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If the sum of the square roots of two integers is

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If the sum of the square roots of two integers is  [#permalink]

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New post 30 Oct 2016, 06:36
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Re: If the sum of the square roots of two integers is  [#permalink]

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New post 08 Nov 2017, 16:35
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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


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

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

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New post 30 Oct 2016, 09:00
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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
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Re: If the sum of the square roots of two integers is  [#permalink]

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New post 30 Oct 2016, 11:18
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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

\(a+b= 9\) [1]

\(2\sqrt{a}\sqrt{b}=6\sqrt{2}\) [2]

[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 \(a^2+b^2=9+36 = 45\)

So the answer is C.
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If the sum of the square roots of two integers is  [#permalink]

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New post 17 Jul 2017, 10:57
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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

\(a+b= 9\) [1]

\(2\sqrt{a}\sqrt{b}=6\sqrt{2}\) [2]

[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 \(a^2+b^2=9+36 = 45\)

So the answer is C.



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

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New post 02 Aug 2017, 15:17
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.
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Re: If the sum of the square roots of two integers is  [#permalink]

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New post 02 Aug 2017, 15:36
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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
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Re: If the sum of the square roots of two integers is  [#permalink]

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New post 02 Aug 2017, 21:00
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.
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If the sum of the square roots of two integers is  [#permalink]

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New post 05 Aug 2017, 16:54
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
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Re: If the sum of the square roots of two integers is  [#permalink]

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New post 30 Sep 2017, 19:08
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..?

thanks in advance, man
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Re: If the sum of the square roots of two integers is  [#permalink]

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New post 01 Oct 2017, 02:59
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..?

thanks in advance, man


Roots DS Questions
Roots PS Questions

Hope it helps.
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If the sum of the square roots of two integers is  [#permalink]

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New post 08 Mar 2018, 14:53
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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New post 24 Apr 2018, 11:21
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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


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

The following property is applied.
\(\sqrt{a+b+2\sqrt{ab}} = \sqrt{a} + \sqrt{b}\)
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Re: If the sum of the square roots of two integers is  [#permalink]

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New post 25 Oct 2018, 00:08
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 the two integers be a and b.

\(\sqrt{a} + \sqrt{b} = \sqrt{9+6\sqrt{2}}\)

Squaring both sides, we get

\(a + b + 2\sqrt{ab} = 9 + 6\sqrt{2}\)

Since a and b are integers, so \(2\sqrt{ab} = 6\sqrt{2}\)

\(\sqrt{ab} = 3*\sqrt{2} = \sqrt{3*3*2}\)

So values of a and b such that ab = 3*3*2 and sum is 9 is 6 and 3.

\(a^2 + b^2 = 6^2 + 3^2 = 45\)

Answer (C)
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Re: If the sum of the square roots of two integers is   [#permalink] 13 May 2020, 06:45

If the sum of the square roots of two integers is

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