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SQ. Roots

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SQ. Roots

by rosh26 » Sun Jun 22, 2008 5:45 pm
I think i saw this solved in an earlier post but I can't seem to find it...
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by Nycgrl » Sun Jun 22, 2008 6:00 pm
Its E =20

use (a+b)^2 = a^2 +b^2+2ab
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by rosh26 » Mon Jun 23, 2008 3:08 pm
Would someone mind working this problem out? Can't seem to solve...

Thanks!
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by Nycgrl » Mon Jun 23, 2008 7:11 pm
Sqrt(9+srq80) + sqt(9-sqrt80)}^2

This is (a+b)^2 = a^2 +b^2+2ab

[sqrt(9+sqrt80)]^2 +[sqrt(9-sqt 80)]^2 - 2*{sqrt(9+sqt80)*sqt(9-sqt80)

=> 9+sqt80 +9- sqt80 - 2
=20

I hope its clear ,its hard to explain without using proper symbols.
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by ksh » Mon Jun 23, 2008 9:40 pm
Hi,

I think nycgrl has done some basic mistakes although answer quoted is correct.

It may be seen that 9+9-2=/20 but,
it would be 9+9+2Sqrt1=20
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by Nycgrl » Tue Jun 24, 2008 3:02 am
Sorry I missed writing sqt......
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by Xia_devil » Tue Jun 24, 2008 3:17 am
Sorry but how is

sqrt(9+sqrt(80)) * sqrt(9- sqrt(80)) =1 ?
With regards,
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by rkotha » Tue Jun 24, 2008 7:50 am
sqrt(9+sqrt(80)) * sqrt(9- sqrt(80)) =1 ?
= Sqrt (9^2 - [sqrt 80]^2); since it is (a+b)*(a-b)=a^2-b^2
= Sqrt(81-80)
=sqrt(1)
=1
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