Official Solution:

What is the value of \(\sqrt{25+10\sqrt{6}}+\sqrt{25-10\sqrt{6}}\)?

A. \(2\sqrt{5}\)
B. \(\sqrt{55}\)
C. \(2\sqrt{15}\)
D. 50
E. 60


Square the given expression to get rid of the roots, though don't forget to un-square the value you get at the end to balance this operation and obtain the right answer:

Must know for the GMAT: \((x+y)^2=x^2+2xy+y^2\) (while \((x-y)^2=x^2-2xy+y^2\)).

So we get:

\((\sqrt{25+10\sqrt{6}}+\sqrt{25-10\sqrt{6}})^2 =\) \(=(\sqrt{25+10\sqrt{6}})^2+2(\sqrt{25+10\sqrt{6}})(\sqrt{25-10\sqrt{6}})+(\sqrt{25-10\sqrt{6}})^2=\)

\(=(25+10\sqrt{6})+2(\sqrt{25+10\sqrt{6}})(\sqrt{25-10\sqrt{6}})+(25-10\sqrt{6})\).

Note that sum of the first and the third terms simplifies to \((25+10\sqrt{6})+(25-10\sqrt{6})=50\), so we have

\(50+2(\sqrt{25+10\sqrt{6}})(\sqrt{25-10\sqrt{6}})\) therefore:

\(50+2(\sqrt{25+10\sqrt{6}})(\sqrt{25-10\sqrt{6}})=\) \(50+2\sqrt{(25+10\sqrt{6})(25-10\sqrt{6})}\).

Also must know for the GMAT: \((x+y)(x-y)=x^2-y^2\), thus

\(50+2\sqrt{(25+10\sqrt{6})(25-10\sqrt{6})}=50+2\sqrt{25^2-(10\sqrt{6})^2)} =\) \(= 50+2\sqrt{625-600}=50+2\sqrt{25}=60\).

Recall that we should un-square this value to get the right answer: \(\sqrt{60}=2\sqrt{15}\).


Answer: C
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Its done so as to express the final answer in the simplest form, something that GMAT will always ask you to do.

It is done becaue Bunuel starts with a squared value of the expression that we need to evalute (look below). This is the reason why you then have to take the square root to get the value of the question asked.


You need to find the value of \((\sqrt{25+10\sqrt{6}}+\sqrt{25-10\sqrt{6}})\) and NOT \((\sqrt{25+10\sqrt{6}}+\sqrt{25-10\sqrt{6}})^2\)



Hope this helps.

Alternative solution (much faster in my opinion)

We factor 5 => \(\sqrt{25+10\sqrt{6}\)+\(\sqrt{25-10\sqrt{6}\)=\(\sqrt{5(5+2\sqrt{6})\)+\(\sqrt{5(5-2\sqrt{6})\)

Now we write \(\sqrt{5(5+2\sqrt{6})\) as \(\sqrt{5(3+2+2\sqrt{6})\)=\(\sqrt{5(\sqrt{3}+\sqrt{2})^2\)
and \(\sqrt{5(5-2\sqrt{6})\) as \(\sqrt{5(3+2-2\sqrt{6})\)=\(\sqrt{5(\sqrt{3}-\sqrt{2})^2\)

=> \(\sqrt{5(\sqrt{3}+\sqrt{2})^2\)+\(\sqrt{5(\sqrt{3}-\sqrt{2})^2\) = \(\sqrt{5}(\sqrt{3}+\sqrt{2})\)+\(\sqrt{5}(\sqrt{3}-\sqrt{2})\)=\(\sqrt{5}*\sqrt{3}\) +\(\sqrt{5}*\sqrt{2}\) +\(\sqrt{5}*\sqrt{3}\) -\(\sqrt{5}*\sqrt{2}\) =\(\sqrt{15}+\sqrt{15}\) =\(2\sqrt{15}\) => Answer C

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Yes, that's correct.
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Let \(25 + 10\sqrt{6}\ = A\) &
Let \(25 - 10\sqrt{6}\ = B =>\)

\(\sqrt{A}\ + \sqrt{B}\ = x\)
\(A + B + 2\sqrt{AB}\ = x^2 =>\)
\(25 + 10\sqrt{6}\ + 25 - 10\sqrt{6}\ + 2\sqrt{(25 + 10\sqrt{6})(25 - 10\sqrt{6})}\ = x^2\)
\(50 + 2\sqrt{25^2 - (10\sqrt{6})^2}\ = x^2 =>\)
\(x^2 = 60\)
\(x = 2\sqrt{15}\)

Same approach as Bunuel, but well laid out. Thank you!