Official Solution:

\(\sqrt{7 + \sqrt{48}} - \sqrt{3} = ?\)

A. 1.0
B. 1.7
C. 2.0
D. 2.4
E. 3.0

The expression \(7 + \sqrt{48}\) can be rewritten as \(4 + 4\sqrt{3} + 3\) which is the same as \((2 + \sqrt{3})^2\). Therefore,\(\sqrt{7 + \sqrt{48}} - \sqrt{3} = 2 + \sqrt{3} - \sqrt{3} = 2\).

Answer: C
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I guess such kind of tricks should come with practice.
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Another smart way of doing this is like this:
\(\sqrt{7 + \sqrt{48}} - \sqrt{3} = X\)
\(\sqrt{7 + \sqrt{48}} = X + \sqrt{3}\)
\(7 + \sqrt{48} = 7 + 4*\sqrt{3} = (X^2+3) + 2*X*\sqrt{3}\)
Its not hard to figure out that X = 2

Hi,
No you don't have to, whenever you encounter something of this sort, there would be better ways to do it, as GMAT is not likely to test you \(\sqrt{14}\) etc..

But say you get this kind of problems, you can always approximate..
\(\sqrt{14}=\sqrt{3}*\sqrt{5}\) = 1.7*2.2 = 3.8
so you can work with lower numbers too to get the answer
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I think this is a high-quality question and I agree with explanation.

Form an equation. Square it both the side. You will get answer as 2.

Or, Use ball park figure.. it will give u approx. 1.9-2.2 range. answer 2 satisfies this. sqrt(3)=1.73

***UPDATE*** (scroll down past the following until the next "***UPDATE***" tag)

I came up with a generalized approach.

Visually superimpose the following form onto your expression (or sub-expression):
\(a±b\sqrt{c}\)

Then test to see if
\(\sqrt{a-c} = \frac{b}{2}\)

If so then you can rearrange your initial expression as
\((a-c) ± b\sqrt{c} + c\)

Try this with the problem example, where a = 7; b = 4; and c = 3

Also see with another example here:

\(14 + 6\sqrt{5}\)

a = 14
b = 6
c = 5

Test it:
\(\sqrt{14-5} = \frac{6}{2}\)

3 = 3

It passes the test, so we can rewrite the original expression as:

\(9 + 6\sqrt{5} + 5\)

Which can then be factored into
\((3+\sqrt{5})^2\)


I'm not sure if there are enough occurrences of this to justify memorizing this trick, but it's a tool you can use nonetheless.

I personally like Magsy's method.

***UPDATE***
There is a similar opportunity to apply this in M26-01... and in light of the example there, I am restating a method that works better, which effectively boils down to just pulling out a \(2\) from the square root (but leave everything else in there!)

.....

M26-01 has part where you get to an expression of \(5 + 2\sqrt{6}\)

Here 5-6 = -1... so my first thought was to adjust my formula to be absolute value (|a-c|), but what seems to be the best heuristic is to manipulate a number to get a 2*square root or just look for a 2 * square root... (a more visual solution)

e.g. let's go back to the original example


\(14 + 6\sqrt{5}\)
Re-write as

\(14 + 2\sqrt{5}(3)\)

Get the \(3\) inside the square root:

\(14 + 2\sqrt{5*9}\)

Well, now you can recognize your 2 * square root expression to be reminding you of

\(2\sqrt{ab}\)

from

\(a +\) \(2\sqrt{ab}\) + \(b\)

so split up your 14 into: 5 + 9

And your overall expression is now:

\(5+ 2\sqrt{5*9} + 9\)

which factors to:

\((\sqrt{5} + 3)^2\)

e.g. #2; now visually, the M26-01 example should be come clear

\(5 + 2\sqrt{6}\)

\((3+2) + 2\sqrt{(3*2)}\)

\((3+2) + 2\sqrt{(3*2)}\)

\((\sqrt{3} + \sqrt{2})^2\)

....

Finally, e.g. 3; the example from this problem... perhaps even more useful than how Bunuel broke it down, let's do this:

\(7 + \sqrt{48}\) = \(7 + 2\sqrt{12}\)

And intuitively, we should immediately be able to see the numbers 3 and 4 (3+4 = 7; and 3*4 = 12)

so \(\sqrt{3}\) and \(\sqrt{4}\) (or 2) become our numbers...

\((\sqrt{3} + 2)^2\)


Note that the trick here is that for \(\sqrt{48}\) we are ONLY trying to pull out a "2" on the outside rather than trying to maximally simplify the square root ( \(2\sqrt{12}\) instead of. \(4\sqrt{3}\) )

This may be counter-intuitive, but if you simply remember that "I need to pull out a 2 that touches the outside of the square root sign directly" that would be a very useful real-time way of visually thinking about it.