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Math Expert V
Joined: 02 Sep 2009
Posts: 59624
If the sum of the square roots of two integers is  [#permalink]

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Difficulty:   45% (medium)

Question Stats: 69% (02:30) correct 31% (02:45) wrong based on 681 sessions

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If the sum of the square roots of two integers is $$\sqrt{9+6\sqrt{2}}$$, what is the sum of the squares of these two integers?

(A) 40
(B) 43
(C) 45
(D) 48
(C) 52

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Target Test Prep Representative G
Affiliations: Target Test Prep
Joined: 04 Mar 2011
Posts: 2809
Re: If the sum of the square roots of two integers is  [#permalink]

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5
Bunuel wrote:
If the sum of the square roots of two integers is $$\sqrt{9+6\sqrt{2}}$$, what is the sum of the squares of these two integers?

(A) 40
(B) 43
(C) 45
(D) 48
(C) 52

We can let a = the first integer and b = the second integer. Thus:

√a + √b = √(9 + 6√2)

We are asked to find a^2 + b^2.

Let’s square both sides of the equation above.

(√a + √b)^2 = [√(9 + 6√2)]^2

a + 2√ab + b = 9 + 6√2

Since a and b are integers, we must have:

a + b = 9 and 2√ab = 6√2

If we square both sides of a + b = 9, we have:

a^2 + 2ab + b^2 = 81

If we square both sides of 2√ab = 6√2, we have:

4ab = 36(2)

2ab = 36

We can now substitute 36 for 2ab in a^2 + 2ab + b^2 = 81 to obtain:

a^2 + 36 + b^2 = 81

a^2 + b^2 = 45

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Joined: 05 Mar 2015
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Re: If the sum of the square roots of two integers is  [#permalink]

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Bunuel wrote:
If the sum of the square roots of two integers is $$\sqrt{9+6\sqrt{2}}$$, what is the sum of the squares of these two integers?

(A) 40
(B) 43
(C) 45
(D) 48
(C) 52

Let nos be x &y
√x + √y= $$\sqrt{9+6\sqrt{2}}$$
sq both sides.
x+y+2√xy=9+6√2
since x & y are integers
x+y=9----------(1)
and 2√xy=6√2
or √xy=3√2
sq both sides to get xy=18-----(2)

sq . both sides (1)
x^2+y^2+2xy=81
x^2+y^2=81-2xy
x^2+y^2=81-36=45---(as xy=18 from (2))

Ans C
##### General Discussion
Manager  P
Status: Quant Expert Q51
Joined: 02 Aug 2014
Posts: 103
Re: If the sum of the square roots of two integers is  [#permalink]

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12
5
Let a and b be both of the integers.

$$\sqrt{a}+\sqrt{b}=\sqrt{9+6\sqrt{2}}$$

Lets square both sides of the equation

we get

$$a+b+2\sqrt{a}\sqrt{b}=9+6\sqrt{2}$$

Then

$$a+b= 9$$ 

$$2\sqrt{a}\sqrt{b}=6\sqrt{2}$$ 

 $$\sqrt{ab}=3\sqrt{2}$$ lets square both sides $$ab=18$$

so we get a system

$$a+b=9$$
$$ab=18$$

Combining both equations we get : $$a^2-9a+18=0$$

Solving this second degree equation we get : $$a = 3$$ and $$b = 6$$

We are searching for the sum of the squares of these two integers.

so $$a^2+b^2=9+36 = 45$$

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Joined: 30 Jun 2017
Posts: 7
Location: India
Schools: Babson '20 (A)
If the sum of the square roots of two integers is  [#permalink]

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3
AnisMURR wrote:
Let a and b be both of the integers.

$$\sqrt{a}+\sqrt{b}=\sqrt{9+6\sqrt{2}}$$

Lets square both sides of the equation

we get

$$a+b+2\sqrt{a}\sqrt{b}=9+6\sqrt{2}$$

Then

$$a+b= 9$$ 

$$2\sqrt{a}\sqrt{b}=6\sqrt{2}$$ 

 $$\sqrt{ab}=3\sqrt{2}$$ lets square both sides $$ab=18$$

so we get a system

$$a+b=9$$
$$ab=18$$

Combining both equations we get : $$a^2-9a+18=0$$

Solving this second degree equation we get : $$a = 3$$ and $$b = 6$$

We are searching for the sum of the squares of these two integers.

so $$a^2+b^2=9+36 = 45$$

I don't think this method will be helpful in GMAT - where we target a problem not more than 2 min.
Just try this one..
we know that sqaure of integers can only be from terms of the series of 1,4,9,16,25,36,49,64,.......
Further, summation of any two terms from the series should be equal to the one of the options given. It comes out that only 40 (36+4) and 45 (36+9) can be formed from the series of square of integers. By ballparking sqaure root of complex number given comes out to be square root of 18 i.e. slightly more than 4. whereas the summation of sqaure root of 2 & 6 is slightly less than 4 and the summation of sqaure root of 3 & 6 is slightly more than 4. Hence answer is C.
Manager  P
Status: Quant Expert Q51
Joined: 02 Aug 2014
Posts: 103
Re: If the sum of the square roots of two integers is  [#permalink]

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Hello Metwing Nice analysis But beleive me it took me less than 2 minutes.

