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# Square root

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Senior Manager
Joined: 20 Mar 2008
Posts: 457
Followers: 1

Kudos [?]: 54 [0], given: 5

If each expression under the square root is greater than or [#permalink]  04 Sep 2009, 17:33
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0% (00:00) correct 0% (00:00) wrong based on 0 sessions
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 ?

* \sqrt{2 - x}
* 2x - 6 + \sqrt{2 - x}
* \sqrt{2 - x} + {x - 3}
* 2x - 6 + \sqrt{x - 2}
* x + \sqrt{x - 2}
Manager
Joined: 25 Aug 2009
Posts: 180
Followers: 1

Kudos [?]: 54 [0], given: 12

Re: Square root [#permalink]  04 Sep 2009, 18:39
Given, all expressions under square roots are grater than 0.

=> 2 - x > 0
=> x < 2

Now, expression is \sqrt{(x-3)^2} + \sqrt{(2-x)} + x-3

=> |x-3| + \sqrt{(2-x)} + x-3
=> -(x-3) + \sqrt{(2-x)} + x-3 ---- x < 2 , hence, modulus will open with negative sign
=> 3 - x + \sqrt{(2-x)} + x-3
=> \sqrt{(2-x)}
Ans A
Senior Manager
Joined: 20 Mar 2008
Posts: 457
Followers: 1

Kudos [?]: 54 [0], given: 5

Re: Square root [#permalink]  04 Sep 2009, 19:26
gmate2010 wrote:
Given, all expressions under square roots are grater than 0.

=> 2 - x > 0
=> x < 2

Now, expression is \sqrt{(x-3)^2} + \sqrt{(2-x)} + x-3

=> |x-3| + \sqrt{(2-x)} + x-3
=> -(x-3) + \sqrt{(2-x)} + x-3 ---- x < 2 , hence, modulus will open with negative sign
=> 3 - x + \sqrt{(2-x)} + x-3
=> \sqrt{(2-x)}
Ans A

I don't get the bold part?
Manager
Joined: 30 Aug 2009
Posts: 65
Location: LA
Followers: 2

Kudos [?]: 9 [0], given: 7

Re: Square root [#permalink]  04 Sep 2009, 20:56
gmate2010:

Why are you taking the modulus |x-3|, why not the positive root (x-3)?
Can you please explain? Thank you.
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Manager
Joined: 25 Aug 2009
Posts: 180
Followers: 1

Kudos [?]: 54 [0], given: 12

Re: Square root [#permalink]  05 Sep 2009, 07:08
sharkk wrote:
gmate2010:

Why are you taking the modulus |x-3|, why not the positive root (x-3)?
Can you please explain? Thank you.

\sqrt{X^2} is always equal to |x|.

We can not write \sqrt{X^2} equal to x unless and until x > 0
property of |x|:-
|x| = x for x > 0 and -x for x < 0

here x-3 < 0 because x < 2 , |x-3| = -(x-3) = 3 - x
Manager
Joined: 22 Aug 2009
Posts: 100
Location: UK
Followers: 2

Kudos [?]: 12 [0], given: 25

Re: Square root [#permalink]  05 Sep 2009, 17:26

Jivana I assume the last (x-3 ) is not under the square root

- solving the expression under 1st square roots gives (x-3 ) ( x-3)
- take the square root and you have (x-3)
- simply the expression and you wil get (B)

note : how do u guys draw that square root and write square ..am not able to explain, as I cant type tat .
Edit:

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Re: Square root   [#permalink] 05 Sep 2009, 17:26
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