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Senior Manager
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Looks simple problem, but could you elaborate on solution, thank you!
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Senior Manager
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mirzohidjon wrote: Looks simple problem, but could you elaborate on solution, thank you! \sqrt{x^2 +6x +9} = \sqrt{(x+3)^2} = (x+3) \sqrt{y^2 -2y +1} = \sqrt{(y-1)^2} = (y-1) so the expression become (x+3) - (y-1) = x-y+4 = 3/4 -2/5 + 4 = (15 - 8 + 80)/20 = 87/20
Last edited by kp1811 on 24 Dec 2009, 11:34, edited 2 times in total.
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Senior Manager
Joined: 18 Aug 2009
Posts: 441
Schools: UT at Austin, Indiana State University, UC at Berkeley
WE 1: 5.5
WE 2: 5.5
WE 3: 6.0
Followers: 4
Kudos [?]:
37
[0], given: 16
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But that is the problem, I got the same answer as you, but official answer says the solution is B 63/20
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Senior Manager
Joined: 30 Aug 2009
Posts: 296
Location: India
Concentration: General Management
Followers: 2
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mirzohidjon wrote: But that is the problem,
I got the same answer as you, but official answer says the solution is B
63/20  well well well .....if we dont simplify the expression then we get 63/20 \sqrt{x^2 + 6x + 9} = \sqrt{(9/16 + 18/4 + 9)} = \sqrt{(9+72+144)/16} = 15/4 \sqrt{y^2 -2y +1} = \sqrt{4/25 - 4/5 + 1} = \sqrt{(4 - 20 + 25)/25} = 3/5 so we have 15/4 - 3/5 = (75 - 12)/ 20 = 63/20 Can someone please let me know what is wrong with the solution provided by simplifying the square roots??
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Senior Manager
Joined: 30 Aug 2009
Posts: 296
Location: India
Concentration: General Management
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Bunuel wrote: kp1811 wrote: mirzohidjon wrote: Looks simple problem, but could you elaborate on solution, thank you! \sqrt{x^2 +6x +9} = \sqrt{(x+3)^2} = (x+3) \sqrt{y^2 -2y +1} = \sqrt{(y-1)^2} = (y-1)so the expression become (x+3) - (y-1) = x-y+4 = 3/4 -2/5 + 4 = (15 - 8 + 80)/20 = 87/20 Thanks Bunuel for the clarification
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Senior Manager
Joined: 18 Aug 2009
Posts: 441
Schools: UT at Austin, Indiana State University, UC at Berkeley
WE 1: 5.5
WE 2: 5.5
WE 3: 6.0
Followers: 4
Kudos [?]:
37
[0], given: 16
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I assumed that square root sign does not mean that it is module, from now on i will know. thank u for clarification
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