M26-01

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Bunuel

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Re M26-01 [#permalink]

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
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Re: M26-01 [#permalink]

12 Jun 2017, 09:50

Kudos

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Bunuel wrote:

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

Can it be solved the following way?
\(\sqrt{25+10\sqrt{6}}+\sqrt{25-10\sqrt{6}}=\sqrt{(\sqrt{15}+\sqrt{10})^2}+\sqrt{(\sqrt{15}-\sqrt{10})^2}\)

then it would turn into \(|\sqrt{15}+\sqrt{10}| +|\sqrt{15}-\sqrt{10}|\)
then \(\sqrt{15}+\sqrt{10}+\sqrt{15}-\sqrt{10}=2\sqrt{15}\)

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pkappaga

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Re: M26-01 [#permalink]

16 Mar 2016, 07:49

I understand the simplification part but Isn`t the final value 60? I am not sure why you are doing the square root of 60

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Re: M26-01 [#permalink]

16 Mar 2016, 07:53

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pkappaga wrote:

I understand the simplification part but Isn`t the final value 60? I am not sure why you are doing the square root of 60

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

Bunuel wrote:

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:

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

Hope this helps.
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Re: M26-01 [#permalink]

13 Sep 2016, 02:36

Kudos

Alternative solution: 10*\(\sqrt{6}\) = \(\sqrt{600}\), which is approximately 24.5 as 600 lies between 24*24=576 and 25*25=625
So, \(\sqrt{25+24.5}\) + \(\sqrt{25-24.5}\) = \(\sqrt{49,5}\) + \(\sqrt{0.5}\) ≈ slightly more than 7+ around 0.7*≈ 8

* 0.7*0.7 =0.49 ≈ 0,5

So, we can eliminate A, D and E

B and C
2*\(\sqrt{15}\)≈ 2*4 ≈ 8 - contender

\(\sqrt{55}\) - loser.
First, 55 is closer to 49 than to 64, so \(\sqrt{55}\) is closer to 7.
Second, using guessing technique, we see that 55 has no pair with any other number in answer choices, whilst there is a pair of 60s - 60 and \(\sqrt{60}\).

Hope this helps!
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AlexBunea

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Re: M26-01 [#permalink]

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

Bunuel

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Re: M26-01 [#permalink]

12 Jun 2017, 13:37

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MikeMighty wrote:

Bunuel wrote:

_______________
Yes, that's correct.
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Milano2017

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Re: M26-01 [#permalink]

13 Jul 2017, 18:57

Kudos

I find the official solution to be very long and it would take me probably 4-5 minutes to finish the answer. By then I would probably pick 60 since i would forget to take the square root.

My approach was using the approximate numbers for each. First I recognized that sqroot of 6 is 2.5 so the second part was equal to 0. The first part was 25+25=50. Square root of 50 is just over 7. Then I looked that answers and found the number closest to the one I got. It will take you probably less than 1:30 to solve using this method.

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Re: M26-01 [#permalink]

11 Oct 2017, 19:50

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Let \(25 + 10\sqrt{6}\ = A\) &
Let \(25 - 10\sqrt{6}\ = B =>\)

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

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Re: M26-01 [#permalink]

20 Jul 2019, 11:07

Kudos

Imo, it's much easier/faster to estimate, especially since D, E are so far apart from A,B,C:
I knew that √5 ≈ 2.25 so let's call √6 ≈ 2.4 (actual is 2.449...)

\(\sqrt{25+10*2.4}+\sqrt{25-10*2.4}\)
\(\sqrt{25+24}+\sqrt{25-24}\)
\(\sqrt{49}+\sqrt{1}\)
\(7 + 1\)
So our answer should be close to 8, D,E are out immediately
C can be rewritten as \(\sqrt{60}\) which is closer than B \(\sqrt{55}\) , so it's C.

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Re: M26-01 [#permalink]

25 Sep 2019, 18:35

I think this is a high-quality question and I agree with explanation.

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Re: M26-01 [#permalink]

03 Oct 2020, 10:32

I think this is a high-quality question and I agree with explanation.

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Re: M26-01 [#permalink]

01 Aug 2021, 20:59

I think this is a high-quality question and I agree with explanation.

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Re: M26-01 [#permalink]

26 Mar 2023, 23:21

My 2 cents : I guess since options are spread pretty far apart except B and C, We can try approximating the answer
\(\sqrt{6}\) is approx 2.4 and \(10\sqrt{6}\) is approx 24. so 1st sqr root can be approximate to \(\sqrt{25+24}\) i.e aprrox \(\sqrt{49}\) i.e 7 and 2nd sqr root can be approximate to \(1\) similarly. Total is \(8\) (approx).

option A : \(2\sqrt{5}\) is \(2*2.2\) = 4.4
option B : \(\sqrt{55}\) is between 7 and 8 as \(\sqrt{49}\) is 7 and \(\sqrt{64}\) is 8 but since 55 is closer to 49 than 64 so value is less than 7.5.
option C : \(2*\sqrt{15}\) , \(\sqrt{15}\) pretty close to 4 , and hence 2*4 =8 . Our required answer.
option D and E can be ignored as they are pretty far away from required answer.

lotif

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Re: M26-01 [#permalink]

06 May 2023, 05:12

In the exam how is one meant to know this is the approach to take? It is not one that for me was intuitive, to take the square then unsquare in the end. If one takes the wrong approach, then you will hve to move on as not enough time to take more than 1 approach. How cn oyu know by just looking at it to do this?

Bunuel

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Re: M26-01 [#permalink]

06 May 2023, 12:57

Expert Reply

rezalotif wrote:

Squaring an expression containing square roots is a common algebraic technique used to eliminate the square roots and simplify the expression. Conversely, taking the square root is an operation performed to balance the initial squaring. Gaining proficiency in these techniques often comes with practice.
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Bunuel

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Re: M26-01 [#permalink]

13 Jul 2023, 01:45

Expert Reply

I have edited the question and the solution by adding more details to enhance its clarity. I hope it is now easier to understand.
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Re M26-01 [#permalink]

09 Aug 2023, 06:23

I think this is a high-quality question and I agree with explanation.
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