The number sqrt(63-36sqrt(3)) can be expressed as x + y sqrt(3) for some integers….

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Question: The number sqrt(63-36sqrt(3)) can be expressed as x + y sqrt(3) for some integers x and y. What is the value of xy ?

Options:

• A. – 18
• B. -6
• C. 6
• D. 18
• E. 27

Solution

We found that (\sqrt{63-36\sqrt{3}} = 6 + (-3)\sqrt{3}), so (x = 6) and (y = -3).

Therefore: (xy = (6)(-3) = -18)

Answer: A. -18

Execution of Strategic Approach

1. Translate the problem requirements

Let’s start by understanding what we’re being asked to do. We have this complicated-looking expression under a square root: (\sqrt{63-36\sqrt{3}}). The problem tells us this can be written in a much simpler form: (x + y\sqrt{3}), where x and y are regular whole numbers (integers).

Think of it like unwrapping a present – we know there’s something simpler inside this complex square root expression, and we need to find what it is. Once we find the values of x and y, we need to multiply them together to get our final answer.

What we’re looking for: Find integers x and y such that (\sqrt{63-36\sqrt{3}} = x + y\sqrt{3}), then calculate xy.

Process Skill: TRANSLATE – Converting the complex radical expression into a clear mathematical goal

2. Set up the denesting assumption

Here’s our strategy: instead of trying to simplify the square root directly (which would be very difficult), let’s work backwards. We’ll assume that (\sqrt{63-36\sqrt{3}}) equals some expression (a + b\sqrt{3}), and then see what values of a and b make this true.

If (\sqrt{63-36\sqrt{3}} = a + b\sqrt{3}), then what happens when we square both sides? The square root and the square will cancel on the left side, giving us:

(63 – 36\sqrt{3} = (a + b\sqrt{3})^2)

Now let’s expand the right side using the formula ((u + v)^2 = u^2 + 2uv + v^2):

((a + b\sqrt{3})^2 = a^2 + 2ab\sqrt{3} + (b\sqrt{3})^2)

(= a^2 + 2ab\sqrt{3} + b^2(3))

(= a^2 + 3b^2 + 2ab\sqrt{3})

So our equation becomes: (63 – 36\sqrt{3} = a^2 + 3b^2 + 2ab\sqrt{3})

3. Create and solve the system of equations

Now comes the clever part! On the left side, we have (63 – 36\sqrt{3}), and on the right side, we have (a^2 + 3b^2 + 2ab\sqrt{3}). For these to be equal, the parts without (\sqrt{3}) must be equal, and the parts with (\sqrt{3}) must be equal.

Comparing the rational parts (no (\sqrt{3})): (63 = a^2 + 3b^2)

Comparing the irrational parts (with (\sqrt{3})): (-36 = 2ab)

From the second equation: (2ab = -36), so (ab = -18)

This means either a is positive and b is negative, or a is negative and b is positive.

From the first equation: (a^2 + 3b^2 = 63)

Let’s try some integer values. Since (ab = -18), possible pairs are: (18,-1), (-18,1), (9,-2), (-9,2), (6,-3), (-6,3), (3,-6), (-3,6), (2,-9), (-2,9), (1,-18), (-1,18)

Testing (a = 9, b = -2): (a^2 + 3b^2 = 81 + 3(4) = 81 + 12 = 93 \neq 63)

Testing (a = 6, b = -3): (a^2 + 3b^2 = 36 + 3(9) = 36 + 27 = 63) ✓

Testing (a = -6, b = 3): (a^2 + 3b^2 = 36 + 3(9) = 36 + 27 = 63) ✓

Process Skill: MANIPULATE – Using algebraic techniques to separate rational and irrational components

4. Determine the correct signs and calculate the product

We found two possibilities: ((a,b) = (6,-3)) or ((a,b) = (-6,3)). But remember, we need (\sqrt{63-36\sqrt{3}}) to be positive since it’s a square root of a positive number.

Let’s check which one gives a positive result:

For (6,-3): (6 + (-3)\sqrt{3} = 6 – 3\sqrt{3})

Since (\sqrt{3} \approx 1.73), we have (3\sqrt{3} \approx 5.19), so (6 – 3\sqrt{3} \approx 6 – 5.19 = 0.81 > 0) ✓

For (-6,3): (-6 + 3\sqrt{3} = -6 + 3\sqrt{3})

Since (3\sqrt{3} \approx 5.19), we have (-6 + 3\sqrt{3} \approx -6 + 5.19 = -0.81 < 0) ✗

Therefore, the correct form is (\sqrt{63-36\sqrt{3}} = 6 – 3\sqrt{3}), which means (x = 6) and (y = -3).

Verification: Let’s double-check by squaring (6 – 3\sqrt{3}):

((6 – 3\sqrt{3})^2 = 36 – 36\sqrt{3} + 9(3) = 36 – 36\sqrt{3} + 27 = 63 – 36\sqrt{3}) ✓

Process Skill: APPLY CONSTRAINTS – Using the requirement that square roots must be positive to choose the correct solution

4. Final Answer

We found that (\sqrt{63-36\sqrt{3}} = 6 + (-3)\sqrt{3}), so (x = 6) and (y = -3).

Therefore: (xy = (6)(-3) = -18)

Answer: A. -18

Common Faltering Points

Errors while devising the approach

Faltering Point 1: Attempting direct simplification instead of denesting

Many students try to simplify (\sqrt{63-36\sqrt{3}}) directly by factoring or using basic radical properties, which leads nowhere. The key insight is to assume the answer has the form (a + b\sqrt{3}) and work backwards by squaring both sides. Students often miss this denesting technique because it’s not immediately obvious.

Faltering Point 2: Misunderstanding the constraint that x and y must be integers

The problem specifically states that x and y are integers, but students might overlook this constraint and attempt to find any real number solutions. This constraint is crucial because it limits the possible values we need to test when solving the system of equations.

Errors while executing the approach

Faltering Point 1: Incorrectly expanding ((a + b\sqrt{3})^2)

When expanding ((a + b\sqrt{3})^2), students commonly make algebraic errors such as forgetting the middle term (2ab\sqrt{3}) or incorrectly computing ((b\sqrt{3})^2 = b^2\sqrt{3}) instead of (3b^2). The correct expansion should be (a^2 + 2ab\sqrt{3} + 3b^2).

Faltering Point 2: Failing to properly separate rational and irrational parts

After setting (63 – 36\sqrt{3} = a^2 + 3b^2 + 2ab\sqrt{3}), students might not recognize that they need to equate the coefficients separately. The rational parts ((63 = a^2 + 3b^2)) and irrational parts ((-36 = 2ab)) must be treated as separate equations.

Faltering Point 3: Arithmetic errors when testing factor pairs

When finding integer solutions to (ab = -18) and (a^2 + 3b^2 = 63), students often make calculation errors. For example, when testing (a = 6, b = -3), they might incorrectly compute (6^2 + 3(-3)^2) or make sign errors in the multiplication.

Errors while selecting the answer

Faltering Point 1: Choosing the negative square root solution

After finding both (6, -3) and (-6, 3) as valid algebraic solutions, students might forget that (\sqrt{63-36\sqrt{3}}) must be positive since it represents a principal square root. They need to verify which solution gives a positive result: (6 – 3\sqrt{3} \approx 0.81 > 0) (correct) vs. (-6 + 3\sqrt{3} \approx -0.81 < 0) (incorrect).