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help! tricky exponents question

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by b.kurland » Mon Jul 28, 2008 7:46 pm
(#(9+(#80))+#(9-(#80)))^2

#=square root

so it reads "(square root of (9 plus square root of 80) + square root of (9 minus square root of 80)) squared"

A=20

can anyone explain how to do this problem?
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by sudhir3127 » Mon Jul 28, 2008 8:37 pm
My answer is 20.

U need it understand that its in the form of (a+b)^2 = a^2+b^2+2ab

once u get it ur done!!!

sq root( 9+sqrt80)^2 + sq root( 9-sqrt80)^2 + 2sq root( 9+sqrt80)*sq root( 9-sqrt80,,,( a^2+b^2+2ab)

solve this u will get

sqroot ans square get cancelled...
9+sqrt80+9-sqrt80 +2 ( 81- 9sqrt80+9sqrt80-80)

18 +2( sqroot 1)

hence 20

do let me know if u still have a doubt.
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by ricky » Tue Jul 29, 2008 2:46 am
CAN u plz give all answer choices?
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by b.kurland » Tue Jul 29, 2008 8:10 am
the answer choices are:

A. 1
B. 9-(4#5)
C. 18-(4#5)
D. 18
E. 20
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by b.kurland » Tue Jul 29, 2008 8:24 am
sudhir3127 wrote:My answer is 20.

U need it understand that its in the form of (a+b)^2 = a^2+b^2+2ab

once u get it ur done!!!

sq root( 9+sqrt80)^2 + sq root( 9-sqrt80)^2 + 2sq root( 9+sqrt80)*sq root( 9-sqrt80,,,( a^2+b^2+2ab)

solve this u will get

sqroot ans square get cancelled...
9+sqrt80+9-sqrt80 +2 ( 81- 9sqrt80+9sqrt80-80)

18 +2( sqroot 1)

hence 20

do let me know if u still have a doubt.
i suppose i'm getting caught up on how to distribute the 2ab portion of the equation. could you go through your calculations to get from 2sqrt(9+sqrt80)*sqrt(9-sqrt80) to 2(81-.....+80)?
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by b.kurland » Tue Jul 29, 2008 8:31 am
ohhh nevermind i see now! since both a and b are raised to the 1/2, you take that out and 2ab converts to 2((9+sqrt80)(9-sqrt80))^1/2 which =2sqrt1! got it thanks sudhir
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