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M08 #16 - squares

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Re: M08 #16 - squares [#permalink] New post 22 Feb 2014, 08:37
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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}.

Answer: A.



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?
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Re: M08 #16 - squares [#permalink] New post 22 Feb 2014, 08:46
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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}.

Answer: A.



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}.
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Re: M08 #16 - squares [#permalink] New post 22 Feb 2014, 10:14
I get it. I assumed by mistake that (x-3)^2 > 0 implied x-3>0. Thanks!
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M08 #16 - squares [#permalink] New post 07 Aug 2014, 00:42
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}.

Answer: A.


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?
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Re: M08 #16 - squares [#permalink] New post 12 Aug 2014, 01:01
Expert's post
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}.

Answer: A.


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.
_________________

NEW TO MATH FORUM? PLEASE READ THIS: ALL YOU NEED FOR QUANT!!!

PLEASE READ AND FOLLOW: 11 Rules for Posting!!!

RESOURCES: [GMAT MATH BOOK]; 1. Triangles; 2. Polygons; 3. Coordinate Geometry; 4. Factorials; 5. Circles; 6. Number Theory; 7. Remainders; 8. Overlapping Sets; 9. PDF of Math Book; 10. Remainders; 11. GMAT Prep Software Analysis NEW!!!; 12. SEVEN SAMURAI OF 2012 (BEST DISCUSSIONS) NEW!!!; 12. Tricky questions from previous years. NEW!!!;

COLLECTION OF QUESTIONS:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS ; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.


What are GMAT Club Tests?
25 extra-hard Quant Tests

Get the best GMAT Prep Resources with GMAT Club Premium Membership

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