Understanding the Problem: Simplifying Square Roots

The question asks us to find the value of A - B, where A is the positive square root of \(41 + 24\sqrt{2}\) and B is the positive square root of \(36 - 16\sqrt{2}\). To solve this, we need to simplify the expressions inside the square roots. We look for expressions that are perfect squares of the form \((x \pm y\sqrt{z})^2 = x^2 \pm 2xy\sqrt{z} + y^2z\).

Step 1: Simplify A = \(\sqrt{41 + 24\sqrt{2}}\)

We want to express \(41 + 24\sqrt{2}\) in the form \((x + y\sqrt{2})^2 = x^2 + (y\sqrt{2})^2 + 2 \cdot x \cdot y\sqrt{2} = x^2 + 2y^2 + 2xy\sqrt{2}\).

Comparing the coefficients, we need:

• \(x^2 + 2y^2 = 41\)
• \(2xy = 24 \implies xy = 12\)

We need to find positive integers \(x\) and \(y\) such that their product is 12. Possible pairs for \((x, y)\) are \((1, 12), (2, 6), (3, 4), (4, 3), (6, 2), (12, 1)\).

Let's test these pairs in the equation \(x^2 + 2y^2 = 41\):

• For \((x, y) = (1, 12)\): \(1^2 + 2(12^2) = 1 + 2(144) = 1 + 288 = 289 \neq 41\)
• For \((x, y) = (2, 6)\): \(2^2 + 2(6^2) = 4 + 2(36) = 4 + 72 = 76 \neq 41\)
• For \((x, y) = (3, 4)\): \(3^2 + 2(4^2) = 9 + 2(16) = 9 + 32 = 41\). This pair works!

So, \(41 + 24\sqrt{2} = (3 + 4\sqrt{2})^2\). Since A is the positive square root, A = \(\sqrt{(3 + 4\sqrt{2})^2} = |3 + 4\sqrt{2}|\). Since \(3 + 4\sqrt{2}\) is positive, A = \(3 + 4\sqrt{2}\).

Step 2: Simplify B = \(\sqrt{36 - 16\sqrt{2}}\)

We want to express \(36 - 16\sqrt{2}\) in the form \((x - y\sqrt{2})^2 = x^2 + (y\sqrt{2})^2 - 2 \cdot x \cdot y\sqrt{2} = x^2 + 2y^2 - 2xy\sqrt{2}\).

Comparing the coefficients, we need:

• \(x^2 + 2y^2 = 36\)
• \(2xy = 16 \implies xy = 8\)

We need to find positive integers \(x\) and \(y\) such that their product is 8. Possible pairs for \((x, y)\) are \((1, 8), (2, 4), (4, 2), (8, 1)\).

Let's test these pairs in the equation \(x^2 + 2y^2 = 36\):

• For \((x, y) = (1, 8)\): \(1^2 + 2(8^2) = 1 + 2(64) = 1 + 128 = 129 \neq 36\)
• For \((x, y) = (2, 4)\): \(2^2 + 2(4^2) = 4 + 2(16) = 4 + 32 = 36\). This pair works!

So, \(36 - 16\sqrt{2} = (2 - 4\sqrt{2})^2\). Since B is the positive square root, B = \(\sqrt{(2 - 4\sqrt{2})^2} = |2 - 4\sqrt{2}|\).

To evaluate \(|2 - 4\sqrt{2}|\), we compare 2 and \(4\sqrt{2}\). \(2 = \sqrt{4}\) and \(4\sqrt{2} = \sqrt{16 \times 2} = \sqrt{32}\). Since \(\sqrt{4} < \sqrt{32}\), \(2 < 4\sqrt{2}\). Therefore, \(2 - 4\sqrt{2}\) is negative.

The absolute value is \(|2 - 4\sqrt{2}| = -(2 - 4\sqrt{2}) = -2 + 4\sqrt{2} = 4\sqrt{2} - 2\). So B = \(4\sqrt{2} - 2\).

