OMG .. there is a much elegant and easier method to solve this
\(\sqrt{6} = ~2.4 ==> 10\sqrt{6}=10*2.4==> 24\)
25+24=49
\(\sqrt{49} =7\)
25-24=1
\(\sqrt{1} = 1\)
So our question is 7+1 = 8
See option \(2 \sqrt{15} = 2*4= 8\)
HENCE C
Bunuel wrote:
SOLUTIONS:
1. 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 fro 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 \(50+2(\sqrt{25+10\sqrt{6}})(\sqrt{25-10\sqrt{6}})\) --> \(50+2(\sqrt{25+10\sqrt{6}})(\sqrt{25-10\sqrt{6}})=50+2\sqrt{(25+10\sqrt{6})(25-10\sqrt{6})}\).
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 the answer: \(\sqrt{60}=2\sqrt{15}\).
Answer: C.
OMG .. there is a much elegant and easier method to solve this
\(\sqrt{6} = ~2.4 ==> 10\sqrt{6}=10*2.4==> 24\)
25+24=49
\(\sqrt{49} =7\)
25-24=1
\(\sqrt{1} = 1\)
So our question is 7+1 = 8
See option \(2 \sqrt{15} = 2*4= 8\)
HENCE C
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