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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
A. \(2\sqrt{5}\)
B. \(\sqrt{55}\)
C. \(2\sqrt{15}\)
D. 50
E. 60
16 Sep 2014, 01:24
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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
First, square the given expression to eliminate the square roots, but remember to take the square root of the result at the end to balance the operation and obtain the correct answer.
Important for the GMAT: \((x+y)^2=x^2+2xy+y^2\) and \((x-y)^2=x^2-2xy+y^2\).
Following these rules, 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 the sum of the first and third terms simplifies to \((25+10\sqrt{6})+(25-10\sqrt{6})=50\), so we have:
\(50+2(\sqrt{25 + 10\sqrt{6} })(\sqrt{25 - 10\sqrt{6} })=\)
\(=50+2\sqrt{(25 + 10\sqrt{6})(25 - 10\sqrt{6}) }\).
Another important concept for the GMAT: \((x+y)(x-y)=x^2-y^2\). Using this, we can simplify further:
\(50+2\sqrt{(25 + 10\sqrt{6})(25 - 10\sqrt{6})}=\)
\(=50+2\sqrt{25^2-(10\sqrt{6})^2)} = \)
\( = 50+2\sqrt{625-600}=\)
\(=50+2\sqrt{25}=\)
\(=60\).
Finally, remember to take the square root of this value to obtain the correct answer: \(\sqrt{60}=2\sqrt{15}\).
Answer: C
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
First, square the given expression to eliminate the square roots, but remember to take the square root of the result at the end to balance the operation and obtain the correct answer.
Important for the GMAT: \((x+y)^2=x^2+2xy+y^2\) and \((x-y)^2=x^2-2xy+y^2\).
Following these rules, 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 the sum of the first and third terms simplifies to \((25+10\sqrt{6})+(25-10\sqrt{6})=50\), so we have:
\(50+2(\sqrt{25 + 10\sqrt{6} })(\sqrt{25 - 10\sqrt{6} })=\)
\(=50+2\sqrt{(25 + 10\sqrt{6})(25 - 10\sqrt{6}) }\).
Another important concept for the GMAT: \((x+y)(x-y)=x^2-y^2\). Using this, we can simplify further:
\(50+2\sqrt{(25 + 10\sqrt{6})(25 - 10\sqrt{6})}=\)
\(=50+2\sqrt{25^2-(10\sqrt{6})^2)} = \)
\( = 50+2\sqrt{625-600}=\)
\(=50+2\sqrt{25}=\)
\(=60\).
Finally, remember to take the square root of this value to obtain the correct answer: \(\sqrt{60}=2\sqrt{15}\).
Answer: C
17 Jan 2017, 02:05
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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
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
