# Question: Root in the Denominator

## Comment on Root in the Denominator

### I solved this in a slightly

I solved this in a slightly different way,and just wanted to check if you could tell me which one would work faster in multiple scenarios.
So instead of trying to get the same denominator as 1 by using two conjugates, I multiplied both the denominators with each other to get a common multiple in the denominator, and then in the numerators, I multiplied the first fraction's numerator by the second denominator, and similarly for the other one.

Do let me know.
Thanks ### That works too.

That works too.

Your method works because the each of the two denominators just happens to be the conjugate of the other.

### Hey!

Hey!
Thanks for the response.
Just a follow up question. Will my method not work, in case they are not conjugates? ### Yes, yours is still a valid

Yes, yours is still a valid approach, but you may have to perform a little extra work (not much though).

### hi brent,

hi brent,

so if the right part of the question is the conjugate of the left part then we don't have to solve for it as we can assume that it would be the opposite of what we get on the left part of the question.

in other words, we got -1 + root 3/1 for the left part so is it safe to assume that the right part will be -1 - root 3/1 ### Sorry, but I'm not 100% sure

Sorry, but I'm not 100% sure what you're asking when you write "we can assume that it would be the opposite of what we get on the left part of the question."

Aside: If A is the conjugate of B, then we also know that B is the conjugate of A.
For example, 5 - √3 is the conjugate of 5 + √3, and we can also say that 5 + √3 is the conjugate of 5 - √3

### what I mean is, for the

what I mean is, for the question:

1 + √3 1 - √3
------ - ------
2 + √3 2 - √3

In the first step of your solution:

1 + √3
------
2 + √3

you multiplied the denominator and the numerator by 2 - √3 to remove the root. And by solving you got the answer:-1+√3
-----
1
so I asked whether it was necessary to perform the same operations to the other fraction as the answer will also be the conjugate since the two fractions are conjugate of each other. ### Thanks for clarifying.

Thanks for clarifying.
You're absolutely right; since each part of the second fraction is the conjugate of each part of the first fraction, the resulting numerator (after we multiply top and bottom by their respective conjugates) of the second fraction will be the conjugate of the numerator of the first fraction.

Cheers,
Brent

### Hi Brent!

Hi Brent!
Does the color beside the question videos indicate the difficulty of the question? ### That's correct. If you hover

That's correct. If you hover your cursor over each color, you'll find that:
- GREEN represents a difficulty level of 350 to 500
- YELLOW represents a difficulty level of 510 to 650
- RED represents a difficulty level of 660 to 800

### https://gmatclub.com/forum/a

https://gmatclub.com/forum/a-club-sold-an-average-arithmetic-mean-of-92-raffle-tickets-per-memb-243665.html

I don't understand this, so 92-84 = 8, but this is for female, 96-92 = 4, this is for male, isn't the ratio of male to female 1:2 then? should I understand it as "Because male's ticket sold more than female that's why it's 2:1? ### Here's my full solution using

Here's my full solution using weighted averages: https://gmatclub.com/forum/a-club-sold-an-average-arithmetic-mean-of-92-...

### https://gmatclub.com/forum/if

https://gmatclub.com/forum/if-the-average-arithmetic-mean-of-x-y-and-z-is-7x-and-x-219222.html

TRICKY! I CHOSED E 