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M06 Q2

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M06 Q2 [#permalink] New post 23 Dec 2009, 13:01
Looks simple problem, but could you elaborate on solution, thank you!
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Re: M06 Q2 [#permalink] New post 24 Dec 2009, 06:39
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, 10:34, edited 2 times in total.
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Re: M06 Q2 [#permalink] New post 24 Dec 2009, 08:05
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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Re: M06 Q2 [#permalink] New post 24 Dec 2009, 10:48
mirzohidjon wrote:
But that is the problem,
I got the same answer as you, but official answer says the solution is B

63/20



:shock:
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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Re: M06 Q2 [#permalink] New post 25 Dec 2009, 04:03
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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


\sqrt{x^2}=|x|

|x+3|-|y-1|=|\frac{3}{4}+3|-|\frac{2}{5}-1|=|\frac{15}{4}|-|-\frac{3}{5}|=\frac{15}{4}-\frac{3}{5}=\frac{63}{20}
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Re: M06 Q2 [#permalink] New post 25 Dec 2009, 06:41
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


\sqrt{x^2}=|x|

|x+3|-|y-1|=|\frac{3}{4}+3|-|\frac{2}{5}-1|=|\frac{15}{4}|-|-\frac{3}{5}|=\frac{15}{4}-\frac{3}{5}=\frac{63}{20}


Thanks Bunuel for the clarification
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Re: M06 Q2 [#permalink] New post 03 Jan 2010, 18:25
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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Re: M06 Q2   [#permalink] 03 Jan 2010, 18:25
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