What is the shortest possible way to solve this?

[Vishnu88]

[spoiler]OA: 3√3[/spoiler]

[cans]

rationalize 2nd part (i.e. multiply divide by 9-4*root(5))
2nd part becomes 3*(9-4*root(5))/1 = 27-12*root(5)
1st part is 12*root(5)
adding both, we get 27
thus root(27) = 3*root(3)

[GMATGuruNY]

[spoiler]OA: 3√3[/spoiler]

The GMAT would provide answer choices.
If the answer choices were not too close, the quickest solution would be to ballpark:

√(3√80 + 3/(9+4√5))
≈ √(3*9 + fraction)
≈ √27
≈ 5+.

The correct answer would be a little more than 5.
3√3 works:
3√3 ≈ 3*(1.7) ≈ 5.1.

[pemdas]

4*sq.root(5)=sq.root(80), let sq.root(80) be a, then
3a + 3/(9+a)
by use of ball-parking 3a=27 as sq.root(80)=~9 and
3/(9+a)=~3/18=1/6, hence 27+1/6
So sq.root(27+1/6) is approximate of 3*sq.root(3) or sq.root(27)

[spoiler]OA: 3√3[/spoiler]

[Brian@VeritasPrep]

Hey everyone,

I just want to draw a little more attention to cans' post and the device she used to solve this one: the Difference of Squares rule.

Of all the rules that you could (or may have to) memorize for the GMAT, Difference of Squares is probably among the top 2-3 most useful (Pythagorean Theorem is up there, too). The Difference of Squares rule is that:

(x + y)(x - y) = x^2 - y^2

and, by the same token:

x^2 - y^2 = (x + y)(x - y)

I've always seen Difference of Squares as something like a Secret Decoder Ring or one of those spy-type toys you got in a cereal box as a kid (or like invisible ink). Difference of Squares can completely transform an algebraic expression into one that's infinitely more useful or convenient for you to use.

Here, that denominator is messy...the only way to get the radical out of that denominator is to use Difference of Squares. If you try multiplying by (9 + 4sqrt5) or by just 4sqrt5, you'll still have a radical down there, and the GMAT won't list any answer choices that way. But if you see that you have an x + y setup there that you need to transform, multiplying by x - y (on both top and bottom) will let you square out that radical:

3/(9 + 4sqrt5) multiplied by (9 - 4sqrt 5)/(9 - 4sqrt 5) = (27 - 12 sqrt 5) / (9^2 - 4^2 * sqrt 5^2)

And then you can rewrite the denominator so that you have:

(27 - 12 sqrt 5) / (81 - 16*5)

Then look at what happens: 16*5 = 80, so that denominator is 81-80 or just 1. So the denominator is redundant and that whole term is just the numerator: 27 - 12 sqrt 5.

So now we have: 3 sqrt 80 + 27 - 12 sqrt 5, and since sqrt 5 can't be broken down any further we want to try to factor out sqrt 80 to make it close. 80 = 16 * 5, so we can take 4 out of 16 and we have: 12 sqrt 5 + 27 - 12 sqrt 5. The two square root terms cancel (one positive, one negative), so we're left with just 27. (And I didn't notice it at first but that whole thing is under a radical, right? So that's where the official answer comes from...)


Now, writing out the steps looks pretty long and if the answer choices are spread far enough apart you may be able to get away with an estimate. But if you recognize the Difference of Squares right away, this is a lot quicker done by hand than it is by typing with explanations (as cans proved!). And spending time with the Difference of Squares can be extremely helpful - if you need to rationalize a compound denominator like we did here, it's huge. If you need to factor out some multiple-variable algebra, it's important. If you get a problem with massive numbers and exponents (I saw one the other day that started with 1001^2 - 999^2), it's also helpful. So consider this my advertising message - learn to love the Difference of Squares!