If the sum of the square roots of two integers is

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

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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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.

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

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

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

The following property is applied.
\(\sqrt{a+b+2\sqrt{ab}} = \sqrt{a} + \sqrt{b}\)

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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Here is my solution

Could you elaborate on this JeffTargetTestPrep?

Questions like these are great tools to get better at translations of word problems; there is enough twists and turns in the question to keep you interested till the end.

We have two integers, say x and y.

Sum of the square roots of the two integers = \(\sqrt{x}\) + \(\sqrt{y}\)

Sum of the squares of the two integers =\( x^2\) + \(y^2\)

It is given that \(\sqrt{x}\) +\( \sqrt{y}\) = \(\sqrt{(9+6√2)}\)

The next obvious step is to square both sides to reduce the terms containing roots. Squaring both sides,
x + y + 2\(\sqrt{xy}\) = 9 + 6\(\sqrt{2}\)

We equate the rational parts and irrational parts respectively, giving us,

x + y = 9 and 2\(\sqrt{xy}\) = 6\(\sqrt{2}\)

2\(\sqrt{xy}\) = 6\(\sqrt{2}\) can be simplified to \(\sqrt{xy}\) = 3\(\sqrt{2}\).

Squaring both sides, xy = 9 * 2 = 18.

The only set of values that satisfy x + y = 9 and xy = 18 are 6 and 3. Note that it’s not important as to which one is 6 and which one is 3, since we have to take their sum.
Therefore, \(x^2\) + \(y^2\) = \(6^2\) +\( 3^2\) = 36 + 9 = 45

The correct answer option is C.
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