Best,
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Intern  B
Joined: 25 Apr 2017
Posts: 14
GMAT 1: 710 Q48 V40 Re: If the sum of the square roots of two integers is  [#permalink]

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AnisMURR wrote:
$$a+b=9$$
$$ab=18$$

Combining both equations we get : $$a^2-9a+18=0$$

Please how do you arrive at the above equation from those 2? Can't seem to figure it out. Seems like a step is missing -- as a expert, it is probably obvious to you. But after 30 minutes, I am still clueless.
Senior Manager  G
Joined: 24 Apr 2016
Posts: 316
Re: If the sum of the square roots of two integers is  [#permalink]

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getitdoneright wrote:
AnisMURR wrote:
$$a+b=9$$
$$ab=18$$

Combining both equations we get : $$a^2-9a+18=0$$

Please how do you arrive at the above equation from those 2? Can't seem to figure it out. Seems like a step is missing -- as a expert, it is probably obvious to you. But after 30 minutes, I am still clueless.

a+b = 9

square both sides

$$(a+b)^2 = 9^2$$

$$a^2 + b^2 + 2ab = 81$$

Substituting the value of ab (18) in the above equation

$$a^2 + b^2 + (2*18) = 81$$

$$a^2 + b^2 = 81 - 36 = 45$$

Hope this helps
Manager  B
Joined: 07 Jun 2017
Posts: 100
Re: If the sum of the square roots of two integers is  [#permalink]

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rohit8865 wrote:
Bunuel wrote:
If the sum of the square roots of two integers is $$\sqrt{9+6\sqrt{2}}$$, what is the sum of the squares of these two integers?

(A) 40
(B) 43
(C) 45
(D) 48
(C) 52

Let nos be x &y
√x + √y= $$\sqrt{9+6\sqrt{2}}$$
sq both sides.
x+y+2√xy=9+6√2
since x & y are integers
x+y=9----------(1)
and 2√xy=6√2
or √xy=3√2
sq both sides to get xy=18-----(2)

sq . both sides (1)
x^2+y^2+2xy=81
x^2+y^2=81-2xy
x^2+y^2=81-36=45---(as xy=18 from (2))

Ans C

Dear,
How do you get "x^2+y^2+2xy=81"?
Where is this 81 from?

Thank you so much.
Manager  B
Joined: 19 Aug 2016
Posts: 73
If the sum of the square roots of two integers is  [#permalink]

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1
pclawong wrote:
rohit8865 wrote:
Bunuel wrote:
If the sum of the square roots of two integers is $$\sqrt{9+6\sqrt{2}}$$, what is the sum of the squares of these two integers?

(A) 40
(B) 43
(C) 45
(D) 48
(C) 52

Let nos be x &y
√x + √y= $$\sqrt{9+6\sqrt{2}}$$
sq both sides.
x+y+2√xy=9+6√2
since x & y are integers
x+y=9----------(1)
and 2√xy=6√2
or √xy=3√2
sq both sides to get xy=18-----(2)

sq . both sides (1)
x^2+y^2+2xy=81
x^2+y^2=81-2xy
x^2+y^2=81-36=45---(as xy=18 from (2))

Ans C

Dear,
How do you get "x^2+y^2+2xy=81"?
Where is this 81 from?

Thank you so much.

In the above equation, we have got x+y=9 (eqn 1)so when u square on both sides u will get
x^2+y^2+2xy=81
Senior Manager  G
Status: love the club...
Joined: 24 Mar 2015
Posts: 265
Re: If the sum of the square roots of two integers is  [#permalink]

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Bunuel wrote:
If the sum of the square roots of two integers is $$\sqrt{9+6\sqrt{2}}$$, what is the sum of the squares of these two integers?

(A) 40
(B) 43
(C) 45
(D) 48
(C) 52

hi Bunuel

very high quality question this one is indeed. Can you please provide some links to such questions to practice..?

