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# M08 #16 - squares

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Intern
Joined: 20 Jan 2013
Posts: 47

Kudos [?]: 43 [0], given: 70

Location: India
Concentration: Finance, Entrepreneurship
GMAT 1: 740 Q49 V41
GPA: 3.42
Re: M08 #16 - squares [#permalink]

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22 Feb 2014, 09:37
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Bunuel wrote:
ConkergMat wrote:
If each expression under the square root is greater than or equal to 0, what is $$\sqrt{x^2 - 6x + 9} + \sqrt{2 - x} + x - 3$$?

(A) $$\sqrt{2-x}$$
(B) $$2x-6 + \sqrt{2-x}$$
(C) $$\sqrt{2-x} + x-3$$
(D) $$2x-6 + \sqrt{x-2}$$
(E) $$x + \sqrt{x-2}$$

[Reveal] Spoiler: OA
A

Source: GMAT Club Tests - hardest GMAT questions

One important note: $$\sqrt{x^2}=|x|$$

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

Now, as the expressions under the square roots are more than or equal to zero than $$2-x>0$$ --> $$x<2$$. Next: as $$x<2$$ then $$|x-3|$$ becomes $$|x-3|=-(x-3)=-x+3$$.

$$|x-3|+\sqrt{2-x}+x-3=-x+3+\sqrt{2-x}+x-3=\sqrt{2-x}$$.

But the questions says that EACH expression under root is greater than or equal to 0, can't we then directly assume the first expression sqrt{x^2 - 6x + 9} to also be greater than 0?
_________________

http://gmatclub.com/forum/collection-of-the-best-gmat-resources-167295.html#p1329720

May we all emerge victorious

Kudos [?]: 43 [0], given: 70

Math Expert
Joined: 02 Sep 2009
Posts: 42249

Kudos [?]: 132626 [1], given: 12326

Re: M08 #16 - squares [#permalink]

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22 Feb 2014, 09:46
1
KUDOS
Expert's post
anindame wrote:
Bunuel wrote:
ConkergMat wrote:
If each expression under the square root is greater than or equal to 0, what is $$\sqrt{x^2 - 6x + 9} + \sqrt{2 - x} + x - 3$$?

(A) $$\sqrt{2-x}$$
(B) $$2x-6 + \sqrt{2-x}$$
(C) $$\sqrt{2-x} + x-3$$
(D) $$2x-6 + \sqrt{x-2}$$
(E) $$x + \sqrt{x-2}$$

[Reveal] Spoiler: OA
A

Source: GMAT Club Tests - hardest GMAT questions

One important note: $$\sqrt{x^2}=|x|$$

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

Now, as the expressions under the square roots are more than or equal to zero than $$2-x>0$$ --> $$x<2$$. Next: as $$x<2$$ then $$|x-3|$$ becomes $$|x-3|=-(x-3)=-x+3$$.

$$|x-3|+\sqrt{2-x}+x-3=-x+3+\sqrt{2-x}+x-3=\sqrt{2-x}$$.

But the questions says that EACH expression under root is greater than or equal to 0, can't we then directly assume the first expression sqrt{x^2 - 6x + 9} to also be greater than 0?

$$\sqrt{x^2 - 6x + 9}=\sqrt{(x-3)^2}$$. Yes, the expression under this square root is also greater than 0, but $$(x-3)^2>0$$ just means that $$x\neq{3}$$.
_________________

Kudos [?]: 132626 [1], given: 12326

Intern
Joined: 20 Jan 2013
Posts: 47

Kudos [?]: 43 [0], given: 70

Location: India
Concentration: Finance, Entrepreneurship
GMAT 1: 740 Q49 V41
GPA: 3.42
Re: M08 #16 - squares [#permalink]

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22 Feb 2014, 11:14
I get it. I assumed by mistake that (x-3)^2 > 0 implied x-3>0. Thanks!
_________________

http://gmatclub.com/forum/collection-of-the-best-gmat-resources-167295.html#p1329720

