1/2) if each term in this [#permalink]

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20 Jun 2016, 07:07
oloz wrote:
Hi Bunuel,
I have completed the Manhattan GMAT Algebra Guide and it says that sqrt(y), where y>=0, is positive only; while for the expression x^2= y, where y>=0, is mod(y), i.e. y can be positive or negative.

Now you have solved the first term of this expression by considering both positive and negative values.
This is confusing me.

That's not true.

We got that $$\sqrt{x^2 -6x+9}=\sqrt{(x-3)^2}=|x-3|$$. |x-3|, on the other hand due to the fact that $$x\leq{0}$$, equals to -(x - 3), which is positive.
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What is (x^2 - 6x + 9)^(1/2)/(x - 3) + (-x)^(1/2) if each term in this [#permalink]

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21 Jun 2016, 05:40
Bunuel wrote:
naeln wrote:
What is $$\frac{\sqrt{x^2 -6x+9}}{x-3}+\sqrt{-x}$$ if each term in this expression is correctly defined?

A. $$\sqrt{-x}$$
B. $$1 + \sqrt{x}$$
C. $$1 + \sqrt{-x}$$
D. $$-1 + \sqrt{x}$$
E. $$-1 + \sqrt{-x}$$

Source: OptimusPrep

STEP 1:
Even roots from a negative numbers are not defined, thus $$\sqrt{-x}$$ to be defined -x must be more than or equal to 0: $$-x\geq{0}$$. Therefore $$x\leq{0}$$.

STEP 2:
Recall that $$\sqrt{x^2}=|x|$$.

$$\sqrt{x^2 -6x+9}=\sqrt{(x-3)^2}=|x-3|$$. Since from above $$x\leq{0}$$, then x-3 is negative and thus |x - 3| = -(x - 3).

STEP 3:
$$\frac{\sqrt{x^2 -6x+9}}{x-3}+\sqrt{-x}=\frac{-(x-3)}{x-3} +\sqrt{-x}=-1+\sqrt{-x}$$.

I got this explanation fully except for the underlined part. x is negative, but |x-3|should still be positive. I dont understand how we get -(x-3) from |x-3|. Can someone please help me understand this part? 'll be much obliged. Thanks.

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Re: What is (x^2 - 6x + 9)^(1/2)/(x - 3) + (-x)^(1/2) if each term in this [#permalink]

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Re: What is (x^2 - 6x + 9)^(1/2)/(x - 3) + (-x)^(1/2) if each term in this [#permalink]

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26 Sep 2017, 20:55
iqbalfiery wrote:
Bunuel wrote:
naeln wrote:
What is $$\frac{\sqrt{x^2 -6x+9}}{x-3}+\sqrt{-x}$$ if each term in this expression is correctly defined?

A. $$\sqrt{-x}$$
B. $$1 + \sqrt{x}$$
C. $$1 + \sqrt{-x}$$
D. $$-1 + \sqrt{x}$$
E. $$-1 + \sqrt{-x}$$

Source: OptimusPrep

STEP 1:
Even roots from a negative numbers are not defined, thus $$\sqrt{-x}$$ to be defined -x must be more than or equal to 0: $$-x\geq{0}$$. Therefore $$x\leq{0}$$.

STEP 2:
Recall that $$\sqrt{x^2}=|x|$$.

$$\sqrt{x^2 -6x+9}=\sqrt{(x-3)^2}=|x-3|$$. Since from above $$x\leq{0}$$, then x-3 is negative and thus |x - 3| = -(x - 3).

STEP 3:
$$\frac{\sqrt{x^2 -6x+9}}{x-3}+\sqrt{-x}=\frac{-(x-3)}{x-3} +\sqrt{-x}=-1+\sqrt{-x}$$.

I got this explanation fully except for the underlined part. x is negative, but |x-3|should still be positive. I dont understand how we get -(x-3) from |x-3|. Can someone please help me understand this part? 'll be much obliged. Thanks.

Modulus is always positive ie Ix-3I has to be positive and is also=(x-3) as per above.
Now we have also derived that x<=0 ie negative or 0.
on a number line, we have ........-1......0......+1........
let us say x is -1, therefore, we have (-1-3)=-4 which is equal to Ix-3I which is not possible since modulus has to be +ve.
Therefore we put -sign in the front of (x-3) to make it positive ie -(-1-3)=-(-4)=+4
This also works if we take x=0.

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Re: What is (x^2 - 6x + 9)^(1/2)/(x - 3) + (-x)^(1/2) if each term in this [#permalink]

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02 Oct 2017, 00:46
Bunuel
Isn't that if something is under even root on gmat, we only consider the positive solution?

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Re: What is (x^2 - 6x + 9)^(1/2)/(x - 3) + (-x)^(1/2) if each term in this [#permalink]

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02 Oct 2017, 00:48
Bunuel

I am struggling with when to consider both positive and negative solutions and when to consider only the positive solution. Can you please help me on this?

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Re: What is (x^2 - 6x + 9)^(1/2)/(x - 3) + (-x)^(1/2) if each term in this [#permalink]

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02 Oct 2017, 00:48
kaleem765 wrote:
Bunuel
Isn't that if something is under even root on gmat, we only consider the positive solution?

Yes. Even roots from negative numbers are not defined on the GMAT.
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Re: What is (x^2 - 6x + 9)^(1/2)/(x - 3) + (-x)^(1/2) if each term in this [#permalink]

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02 Oct 2017, 00:53
Bunuel wrote:
kaleem765 wrote:
Bunuel
Isn't that if something is under even root on gmat, we only consider the positive solution?

Yes. Even roots from negative numbers are not defined on the GMAT.

Then why did we consider -(x-3) as a solution for \sqrt{x^2-6x+9}?

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Re: What is (x^2 - 6x + 9)^(1/2)/(x - 3) + (-x)^(1/2) if each term in this [#permalink]

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02 Oct 2017, 00:59
kaleem765 wrote:
Bunuel wrote:
kaleem765 wrote:
Bunuel
Isn't that if something is under even root on gmat, we only consider the positive solution?

Yes. Even roots from negative numbers are not defined on the GMAT.

Then why did we consider -(x-3) as a solution for \sqrt{x^2-6x+9}?

Since we got that x must be less than or equal to 0, then -(x - 3) is positive not negative.
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Re: What is (x^2 - 6x + 9)^(1/2)/(x - 3) + (-x)^(1/2) if each term in this [#permalink]

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02 Oct 2017, 01:04
Thanks Bunuel. I got it now.

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Re: What is (x^2 - 6x + 9)^(1/2)/(x - 3) + (-x)^(1/2) if each term in this   [#permalink] 02 Oct 2017, 01:04
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