Wierd sqrt problem...

just4myself wrote:Which of the following is equal to [1/ (sqrt (3)-sqrt(2))]^2
A) 1
B) 5
C) SQRT (6)
D) 5-SQRT (6)
E) 5+2SQRT(6)

Please elaborate...

Thank you...

With roots in the denominator, you will want to rationalize, using the difference of squares. Notice that:

1/ (sqrt (3)-sqrt(2))
= [1/ (sqrt (3)-sqrt(2))]*[(sqrt(3)+sqrt(2))/(sqrt(3)+sqrt(2))]
= sqrt(3)+sqrt(2)

The question asks for the square of this:

(sqrt(3)+sqrt(2))^2 = 3 + 2*sqrt(6) + 2 = 5 + 2sqrt(6)

I don't see anything wrong with using estimates for sqrt(2) and sqrt(3) for this particular question, but I do think it's better to understand the method above- you might instead see a question with much larger roots that you are not able to estimate quickly.

just4myself wrote:But Ian,
In the given question,
(sqrt(3)+sqrt(2))^2 = 3 + 2*sqrt(6) + 2 = 5 + 2sqrt(6)
is the denominator right?
So we cannot get the answer directly...
Now in fact the ans E becomes more ambiguous

I'm not sure what you mean by 'we cannot get the answer directly'; we can, as I outlined above. The denominator is indeed right- the denominator is equal to 1. We multiplied the top and bottom of the fraction by sqrt(3) + sqrt(2). In the denominator, we therefore have:

(sqrt(3) + sqrt(2))(sqrt(3) - sqrt(2))

Notice this is just the factorization of a difference of squares. When you multiply this out, you get:

(sqrt(3))^2 - (sqrt(2))^2 = 3 - 2 = 1