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

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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))

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

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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.

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

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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.

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

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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

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

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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?

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

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05 Aug 2017, 16:54

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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=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

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

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06 Nov 2017, 09:20

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 : \(a^2-9a+18=0\)

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.

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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