What is (x^2 - 6x + 9)^(1/2)/(x - 3) + (-x)^(1/2) if each term in this : Problem Solving (PS)

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03 Jun 2015, 03:55

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√-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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03 Jun 2015, 06:06

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naeln

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

Hi Readers,

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

\(-1+\sqrt{-x}\)

Answer: Option

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03 Jun 2015, 06:18

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naeln

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:

Hi Readers,

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

Answer: Option

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E

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18 Dec 2015, 18:34

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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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21 May 2016, 16:34

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Bunuel

naeln

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

Answer: E.

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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21 May 2016, 20:33

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mdacosta

Bunuel

naeln

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

Answer: E.

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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20 Jun 2016, 06:59

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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.

Bunuel

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20 Jun 2016, 07:07

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oloz

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.

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21 Jun 2016, 05:40

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Bunuel

naeln

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

Answer: E.

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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26 Sep 2017, 20:55

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iqbalfiery

Bunuel

naeln

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

Answer: E.

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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Bunuel
Isn't that if something is under even root on gmat, we only consider the positive solution?

Bunuel

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02 Oct 2017, 00:48

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kaleem765

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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Bunuel

kaleem765

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}?

Bunuel

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02 Oct 2017, 00:59

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kaleem765

Bunuel

kaleem765

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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02 Oct 2017, 01:04

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Thanks Bunuel. I got it now.

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31 Oct 2017, 03:06

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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.

Many thanks!

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01 Jul 2020, 11:58

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Since there is a minus sign inside the root x must be negative or zero.
(√(x2−6x+)/(x-3))+√−x
=(√(x-3)^2/(x-3))+√-x
=((3-x)/(x-3))+√-x
=-1+√x

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

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