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June 4th, 2011, 01:33 AM   #1
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Surds(need help)

1.Find the square root of surdic quantities

73 +40?3 =( ?x + ?y)^2
73+40?3 = x+y + 2?xy
x+y=73
y=73-x ..........(1)

2?xy=40?3
?xy=20?3
xy=1200 ........(2)

substitute (1) into (2)
x(73-x)=1200
73x^2 - x^2 =1200
x^2-73x+1200=0
then i m stuck at here,i have problem in factorize it

2.Show that ?2x-3 - ?x+2 =3?x-1 does not have any real root.

I know how to solve but i dunno how to show it does not have any real root.

3. Show that ?10+2?2 is less than 6 .

This is the provided solution
(?10+2?2)^2 = 10+ 2^2(2)+2(2)?10x2
= 18 + 4?20
= 18 + ?16x20
= 18+ ?320

(?10 + 2?2)^2 < 18 + ?324 ( i don understand how ?320 become ?324 )
< 18+18
< 36
?10 + 2?2 < 6

Thank you .
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June 4th, 2011, 01:51 AM   #2
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3. ?10 + 2?2 < 6 is trivial. Square both sides and we get 18 + 8?5 < 36, i.e. ?5 < 9/4 = ?(81/16).
?5 = ?(80/16) < ?(81/16), hence ?5 < 9/4 is correct, i.e. 18 + 8?5 < 36 is correct.
Therefore ?10 + 2?2 < 6 is correct.
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June 4th, 2011, 03:44 AM   #3
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Re: Surds(need help)

1.)





x = 25, y = 48 or x = 48, y = 25

2.) Use parentheses to clearly show what the radicands are. I will ASSUME:

















Neither of these roots are in the domain of the original equation.
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June 4th, 2011, 06:49 AM   #4
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Re: Surds(need help)

Hello, hoyy1kolko!

Quote:
\sqrt{x}\,+\,\sqrt{y})^2" />

You had the right idea!

\,+\,2\sqrt{xy}\,+\,y" />





\:=\:0 \;\;\;\Rightarrow\;\;\;x \:=\:25,\:48 \;\;\;\Rightarrow\;\;\;y \:=\:48,\:25" />




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June 4th, 2011, 06:58 AM   #5
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Re: Surds(need help)

Thank you very much,i still don understand the solution from johnny for question 3 ,i need more help. What is mean by trivial?
and isn't he use comparison to solve the question?
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June 4th, 2011, 02:59 PM   #6
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What I mean by trivial is ?10 + 2?2 ?< 6, in another words we don't know if the inequality is true.
Afterwards I did some simple mathematics to see that if it's true and then arrived at the correct conclusion.
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June 5th, 2011, 05:16 AM   #7
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Re: Surds(need help)

Thank you very much!!!!!
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