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joemama142000
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which is equal to ^2 a) 1 b) 5 c) sqrt6 d) 5-sqrt6 e) 5 + 18 Mar 2006, 21:39
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which is equal to [ 1/(sqrt3-sqrt2)]^2

a) 1
b) 5
c) sqrt6
d) 5-sqrt6
e) 5 + 2*sqrt6



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which is equal to ^2 a) 1 b) 5 c) sqrt6 d) 5-sqrt6 e) 5 + 18 Mar 2006, 22:01
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joemama142000
which is equal to [ 1/(sqrt3-sqrt2)]^2

a) 1
b) 5
c) sqrt6
d) 5-sqrt6
e) 5 + 2*sqrt6
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are you sure your question is complete?
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joemama142000
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which is equal to ^2 a) 1 b) 5 c) sqrt6 d) 5-sqrt6 e) 5 + 18 Mar 2006, 22:10
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the oa is E

1/(3-sqr2*sqrt3*2+2)=(5+2sqrt3*sqrt2)/( 25-24)= E
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which is equal to ^2 a) 1 b) 5 c) sqrt6 d) 5-sqrt6 e) 5 + 18 Mar 2006, 22:24
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joemama142000
the oa is E

1/(3-sqr2*sqrt3*2+2)=(5+2sqrt3*sqrt2)/( 25-24)= E
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good question..........
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vivek123
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which is equal to ^2 a) 1 b) 5 c) sqrt6 d) 5-sqrt6 e) 5 + 18 Mar 2006, 23:17
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Guys,

There is simple method.

[ 1/(sqrt3-sqrt2)]^2
= [(sqrt3+sqrt2)/(3-2)]^2
= (sqrt3+sqrt2)^2
= 3+2+2*sqrt3*sqrt2
= 5+2*sqrt(6)
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Angela780
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which is equal to ^2 a) 1 b) 5 c) sqrt6 d) 5-sqrt6 e) 5 + 19 Mar 2006, 08:30
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vivek123
Guys,

There is simple method.

[ 1/(sqrt3-sqrt2)]^2
= [(sqrt3+sqrt2)/(3-2)]^2
= (sqrt3+sqrt2)^2
= 3+2+2*sqrt3*sqrt2
= 5+2*sqrt(6)
Show more


Vivek, can you explain this to me?
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which is equal to ^2 a) 1 b) 5 c) sqrt6 d) 5-sqrt6 e) 5 + 19 Mar 2006, 08:39
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vivek123
Guys,

There is simple method.

[ 1/(sqrt3-sqrt2)]^2
= [(sqrt3+sqrt2)/(3-2)]^2
= (sqrt3+sqrt2)^2
= 3+2+2*sqrt3*sqrt2
= 5+2*sqrt(6)

Vivek, can you explain this to me?
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Sure,
[ 1/(sqrt3-sqrt2)]^2 ----------------- (1)

The denominator is (sqrt3-sqrt2), in such problems, if we can get the denominator in simpler form, great!

Multiply & divide (1) by (sqrt3 + sqrt2)

= [(sqrt3 + sqrt2)/((sqrt3 + sqrt2)(sqrt3 - sqrt2))]^2 ------------------------------- (2)

Notice that, now the denominator is in the form, (a-b)(a+b)
we know that, (a-b)(a+b) = a^2-b^2
i.e. (sqrt3 + sqrt2)(sqrt3 - sqrt2) = (sqrt3)^2 - (sqrt2)^2 = 3 - 2 = 1

Put this value in equation (2)

= [(sqrt3+sqrt2)/1]^2
= (sqrt3+sqrt2)^2

Now apply (a+b)^2 = a^2+b^2+2ab

= 3+2+(2*sqrt3*sqrt2)
= 5+2*sqrt6
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which is equal to ^2 a) 1 b) 5 c) sqrt6 d) 5-sqrt6 e) 5 + 19 Mar 2006, 08:54
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Thanks Vivek.. :-D ..

You make it seem so simple :wink:
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which is equal to ^2 a) 1 b) 5 c) sqrt6 d) 5-sqrt6 e) 5 + 19 Mar 2006, 13:18
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vivek123
Guys,

There is simple method.

[ 1/(sqrt3-sqrt2)]^2
= [(sqrt3+sqrt2)/(3-2)]^2
= (sqrt3+sqrt2)^2
= 3+2+2*sqrt3*sqrt2
= 5+2*sqrt(6)
Show more


very good approach...........



Archived Topic
Hi there,
This topic has been closed and archived due to inactivity or violation of community quality standards. No more replies are possible here.
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