I didn't notice this. Can you write down your explanation thoroughly?

a and b are perfect squares as the addition of their square roots equal to an integer (8). In this case, only a=25 and b=9 match the conditions. So, answer is 15 ( after some calculation, right ?

\(\sqrt{a} + \sqrt{b} =8\)

For a under square root and b as well to give 8 a and b must be perfect squares such as \(\sqrt{16} + \sqrt{16} = 4+4=8\)

However, a-b=16 which seems to contradict the statement of before.

BUT this gives us a clue: a and b must be a perfect square. As such, for a-b =16 follow that a must be 25 and b must be 9, which are perfect squares.

\(25 \times 9= 225\) which is a perfect square and is equal to 15

E is the answer

GIVEN: √a + √b = 8


GIVEN: a - b = 16
We can think of a - b as a difference of squares, since (√a)² = a, and (√b)² = b

So, take a - b = 16
Rewrite as: (√a + √b)(√a - √b) = 16
Substitute to get: (8)(√a - √b) = 16
Divide both sides by 8 to get: √a - √b = 2

We now have:
√a + √b = 8
√a - √b = 2
ADD the equations to get: 2√a = 10, which means √a = 5

Take:
√a + √b = 8
√a - √b = 2
SUBTRACT the bottom equation from the top equation to get: 2√b = 6, which means √b = 3

So, √(ab) = (√a)(√b) = (5)(3) = 15

Answer: E

Cheers,
Brent

Here is an algebraic solution to the problem.
First of all, we need to simplify the question.
Let's say: The sum of two numbers/integers is equal to 8 and the difference between the square of the two numbers/integers is equal to 16. Find the two numbers/integers and calculate the value of the square root of the product of the two numbers/integers.

Let's \(\sqrt{a}\) be the first number/integer and \(\sqrt{b}\) the second number/integer

\(\sqrt{a}+\sqrt{b}=8\)
\(\sqrt{a}^2-\sqrt{b}^2=16\)
\((\sqrt{a}+\sqrt{b})(\sqrt{a}-\sqrt{b}) =16\)
\(8(\sqrt{a}-\sqrt{b})=16\)
\(\sqrt{a}-\sqrt{b}=2\)

Then solve the two equations
\(\sqrt{a}+\sqrt{b}=8\)
\(\sqrt{a}-\sqrt{b}=2\)
\(2\sqrt{a}=10\)
\(\sqrt{a}=5\)
\(\sqrt{b}=3\)

Thus, \(a=25\)
and \(b=9\)
\(\sqrt{ab}=\sqrt{25*9}\)

The answer is 15.
I hope that the formula will work

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