If $(x+6\sqrt{2})^{\frac{1}{2}} - (x-6\sqrt{2})^{\frac{1}{2}} = 2\sqrt{2}$, then x equals

Given: $(x+6\sqrt{2})^{\frac{1}{2}} - (x-6\sqrt{2})^{\frac{1}{2}} = 2\sqrt{2}$

$\sqrt{x+6\sqrt{2}} - \sqrt{x-6\sqrt{2}} = 2\sqrt{2}$

---

Squaring both sides helps eliminate the roots.

$[\sqrt{x+6\sqrt{2}} - \sqrt{x-6\sqrt{2}}]^2 = (2\sqrt{2})^2$

---

The right side: $(2\sqrt{2})^2 = 4 \times 2 = 8$

The left side:

Expand using $(a-b)^2 = a^2 - 2ab + b^2$

Here, $a = \sqrt{x+6\sqrt{2}}$ and $b = \sqrt{x-6\sqrt{2}}$

$(\sqrt{x+6\sqrt{2}})^2 - 2\sqrt{x+6\sqrt{2}} \cdot \sqrt{x-6\sqrt{2}} + (\sqrt{x-6\sqrt{2}})^2 = 8$

$(x+6\sqrt{2}) - 2\sqrt{(x+6\sqrt{2})(x-6\sqrt{2})} + (x-6\sqrt{2}) = 8$

---

The first and third terms combine: $(x+6\sqrt{2}) + (x-6\sqrt{2}) = 2x$

For the middle term, we need to find $(x+6\sqrt{2})(x-6\sqrt{2})$.

This is a difference of squares pattern: $(a+b)(a-b) = a^2 - b^2$

$(x+6\sqrt{2})(x-6\sqrt{2})$

$= x^2 - (6\sqrt{2})^2$

$= x^2 - 36 \times 2$

$= x^2 - 72$

---

Our equation becomes:

$2x - 2\sqrt{x^2-72} = 8$

$x - \sqrt{x^2-72} = 4$

$x - 4 = \sqrt{x^2-72}$

Important conditions: For this to be valid, we need:

$x - 4 \geq 0$ (since square root equals a non-negative number)

$x^2 - 72 \geq 0$ (since we can't take square root of negative numbers in real numbers)

---

$(x-4)^2 = (\sqrt{x^2-72})^2$

$x^2 - 8x + 16 = x^2 - 72$

---

The $x^2$ terms cancel:

$-8x + 16 = -72$

$-8x = -72 - 16$

$x = \frac{88}{8} = 11$