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by harsh.champ » Wed Mar 03, 2010 12:20 am
Anchit 777 wrote:IF ( a + broot2 ) x ( p + qroot2 ) = 1;

where a,b,p and q are rational numbers and they are not equal to zero.

Prove that:

p square - 2 q square = 1/( a square - 2 b square)
Multiplying by ( a - broot2 ) x ( p - qroot2 ) in the bold-faced equation:-

we get (p square - 2 q square) x ( a square - 2 b square) = ( a - broot2 ) x ( p - qroot2 )
It takes time and effort to explain, so if my comment helped you please press Thanks button :)



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by Anchit 777 » Wed Mar 03, 2010 2:28 am
But harsh how does that prove that:

( p square - 2 q square ) = 1 / ( a square - 2 b square ).

Can you explain . please.
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by sanju09 » Wed Mar 03, 2010 3:10 am
Anchit 777 wrote:But harsh how does that prove that:

( p square - 2 q square ) = 1 / ( a square - 2 b square ).

Can you explain . please.
If (a + b √2) (p + q √2) = 1, with a, b, p, and q as rationals, then

a p + 2 b q + (a q + b p) √2 = 1, comparing the rational and irrational parts, we have

a p + 2 b q = 1, and a q + b p = 0, now let's take

(a - b √2) (p - q √2)

= a p + 2 b q - (a q + b p) √2

= 1 - 0

= 1

does that help?
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by yeahdisk » Wed Mar 03, 2010 3:26 am
sanju09 wrote:
...comparing the rational and irrational parts, we have

a p + 2 b q = 1, and a q + b p = 0,
Sorry - can you please explain how you can equate the bold part to 0?
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by harsh.champ » Wed Mar 03, 2010 3:29 am
Anchit 777 wrote:But harsh how does that prove that:

( p square - 2 q square ) = 1 / ( a square - 2 b square ).

Can you explain . please.
I think that sanju's post will be helpful.


Also remember that roots of surds occur in pairs.
i.e. if one root is 2 + sqrt(3)
the other root will be 2 - sqrt(3)

Always remember this while solving quadratic or biquadratic equations.


Also,I doubt whether bi-quads are asked in the GMAT??
It takes time and effort to explain, so if my comment helped you please press Thanks button :)



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by harsh.champ » Wed Mar 03, 2010 3:32 am
yeahdisk wrote:
sanju09 wrote:
...comparing the rational and irrational parts, we have

a p + 2 b q = 1, and a q + b p = 0,
Sorry - can you please explain how you can equate the bold part to 0?
Since,no root part is there(in 1 + 0*root(2)) we have aq+bp = 0
Had the no. been 1 + 5root(2) , aq + bp = 5

I hope the explanation would be more clear now. :)
It takes time and effort to explain, so if my comment helped you please press Thanks button :)



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by sanju09 » Wed Mar 03, 2010 3:35 am
yeahdisk wrote:
sanju09 wrote:
...comparing the rational and irrational parts, we have

a p + 2 b q = 1, and a q + b p = 0,
Sorry - can you please explain how you can equate the bold part to 0?

Once, a, b, p, and q are rationals, so is a q + b p, but this multiplied to √2 would make it irrational, and a rational (a p + 2 b q) added to an irrational {(a q + b p) √2} cannot yield a pure rational (1). Only possibility for this to be true is that a q + b p = 0. :|
The mind is everything. What you think you become. -Lord Buddha



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by sanju09 » Wed Mar 03, 2010 3:45 am
harsh.champ wrote:
Anchit 777 wrote:But harsh how does that prove that:

( p square - 2 q square ) = 1 / ( a square - 2 b square ).

Can you explain . please.
I think that sanju's post will be helpful.


Also remember that roots of surds occur in pairs.
i.e. if one root is 2 + sqrt(3)
the other root will be 2 - sqrt(3)

Always remember this while solving quadratic or biquadratic equations.


Also,I doubt whether bi-quads are asked in the GMAT??
Never wonder if that happens, it's not totally out of GMAT concepts though :)
The mind is everything. What you think you become. -Lord Buddha



Sanjeev K Saxena
Quantitative Instructor
The Princeton Review - Manya Abroad
Lucknow-226001

www.manyagroup.com
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by Anchit 777 » Wed Mar 03, 2010 7:56 am
I sincerely appreciate and value the posts of SANJU and HARSH . Thank you very much guys, please reply to my posts in the future as well.

Thanks Again. Cheers!
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