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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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01 Apr 2015, 04:15
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56% (01:31) correct 44% (02:09) wrong based on 154 sessions

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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.
[Reveal] Spoiler: OA

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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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01 Apr 2015, 04:29
2
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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

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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01 Apr 2015, 04:46
2
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1
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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.

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

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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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01 Apr 2015, 22:13
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$$\sqrt{17+\sqrt{264}} = \sqrt{a}+\sqrt{b}$$

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

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

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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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

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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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04 Apr 2015, 23:34
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This post was
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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:
$$11+2\sqrt{11*6}+6=a + 2sqrt{ab}+b$$
$$a - b = 11 - 6 = 5$$

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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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06 Apr 2015, 06:01
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Expert's post
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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.

MAGOOSH OFFICIAL SOLUTION:
Attachment:

rootaplusrootb_text.png [ 12.94 KiB | Viewed 3464 times ]

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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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04 Sep 2017, 07:53
see pic for details
Attachments

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