May we all emerge victorious

Kudos [?]: 43 [0], given: 70

Intern
Joined: 21 Oct 2012
Posts: 36

Kudos [?]: 18 [0], given: 19

Location: United States
Concentration: Marketing, Operations
GMAT 1: 650 Q44 V35
GMAT 2: 600 Q47 V26
GMAT 3: 660 Q43 V38
GPA: 3.6
WE: Information Technology (Computer Software)
M08 #16 - squares [#permalink]

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07 Aug 2014, 01:42
Bunuel wrote:
ConkergMat wrote:
If each expression under the square root is greater than or equal to 0, what is $$\sqrt{x^2 - 6x + 9} + \sqrt{2 - x} + x - 3$$?

(A) $$\sqrt{2-x}$$
(B) $$2x-6 + \sqrt{2-x}$$
(C) $$\sqrt{2-x} + x-3$$
(D) $$2x-6 + \sqrt{x-2}$$
(E) $$x + \sqrt{x-2}$$

[Reveal] Spoiler: OA
A

Source: GMAT Club Tests - hardest GMAT questions

One important note: $$\sqrt{x^2}=|x|$$

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

Now, as the expressions under the square roots are more than or equal to zero than $$2-x>0$$ --> $$x<2$$. Next: as $$x<2$$ then $$|x-3|$$ becomes $$|x-3|=-(x-3)=-x+3$$.

$$|x-3|+\sqrt{2-x}+x-3=-x+3+\sqrt{2-x}+x-3=\sqrt{2-x}$$.

Hi Bunuel could you please explain to me how x<2 ---> |x-3|= -(x-3)?

Also could we follow the method of plugging in values for this question as in:

Suppose choose the value x=1 which satisfies the condition
Hence the given becomes 2+1+1-3=1
1.1
2.3.4.5.not equal

is this method acceptable too?

Kudos [?]: 18 [0], given: 19

Math Expert
Joined: 02 Sep 2009
Posts: 42249

Kudos [?]: 132626 [0], given: 12326

Re: M08 #16 - squares [#permalink]

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12 Aug 2014, 02:01
havoc7860 wrote:
Bunuel wrote:
ConkergMat wrote:
If each expression under the square root is greater than or equal to 0, what is $$\sqrt{x^2 - 6x + 9} + \sqrt{2 - x} + x - 3$$?

(A) $$\sqrt{2-x}$$
(B) $$2x-6 + \sqrt{2-x}$$
(C) $$\sqrt{2-x} + x-3$$
(D) $$2x-6 + \sqrt{x-2}$$
(E) $$x + \sqrt{x-2}$$

[Reveal] Spoiler: OA
A

Source: GMAT Club Tests - hardest GMAT questions

One important note: $$\sqrt{x^2}=|x|$$

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

Now, as the expressions under the square roots are more than or equal to zero than $$2-x>0$$ --> $$x<2$$. Next: as $$x<2$$ then $$|x-3|$$ becomes $$|x-3|=-(x-3)=-x+3$$.

$$|x-3|+\sqrt{2-x}+x-3=-x+3+\sqrt{2-x}+x-3=\sqrt{2-x}$$.

Hi Bunuel could you please explain to me how x<2 ---> |x-3|= -(x-3)?

Also could we follow the method of plugging in values for this question as in:

Suppose choose the value x=1 which satisfies the condition
Hence the given becomes 2+1+1-3=1
1.1
2.3.4.5.not equal

is this method acceptable too?

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: $$|5|=5$$.

BACK TO THE QUESTION:
|x-3|=x-3, when x>3.
|x-3|=-(x-3), when x<3.

Since we know that x<2, then we have the second case (when x<2, x-3<0, thus |x-3|=-(x-3)).

Hope it's clear.
_________________

Kudos [?]: 132626 [0], given: 12326

Re: M08 #16 - squares   [#permalink] 12 Aug 2014, 02:01

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# M08 #16 - squares

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