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gmatjon
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M06 Q2 23 Dec 2009, 14:01
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Looks simple problem, but could you elaborate on solution, thank you!

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M06 Q2 Updated on: 24 Dec 2009, 11:34
Originally posted by kp1811 on 24 Dec 2009, 07:39.
Last edited by kp1811 on 24 Dec 2009, 11:34, edited 2 times in total.
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mirzohidjon
Looks simple problem, but could you elaborate on solution, thank you!
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\(\sqrt{x^2 +6x +9}\) = \(\sqrt{(x+3)^2}\) = (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
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M06 Q2 24 Dec 2009, 09:05
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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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M06 Q2 24 Dec 2009, 11:48
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mirzohidjon
But that is the problem,
I got the same answer as you, but official answer says the solution is B

63/20
Show more


: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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M06 Q2 25 Dec 2009, 05:03
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kp1811
mirzohidjon
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
Show more

\(\sqrt{x^2}=|x|\)

\(|x+3|-|y-1|=|\frac{3}{4}+3|-|\frac{2}{5}-1|=|\frac{15}{4}|-|-\frac{3}{5}|=\frac{15}{4}-\frac{3}{5}=\frac{63}{20}\)
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M06 Q2 25 Dec 2009, 07:41
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Bunuel
kp1811
mirzohidjon
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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Thanks Bunuel for the clarification
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M06 Q2 03 Jan 2010, 19:25
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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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