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If \sqrt{17+\sqrt{264}} can be written in the form \sqrt{a}+\sqrt{b} ,

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If \sqrt{17+\sqrt{264}} can be written in the form \sqrt{a}+\sqrt{b} ,  [#permalink]

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New post 01 Apr 2015, 04:15
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E

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  65% (hard)

Question Stats:

59% (02:15) correct 41% (02:30) wrong based on 243 sessions

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Re: If \sqrt{17+\sqrt{264}} can be written in the form \sqrt{a}+\sqrt{b} ,  [#permalink]

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Re: If \sqrt{17+\sqrt{264}} can be written in the form \sqrt{a}+\sqrt{b} ,  [#permalink]

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New post 01 Apr 2015, 04:29
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Bunuel wrote:
If \(\sqrt{17+\sqrt{264}}\) can be written in the form \(\sqrt{a}+\sqrt{b}\) , where a and b are integers and b < a, then a - b =

A. 1
B. 2
C. 3
D. 4
E. 5


Kudos for a correct solution.


\(16^2\)=256

17+2\(\sqrt{66}\)=a+b+2\(\sqrt{ab}\)
b=6 a=11

a-b=5

Answer: E

a+b=17
a=\(\frac{66}{b}\)
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If \sqrt{17+\sqrt{264}} can be written in the form \sqrt{a}+\sqrt{b} ,  [#permalink]

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New post 01 Apr 2015, 04:46
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1
Bunuel wrote:
If \(\sqrt{17+\sqrt{264}}\) can be written in the form \(\sqrt{a}+\sqrt{b}\) , where a and b are integers and b < a, then a - b =

A. 1
B. 2
C. 3
D. 4
E. 5


Kudos for a correct solution.



IF \(\sqrt{17+\sqrt{264}}\) = \(\sqrt{a}+\sqrt{b}\) THEN
\(17+ \sqrt{264} = a + b +2 \sqrt{ab}\)
\(17+2 * \sqrt{66} = a + b +2 \sqrt{ab}\)
implying that b=6 and a=11

Answer 5. E.
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Re: If \sqrt{17+\sqrt{264}} can be written in the form \sqrt{a}+\sqrt{b} ,  [#permalink]

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New post 01 Apr 2015, 22:13
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Answer = E = 5

\(\sqrt{17+\sqrt{264}} = \sqrt{a}+\sqrt{b}\)

\(17 + \sqrt{264} = a + 2\sqrt{ab} + b\)

\(11 + 2\sqrt{11*6} + 6 = a + 2\sqrt{ab} + b\)

LHS & RHS are similar

a-b = 11-6 = 5
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Re: If \sqrt{17+\sqrt{264}} can be written in the form \sqrt{a}+\sqrt{b} ,  [#permalink]

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New post 04 Apr 2015, 19:45
We can write mentioned equation as; 17+root 264 = (root a+root b)^2= a+b+2 root a*b
Therefore, a+b=17 & 2 root a*b=root 264; square both side we can get value of a*b = 66;
Using above data we can find out a-b; by entering values in (a-b)^2 =(a+b)^2-4ab we get (a-b) =5

Hence answer is E
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Re: If \sqrt{17+\sqrt{264}} can be written in the form \sqrt{a}+\sqrt{b} ,  [#permalink]

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New post 04 Apr 2015, 23:34
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Bunuel wrote:
If \(\sqrt{17+\sqrt{264}}\) can be written in the form \(\sqrt{a}+\sqrt{b}\) , where a and b are integers and b < a, then a - b =

A. 1
B. 2
C. 3
D. 4
E. 5


Kudos for a correct solution.


At first we see that we can't calculate result of this expression \(\sqrt{17+\sqrt{264}}\)
So we should eliminate the root by powering equation
\((\sqrt{17+\sqrt{264}})^2=(\sqrt{a}+\sqrt{b})^2\)
\(17+\sqrt{264}=a + 2sqrt{ab}+b\)

And now we should try to express first part of equation in the form of second equation. As we have 2 before root in right side of equation, we should extract 2 from root in left side of equation:
\(\sqrt{264}=\sqrt{4 * 66}=2\sqrt{66}\)

And as a final step we should find roots for equations \(ab=66\) and \(a+b=17\) this is 6 and 11
And we can write our equation in symmetric view:
\(11+2\sqrt{11*6}+6=a + 2sqrt{ab}+b\)
\(a - b = 11 - 6 = 5\)

So answer is E
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Re: If \sqrt{17+\sqrt{264}} can be written in the form \sqrt{a}+\sqrt{b} ,  [#permalink]

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New post 04 Sep 2017, 07:53
Answer is E
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Re: If \sqrt{17+\sqrt{264}} can be written in the form \sqrt{a}+\sqrt{b} ,  [#permalink]

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New post 23 Nov 2018, 02:30
Bunuel wrote:
If \(\sqrt{17+\sqrt{264}}\) can be written in the form \(\sqrt{a}+\sqrt{b}\) , where a and b are integers and b < a, then a - b =

A. 1
B. 2
C. 3
D. 4
E. 5


Kudos for a correct solution.


The equation can be written as -

\(\sqrt{17+\sqrt{264}}\) = \(\sqrt{a}+\sqrt{b}\)

Squaring on both Sides we get -

a + b = 17, ab = 66

Hence a = 11, b = 6.

a - b = 5
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Re: If \sqrt{17+\sqrt{264}} can be written in the form \sqrt{a}+\sqrt{b} ,   [#permalink] 23 Nov 2018, 02:30
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