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Help on P.S. Question

by GMATJeff » Wed Nov 26, 2008 12:58 pm
Here is the P.S. question:

(Sq Rt of (9 + (Sq Rt of 80)) + Sq Rt of (9 - (Sq Rt of 80))^2

The answer is 20.

Can anyone walk me through the solution to this?

Thanks!
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by Neo2000 » Wed Nov 26, 2008 3:24 pm
if you observe carefully you will notice that this of the form a+b and a-b

So your first step should be to re-write the numbers in a+b form

9 +sqRt(80) = 9 + SqRoot(4x20) = 9 + 2SqRoot(20)


Now this of the form (a^2) +(b^2) +2ab
where (a^2) +(b^2) = 9 and ab = SqRoot(20)
9 + 2SqRoot(20) is therefore = 5+4 +2SqRoot(5x4)
5+4 +2SqRoot(5x4) is now = (Root5 +Root4)^2 = (Root5 +2)^2
Since this already has a Root around it, it reduces to (Root5 +2)

The other part of the equation reduces to 9 - 2SqRoot(20) which, if you follow the same steps as above will reduce to Root5 -2
Adding them both gives you 2Root5
And since the entire equation is to be Squared you get 4x5 = 20

P.S. Please tell me i was clear enough :D
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by anayeri » Fri Nov 28, 2008 11:59 am
I think I totally missed something - where did that new equation come from????
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Neo 2000

by GMATJeff » Fri Nov 28, 2008 1:41 pm
Thanks for the response. However, I'm completely lost by the explanation. I'm not sure how the equation is in the form of (a^2) +(b^2) +2ab to begin with. I cannot follow the rest of the reasoning either.
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by Neo2000 » Fri Nov 28, 2008 4:15 pm
Ignore the Outside Square Root and Square power. Only Consider the expression

You have 9 + SqRt(80) and 9 - SqRt(80)

Now, because you need to eliminate the square root, you start to wonder if the inside expression can be written as the Square of a number since we know that SqRt((a+b)^2) = (a+b)

Now factors of 80 = 2x2x2x2x5 = 4x20 And you know that SqRt(4) = 2
So SqRt(80) now becomes 2SqRt(20)

Your expression has now simplified to 9 + 2SqRt(20)
We have to still express this as the Square of some expression
(a+b)^2 = a^2 + b^2 +2ab Compare this to your expression
a^2 + b^2 = 9
2ab = 2SqRt(20)
You can now simplify to get values for a and b
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by raajan_p » Sat Nov 29, 2008 9:21 am
Neo2000 wrote:if you observe carefully you will notice that this of the form a+b and a-b

So your first step should be to re-write the numbers in a+b form

9 +sqRt(80) = 9 + SqRoot(4x20) = 9 + 2SqRoot(20)


Now this of the form (a^2) +(b^2) +2ab
where (a^2) +(b^2) = 9 and ab = SqRoot(20)
9 + 2SqRoot(20) is therefore = 5+4 +2SqRoot(5x4)
5+4 +2SqRoot(5x4) is now = (Root5 +Root4)^2 = (Root5 +2)^2
Since this already has a Root around it, it reduces to (Root5 +2)

The other part of the equation reduces to 9 - 2SqRoot(20) which, if you follow the same steps as above will reduce to Root5 -2
Adding them both gives you 2Root5
And since the entire equation is to be Squared you get 4x5 = 20

P.S. Please tell me i was clear enough :D
Totally understood your method...

When I was working out, I dint realize that 9 + 2SqRoot(20) is of the form (a+b)^2 :)
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by GMATJeff » Sat Nov 29, 2008 4:14 pm
Still not getting it NEO. Can anyone else please provide an explanation?
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by cramya » Sat Nov 29, 2008 4:27 pm
Jeff,
See if this helps any!

(Sq Rt of (9 + (Sq Rt of 80)) + Sq Rt of (9 - (Sq Rt of 80))^2

a= Sq Rt of (9 + (Sq Rt of 80))
b = Sq Rt of (9 - (Sq Rt of 80))

As Neo pointed out (a+b)^2 = a^2+b^2+2ab

a^2 = Sq Rt of (9 + (Sq Rt of 80)) ^ 2 = 9+sqrt(80)

b^2 = Sq Rt of (9 - (Sq Rt of 80)) = 9-sqrt(80)

2ab = 2 * Sq Rt of (9 + (Sq Rt of 80)) * Sq Rt of (9 - (Sq Rt of 80)

= 2 * SQRT((9+SQRT(80)) * (9-SQRT(80))
(COLLAPSING A AND B UNDER ONE SQRT IN BOLD)

= 2 * SQRT ((81-9SQRT(80)+9SQRT(80)-80))
= 2 * sqrt(81-80)
= 2 * sqrt(1)
= 2

a^2+b^2+2ab

= 9+sqrt(80)+9-sqrt(80)+2
= 20
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by GMATJeff » Sat Nov 29, 2008 10:29 pm
I got it now! Thanks for your help.
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