Question

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

Answer 1

Step 1 : Let $\sqrt{x + 6\sqrt{2}} = a$ and $\sqrt{x - 6\sqrt{2}} = b$

The equation becomes:
$a - b = 2\sqrt{2}. \quad \text{(1)}$

Square both sides:
$(a - b)^2 = (2\sqrt{2})^2.$

Simplify:
$a^2 - 2ab + b^2 = 8. \quad \text{(2)}$

Step 2 : Express $a^2$ and $b^2$ in terms of $x$

From the definitions:
$a^2 = x + 6\sqrt{2}, \quad b^2 = x - 6\sqrt{2}.$

Add $a^2 + b^2$:
$a^2 + b^2 = (x + 6\sqrt{2}) + (x - 6\sqrt{2}) = 2x. \quad \text{(3)}$

Subtract $a^2 - b^2$:
$a^2 - b^2 = (x + 6\sqrt{2}) - (x - 6\sqrt{2}) = 12\sqrt{2}. \quad \text{(4)}$

Step 3 : Substitute into Equation (2)

From Equation (2):
$a^2 + b^2 - 2ab = 8.$

Substitute $a^2 + b^2 = 2x$:
$2x - 2ab = 8.$

Solve for $ab$:
$ab = \frac{2x - 8}{2} = x - 4. \quad \text{(5)}$

Step 4 : Solve for $a + b$ using $(a - b)(a + b) = a^2 - b^2$

From Equation (1), $a - b = 2\sqrt{2}$, and from Equation (4), $a^2 - b^2 = 12\sqrt{2}$:
$(a - b)(a + b) = a^2 - b^2.$

Substitute:
$(2\sqrt{2})(a + b) = 12\sqrt{2}.$

Simplify:
$a + b = 6. \quad \text{(6)}$

Step 5 : Solve for $a$ and $b$

From the equations:
$a - b = 2\sqrt{2}, \quad a + b = 6,$

add the two equations:
$2a = 6 + 2\sqrt{2}.$

Solve for $a$:
$a = 3 + \sqrt{2}.$

Subtract the two equations:
$2b = 6 - 2\sqrt{2}.$

Solve for $b$:
$b = 3 - \sqrt{2}.$

Step 6 : Use $a^2 = x + 6\sqrt{2}$ to find $x$

Substitute $a = 3 + \sqrt{2}$ into $a^2 = x + 6\sqrt{2}$:
$(3 + \sqrt{2})^2 = x + 6\sqrt{2}.$

Expand $a^2$:
$a^2 = 9 + 6\sqrt{2} + 2 = 11 + 6\sqrt{2}.$

Substitute:
$11 + 6\sqrt{2} = x + 6\sqrt{2}.$

Solve for $x$:
$x = 11.$

Final Answer:
$x = 11.$