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Rationalizing Denominator

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Rationalizing Denominator

by ejose » Sun Feb 14, 2010 1:11 pm
Hi -

I tried searching for this one, but the notation makes it very search un-friendly. It's problem 117 from OG 12. I suspect many of you will find this very simple, but I'm fairly early on in my GMAT prepping :)

If n is positive, which of the following is equal to ---- 1 / (sqrt(n+1) - sqrt(n))

A. 1
B. sqrt(2n+1)
C. sqrt(n+1) / sqrt(n)
D. sqrt(n+1) - sqrt(n)
E. sqrt(n+1) + sqrt(n)

The solution in the book recommends rationalizing the denominator by using (sqrt(n+1) + sqrt(n) / sqrt(n+1) + sqrt(n)). My question is ... how would you know to start there? In the same vein, is there an easier way to solve this? The concept of dividing sums of roots throws me a bit.

(Note to self, perhaps I should focus on operations with roots :))
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by ajith » Sun Feb 14, 2010 1:29 pm
ejose wrote:Hi -

I tried searching for this one, but the notation makes it very search un-friendly. It's problem 117 from OG 12. I suspect many of you will find this very simple, but I'm fairly early on in my GMAT prepping :)

If n is positive, which of the following is equal to ---- 1 / (sqrt(n+1) - sqrt(n))

A. 1
B. sqrt(2n+1)
C. sqrt(n+1) / sqrt(n)
D. sqrt(n+1) - sqrt(n)
E. sqrt(n+1) + sqrt(n)

The solution in the book recommends rationalizing the denominator by using (sqrt(n+1) + sqrt(n) / sqrt(n+1) + sqrt(n)). My question is ... how would you know to start there? In the same vein, is there an easier way to solve this? The concept of dividing sums of roots throws me a bit.

(Note to self, perhaps I should focus on operations with roots :))
The underlying rule used here is (a+b) (a-b) = a^2 -b^2

if your root is in the form 1/(sqrt(x)+sqrt(y)) to rationalize the denominator use the above mentioned rule
say a = sqrt(x) and b =sqrt(y)

1/(a+b) is to be rationalized; we know that a^2, b^2 and thus, a^2 -b^2 are rational

to get a^2 -b^2 in the denominator we multiply by (a-b) both denominator and numerator

thus 1/(a+b) becomes (a-b)/(a^2-b^2)

Substituting back 1/(sqrt(x)+sqrt(y)) becomes (sqrt(x) - sqrt(y))/(x-y) [denominator is rational)

Now if your fraction is in the form 1/(sqrt(x) - sqrt(y)) we multiply numerator and denominator by (sqrt(x)+sqrt(y))

Rationalized Denominator remains x-y

-----------------

Now if we apply the same rules in this question

1 / (sqrt(n+1) - sqrt(n)) [ in the form 1/(a-b)]
= 1 / (sqrt(n+1) - sqrt(n)) * (sqrt(n+1) + sqrt(n))/(sqrt(n+1) + sqrt(n))
= (sqrt(n+1) + sqrt(n))/(n+1-n)
= sqrt(n+1) + sqrt(n)

It is not that complicated as it looks to be
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by ajith » Sun Feb 14, 2010 1:31 pm
ejose wrote:Hi -

I tried searching for this one, but the notation makes it very search un-friendly. It's problem 117 from OG 12. I suspect many of you will find this very simple, but I'm fairly early on in my GMAT prepping :)

If n is positive, which of the following is equal to ---- 1 / (sqrt(n+1) - sqrt(n))

A. 1
B. sqrt(2n+1)
C. sqrt(n+1) / sqrt(n)
D. sqrt(n+1) - sqrt(n)
E. sqrt(n+1) + sqrt(n)

The solution in the book recommends rationalizing the denominator by using (sqrt(n+1) + sqrt(n) / sqrt(n+1) + sqrt(n)). My question is ... how would you know to start there? In the same vein, is there an easier way to solve this? The concept of dividing sums of roots throws me a bit.