Math Expert V
Joined: 02 Sep 2009
Posts: 59624
Re: If the sum of the square roots of two integers is  [#permalink]

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gmatcracker2017 wrote:
Bunuel wrote:
If the sum of the square roots of two integers is $$\sqrt{9+6\sqrt{2}}$$, what is the sum of the squares of these two integers?

(A) 40
(B) 43
(C) 45
(D) 48
(C) 52

hi Bunuel

very high quality question this one is indeed. Can you please provide some links to such questions to practice..?

Roots DS Questions
Roots PS Questions

Hope it helps.
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Senior Manager  G
Status: love the club...
Joined: 24 Mar 2015
Posts: 265
Re: If the sum of the square roots of two integers is  [#permalink]

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Bunuel wrote:
gmatcracker2017 wrote:
Bunuel wrote:
If the sum of the square roots of two integers is $$\sqrt{9+6\sqrt{2}}$$, what is the sum of the squares of these two integers?

(A) 40
(B) 43
(C) 45
(D) 48
(C) 52

hi Bunuel

very high quality question this one is indeed. Can you please provide some links to such questions to practice..?

Roots DS Questions
Roots PS Questions

Hope it helps.

thanks man
great you are Current Student G
Joined: 19 Aug 2016
Posts: 145
Location: India
GMAT 1: 640 Q47 V31 GPA: 3.82
Re: If the sum of the square roots of two integers is  [#permalink]

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hi Bunuel

very high quality question this one is indeed. Can you please provide some links to such questions to practice..?

Hello,

I'm still unable to understand the solution. Could you please provide the official solution or another explaination to the question?

Thanks
Senior SC Moderator V
Joined: 22 May 2016
Posts: 3729
If the sum of the square roots of two integers is  [#permalink]

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Bunuel wrote:
If the sum of the square roots of two integers is $$\sqrt{9+6\sqrt{2}}$$, what is the sum of the squares of these two integers?

(A) 40
(B) 43
(C) 45
(D) 48
(C) 52

AnisMURR wrote:
Let a and b be both of the integers.

$$\sqrt{a}+\sqrt{b}=\sqrt{9+6\sqrt{2}}$$

Lets square both sides of the equation

we get

$$a+b+2\sqrt{a}\sqrt{b}=9+6\sqrt{2}$$

Then

$$a+b= 9$$ 

$$2\sqrt{a}\sqrt{b}=6\sqrt{2}$$ 

 $$\sqrt{ab}=3\sqrt{2}$$ lets square both sides $$ab=18$$

so we get a system

$$a+b=9$$
$$ab=18$$

Combining both equations we get : $$a^2-9a+18=0$$

Solving this second degree equation we get : $$a = 3$$ and $$b = 6$$

We are searching for the sum of the squares of these two integers.

so $$a^2+b^2=9+36 = 45$$

AnisMURR , I can follow everything if I accept this part's last line:
Quote:
$$\sqrt{a}+\sqrt{b}=\sqrt{9+6\sqrt{2}}$$

Lets square both sides of the equation

we get

$$a+b+2\sqrt{a}\sqrt{b}=9+6\sqrt{2}$$

It looks as if you've gotten to a version of a square of a sum (?):
$$(a + b)^2 = a^2 + 2ab + b^2$$
Why does (a + b) = 9?
Put another way, why is there not a separate "b" (or analogous b^2?) term?

I think I am missing something really obvious.
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Joined: 27 May 2012
Posts: 945
Re: If the sum of the square roots of two integers is  [#permalink]

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JeffTargetTestPrep wrote:
Bunuel wrote:
If the sum of the square roots of two integers is $$\sqrt{9+6\sqrt{2}}$$, what is the sum of the squares of these two integers?

(A) 40
(B) 43
(C) 45
(D) 48
(C) 52

We can let a = the first integer and b = the second integer. Thus:

√a + √b = √(9 + 6√2)

We are asked to find a^2 + b^2.

Let’s square both sides of the equation above.

(√a + √b)^2 = [√(9 + 6√2)]^2

a + 2√ab + b = 9 + 6√2

Since a and b are integers, we must have:

a + b = 9 and 2√ab = 6√2

If we square both sides of a + b = 9, we have:

a^2 + 2ab + b^2 = 81

If we square both sides of 2√ab = 6√2, we have:

4ab = 36(2)

2ab = 36

We can now substitute 36 for 2ab in a^2 + 2ab + b^2 = 81 to obtain:

a^2 + 36 + b^2 = 81

a^2 + b^2 = 45
=

Just curious if the individual values of the two integers are 6 and 3 or 3 and 6 respectively then of course on squaring both $$6^2 + 3^2 = 45$$
but how on taking square root of 6 and 3 and summing them do we get $$\sqrt {9 + 6\sqrt{2}}$$
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Re: If the sum of the square roots of two integers is  [#permalink]

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stne wrote:
JeffTargetTestPrep wrote:
Bunuel wrote:
If the sum of the square roots of two integers is $$\sqrt{9+6\sqrt{2}}$$, what is the sum of the squares of these two integers?

