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If the sum of the square roots of two integers is [#permalink]
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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=812xy x^2+y^2=8136=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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30 Oct 2016, 12:18
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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 : \(a^29a+18=0\) 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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17 Jul 2017, 11:57
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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 : \(a^29a+18=0\)
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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19 Jul 2017, 23:02
Hello Metwing Nice analysis But beleive me it took me less than 2 minutes. Best,
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Re: If the sum of the square roots of two integers is [#permalink]
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02 Aug 2017, 16:17
AnisMURR wrote: \(a+b=9\) \(ab=18\)
Combining both equations we get : \(a^29a+18=0\)
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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getitdoneright wrote: AnisMURR wrote: \(a+b=9\) \(ab=18\)
Combining both equations we get : \(a^29a+18=0\)
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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02 Aug 2017, 22:00
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=812xy x^2+y^2=8136=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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05 Aug 2017, 17:54
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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=812xy x^2+y^2=8136=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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30 Sep 2017, 20:08
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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01 Oct 2017, 08:50
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..? thanks in advance, man Roots DS QuestionsRoots PS QuestionsHope it helps. thanks man great you are



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Re: If the sum of the square roots of two integers is [#permalink]
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05 Nov 2017, 12:24
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[/quote][/quote] Hello, I'm still unable to understand the solution. Could you please provide the official solution or another explaination to the question? Thanks
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If the sum of the square roots of two integers is [#permalink]
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06 Nov 2017, 10:20
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
\(a+b= 9\) [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 : \(a^29a+18=0\)
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. 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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Re: If the sum of the square roots of two integers is [#permalink]
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08 Nov 2017, 17:35
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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22 Feb 2018, 07:17
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 = Answer: C 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 = Answer: C 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 stneHere 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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22 Feb 2018, 12:25
niks18 wrote: Hi stneHere 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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08 Mar 2018, 15:53
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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