(Note to self, perhaps I should focus on operations with roots :))
The underlying rule used here is (a+b) (a-b) = a^2 -b^2

if your root is in the form 1/(sqrt(x)+sqrt(y)) to rationalize the denominator use the above mentioned rule
say a = sqrt(x) and b =sqrt(y)

1/(a+b) is to be rationalized; we know that a^2, b^2 and thus, a^2 -b^2 are rational

to get a^2 -b^2 in the denominator we multiply by (a-b) both denominator and numerator

thus 1/(a+b) becomes (a-b)/(a^2-b^2)

Substituting back 1/(sqrt(x)+sqrt(y)) becomes (sqrt(x) - sqrt(y))/(x-y) [denominator is rational)

Now if your fraction is in the form 1/(sqrt(x) - sqrt(y)) we multiply numerator and denominator by (sqrt(x)+sqrt(y))

Rationalized Denominator remains x-y

-----------------

Now if we apply the same rules in this question

1 / (sqrt(n+1) - sqrt(n)) [ in the form 1/(a-b)]
= 1 / (sqrt(n+1) - sqrt(n)) * (sqrt(n+1) + sqrt(n))/(sqrt(n+1) + sqrt(n))
= (sqrt(n+1) + sqrt(n))/(n+1-n)
= sqrt(n+1) + sqrt(n)

It is not that complicated as it looks to be
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by harshavardhanc » Mon Feb 15, 2010 6:27 am
ejose wrote:Hi -

My question is ... how would you know to start there? In the same vein, is there an easier way to solve this? The concept of dividing sums of roots throws me a bit.

(Note to self, perhaps I should focus on operations with roots :))

Eric,

I'm not a maths expert, but can advise you on this one.

Just look at the answer options: you are being asked to compare a fraction (with a weird denominator) with 4 options having the simplest of the denominators : 1 :) . You can also view the 4 options as having "no denominator" (for our understanding purpose).

Now the question pops up : If any of the options is equal to our question fraction, where has the the "tail" gone ?

it should give you a hint that you will have to simplify the denominator somehow .

as an observation, in most of the questions where you have subtraction of square roots in the denominator, you will have to rationalize the denominator.

Reason : in this process, you are essentially bringing the denominator's conjugate (same terms, opposite signs ) in the picture. This, in turn, will enable you to apply the formula (A+B)(A-B) {multiplication of conjugates :) }

and we know that this product is equivalent to sq(A) - sq(B).

Now when you square the square root, the net effect is : simple, plain and vanilla-looking numbers :)

sq ( sqrt(A) ) = A :)



hope this helps!
Regards,
Harsha
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by shashank.ism » Mon Feb 15, 2010 11:04 am
ejose wrote:Hi -

I tried searching for this one, but the notation makes it very search un-friendly. It's problem 117 from OG 12. I suspect many of you will find this very simple, but I'm fairly early on in my GMAT prepping :)

If n is positive, which of the following is equal to ---- 1 / (sqrt(n+1) - sqrt(n))

A. 1
B. sqrt(2n+1)
C. sqrt(n+1) / sqrt(n)
D. sqrt(n+1) - sqrt(n)
E. sqrt(n+1) + sqrt(n)

The solution in the book recommends rationalizing the denominator by using (sqrt(n+1) + sqrt(n) / sqrt(n+1) + sqrt(n)). My question is ... how would you know to start there? In the same vein, is there an easier way to solve this? The concept of dividing sums of roots throws me a bit.

(Note to self, perhaps I should focus on operations with roots :))
Its quite simple...1 / (sq.(n+1) - sq.(n)) = 1 / (sq.(n+1) - sq.(n)) * (sq.(n+1) + sq.(n))/(sq.(n+1) + sq.n))
= (sq.(n+1) + sq.(n))/(n+1-n) = sq.(n+1) + sq.(n)
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by vijay_venky » Mon Feb 15, 2010 10:30 pm
Though it is advisable to study these types of things as they are not very difficult to do so, you could start from the options.

Equate each option to 1 / (sqrt(n+1) - sqrt(n)) and find out which concurs. This involves some calculation. But the best method is to rationalize.
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