When I come across ratios that have square roots such as below,

(1+√5)/ (2+√7)

would it be ok for me to unsquare all sides to get an integer on all sides?

=(1+√5)^2/ (2+√7)^2
=1+5 / 2+7
=6 /7

Not quite. Firstly, you provided an expression, not an equation. As a result, there aren't two sides to perform operations on. If you were dealing with an equation, then you could square both sides, add the same value to both sides, multiply both sides by the same value, etc.

Instead, with a fractional expression like this one, the only manipulations you can perform involve multiplying the numerator and the denominator by the same value. It works because essentially all you're doing is multiplying the fraction by 1 and the original value is unchanged.

By squaring a fraction, the value of the fraction changes. Consider the fraction 2/3. That's roughly 0.67. If you square the top and the bottom, now you have 4/9, which is roughly 0.44 -- a different value, so not a valid operation.

Secondly, be careful with how you square binomials (a fancy word for two terms added together). For example, (1+√5)^2 does not equal 1^2 + √5^2. Instead, think of it as (1+√5)(1+√5). That would yield 1 + 2√5 + 5 = 6 + 2√5. Similarly, (2+√7)^2 yields (2+√7)(2+√7) = 4 + 4√7 + 7 = 11 + 4√7.

Rey

Then is there no way to elegantly remove square roots from binomials such as the example below?

2x+√3 = x- √5

Must I always factor out or is there another way to simplify? Thanks.

2x+√3 = x- √5

Hi, I'd like to add another question to this last post by kimd6746:

If I saw this on the test, I would probably keep it safe and square both sides, but can I subtract the Xs to get the below?

x+ radical 3 = - radical 5?

Is leaving the - radical 5 problematic / not correct? Would this be considered undefined or out of bounds for the GMAT?

Thanks...

If the goal is to solve for x (and it usually is!) then the solution is fairly straightforward:

2x + √3 = x - √5
x + √3 = -√5
x = -√3 - √5

There's nothing else to do beyond this point. √3 and √5 are numbers just as 17 and -2 are. There is no way to rewrite -√3 - √5 into an equivalent expression that contains no square roots. The solution may look a little cumbersome, but -√3 - √5 is a number, too!



Whoa! Squaring both sides would NOT be playing it safe. Here's how it would turn out:

2x + √3 = x - √5
(2x + √3)(2x + √3) = (x - √5)(x - √5)
4x^2 + 2x√3 + 2x√3 + 3 = x^2 - x√5 - x√5 + 5
4x^2 + 4x√3 + 3 = x^2 - 2x√5 + 5
3x^2 + 4x√3 = - 2x√5 + 2

I'll stop there because it's just not looking good. But I hope it's clear that squaring both sides puts you in a bad spot. The only time that squaring simplifies things is when you have a variable inside a radical (or root). In that case, you should isolate that radical on one side of the equation, and then square both sides, thereby "liberating" that variable from the radical.

For this problem, however, keep it simple by isolating x on one side of the equation as I show above. Again, radicals are real numbers, so -√5 is totally acceptable. As is -√3 - √5, the solution to this equation.

Rey