(A) 40
(B) 43
(C) 45
(D) 48
(C) 52

We can let a = the first integer and b = the second integer. Thus:

√a + √b = √(9 + 6√2)

We are asked to find a^2 + b^2.

Let’s square both sides of the equation above.

(√a + √b)^2 = [√(9 + 6√2)]^2

a + 2√ab + b = 9 + 6√2

Since a and b are integers, we must have:

a + b = 9 and 2√ab = 6√2

If we square both sides of a + b = 9, we have:

a^2 + 2ab + b^2 = 81

If we square both sides of 2√ab = 6√2, we have:

4ab = 36(2)

2ab = 36

We can now substitute 36 for 2ab in a^2 + 2ab + b^2 = 81 to obtain:

a^2 + 36 + b^2 = 81

a^2 + b^2 = 45
=

Just curious if the individual values of the two integers are 6 and 3 or 3 and 6 respectively then of course on squaring both $$6^2 + 3^2 = 45$$
but how on taking square root of 6 and 3 and summing them do we get $$\sqrt {9 + 6\sqrt{2}}$$

Hi stne

Here it is said that SUM of square root of integer equals $$\sqrt {9 + 6\sqrt{2}}$$

Now a funny thing about the SUM is that you can arrive at a particular SUM by using various combination. For eg. if I say SUM of two integer is 9, then it can be 6+3 also
or 8+1. So we have 6+3=9=8+1 but 6, 8, 3 & 1 are all different. So if $$\sqrt{6}+\sqrt{3}=4.18154055$$, so is $$\sqrt {9 + 6\sqrt{2}}=4.18154055$$

Hence here we will have to look at the totality and not the individual elements.

I also believe that there might be a way to simplify $$\sqrt{6}+\sqrt{3}$$, and get $$\sqrt {9 + 6\sqrt{2}}$$
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Re: If the sum of the square roots of two integers is  [#permalink]

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niks18 wrote:

Hi stne

Here it is said that SUM of square root of integer equals $$\sqrt {9 + 6\sqrt{2}}$$

Now a funny thing about the SUM is that you can arrive at a particular SUM by using various combination. For eg. if I say SUM of two integer is 9, then it can be 6+3 also
or 8+1. So we have 6+3=9=8+1 but 6, 8, 3 & 1 are all different. So if $$\sqrt{6}+\sqrt{3}=4.18154055$$, so is $$\sqrt {9 + 6\sqrt{2}}=4.18154055$$

Hence here we will have to look at the totality and not the individual elements.

I also believe that there might be a way to simplify $$\sqrt{6}+\sqrt{3}$$, and get $$\sqrt {9 + 6\sqrt{2}}$$

Hi niks18,
Really appreciate your reply,thanks a ton,maybe we have to work our way backwards to arrive at our query. Maybe some one will show us the way some day.Thanks again.
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If the sum of the square roots of two integers is  [#permalink]

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Bunuel wrote:
If the sum of the square roots of two integers is $$\sqrt{9+6\sqrt{2}}$$, what is the sum of the squares of these two integers?

(A) 40
(B) 43
(C) 45
(D) 48
(C) 52

Main Idea:Make the LHS correspond to RHS

Details : Let the integers be x and y. We have sqrt(x) + sqrt(y) = sqrt(9+6*sqrt(2))

Squaring both sides, we have

x+y+2 *sqrt(xy) =9+6*sqrt(2).

6*sqrt(2) can be written as 2*sqrt(18)

So we have x+y+2 *sqrt(xy)=9+2*sqrt(18)

LHs and RHS correspond .

We see x+y=9 and xy=18

Solving we have x=3 and y=6

x^2 +y^2 = 36 +9 =45

Hence C.
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Re: If the sum of the square roots of two integers is  [#permalink]

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1
Bunuel wrote:
If the sum of the square roots of two integers is $$\sqrt{9+6\sqrt{2}}$$, what is the sum of the squares of these two integers?

(A) 40
(B) 43
(C) 45
(D) 48
(C) 52

√( 9 + 6√2) = √(9 + 2√18) = √6 + √3
6^2 + 3^2 = 36 + 9 = 45

The following property is applied.
$$\sqrt{a+b+2\sqrt{ab}} = \sqrt{a} + \sqrt{b}$$
_________________ Re: If the sum of the square roots of two integers is   [#permalink] 24 Apr 2018, 12:21

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