Step 3: Calculate A - B

Now we subtract B from A:

\(A - B = (3 + 4\sqrt{2}) - (4\sqrt{2} - 2)\)

\(A - B = 3 + 4\sqrt{2} - 4\sqrt{2} + 2\)

The terms \(4\sqrt{2}\) and \(-4\sqrt{2}\) cancel each other out.

\(A - B = 3 + 2\)

\(A - B = 5\)

The value of A - B is 5.

Revision Table: Key Concepts

Additional Information: General Method for \(\sqrt{a \pm \sqrt{b}}\)

A general formula exists to simplify square roots of the form \(\sqrt{a \pm \sqrt{b}}\) if it can be expressed as \(\sqrt{x} \pm \sqrt{y}\). The formula is:

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

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

This formula works if \(a^2 - b\) is a perfect square. In our problem, the expressions are \(41 + 24\sqrt{2}\) and \(36 - 16\sqrt{2}\). We can rewrite them to fit the form \(\sqrt{a \pm \sqrt{b}}\) by moving the coefficient inside the inner square root:

• \(41 + 24\sqrt{2} = 41 + \sqrt{24^2 \times 2} = 41 + \sqrt{576 \times 2} = 41 + \sqrt{1152}\). Here \(a=41, b=1152\). Check \(a^2 - b = 41^2 - 1152 = 1681 - 1152 = 529 = 23^2\). This works.
• \(36 - 16\sqrt{2} = 36 - \sqrt{16^2 \times 2} = 36 - \sqrt{256 \times 2} = 36 - \sqrt{512}\). Here \(a=36, b=512\). Check \(a^2 - b = 36^2 - 512 = 1296 - 512 = 784 = 28^2\). This also works.

Let's apply the formula for A:

\(\sqrt{41 + \sqrt{1152}} = \sqrt{\frac{41 + \sqrt{41^2 - 1152}}{2}} + \sqrt{\frac{41 - \sqrt{41^2 - 1152}}{2}}\)

\(= \sqrt{\frac{41 + \sqrt{1681 - 1152}}{2}} + \sqrt{\frac{41 - \sqrt{1681 - 1152}}{2}}\)

\(= \sqrt{\frac{41 + \sqrt{529}}{2}} + \sqrt{\frac{41 - \sqrt{529}}{2}}\)

\(= \sqrt{\frac{41 + 23}{2}} + \sqrt{\frac{41 - 23}{2}}\)

\(= \sqrt{\frac{64}{2}} + \sqrt{\frac{18}{2}}\)

\(= \sqrt{32} + \sqrt{9} = \sqrt{16 \times 2} + 3 = 4\sqrt{2} + 3\)

This matches our previous result for A: \(3 + 4\sqrt{2}\).

Now apply the formula for B:

\(\sqrt{36 - \sqrt{512}} = \sqrt{\frac{36 + \sqrt{36^2 - 512}}{2}} - \sqrt{\frac{36 - \sqrt{36^2 - 512}}{2}}\)

\(= \sqrt{\frac{36 + \sqrt{1296 - 512}}{2}} - \sqrt{\frac{36 - \sqrt{1296 - 512}}{2}}\)

\(= \sqrt{\frac{36 + \sqrt{784}}{2}} - \sqrt{\frac{36 - \sqrt{784}}{2}}\)

\(= \sqrt{\frac{36 + 28}{2}} - \sqrt{\frac{36 - 28}{2}}\)

\(= \sqrt{\frac{64}{2}} - \sqrt{\frac{8}{2}}\)

\(= \sqrt{32} - \sqrt{4} = \sqrt{16 \times 2} - 2 = 4\sqrt{2} - 2\)

This matches our previous result for B: \(4\sqrt{2} - 2\). Note that \(\sqrt{32} - \sqrt{4} = 4\sqrt{2} - 2\) is positive since \(4\sqrt{2} = \sqrt{32}\) is greater than \(2 = \sqrt{4}\).

Both methods yield the same simplified forms for A and B, confirming our steps.