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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

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).

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, 02:15

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).

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, 02:55

√-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.

Prosper!!! GMATinsight Bhoopendra Singh and Dr.Sushma Jha e-mail: info@GMATinsight.com I Call us : +91-9999687183 / 9891333772 Online One-on-One Skype based classes and Classroom Coaching in South and West Delhi http://www.GMATinsight.com/testimonials.html

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

Prosper!!! GMATinsight Bhoopendra Singh and Dr.Sushma Jha e-mail: info@GMATinsight.com I Call us : +91-9999687183 / 9891333772 Online One-on-One Skype based classes and Classroom Coaching in South and West Delhi http://www.GMATinsight.com/testimonials.html

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, 17:34

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.

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, 15:34

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).

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).

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
_________________

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, 05:59

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.

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.

Please, provide some insight.

Thanks in advance.

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.
_________________

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, 04:40

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).

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.

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, 19:55

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).

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.

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

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31 Oct 2017, 02:06

I'm sorry - I still don't get it. Quant is definitely my weak point, but I simply cannot understand how we get to -(x-3)... I'd hugely appreciate anyone spelling it out for me in layman's terms (as much as that is possible) - or pointing me to a relevant article where I can get into it there.