Arithmetic

Roland2rule 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
(E) 52

OA is c.

Can any Expert help me out on how to solve this question correctly? i will gladly appreciate your contribution. Thanks

Roland2rule 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
(E) 52

Scott@TargetTestPrep wrote:

Roland2rule 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
(E) 52

The correct version of this problem is above.

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