Question

If \(\sqrt{x + 6\sqrt{2}} - \sqrt{x - 6\sqrt{2}} = 2\sqrt{2}\), then \(x\) equals:

• A. 2
• B. 17
• C. 11
• D. 13
• E. 12

Hint:

For equations of the form $\sqrt{a+\sqrt{b}}$, try to express the term under the square root as a perfect square of the form $(m+\sqrt{n})^2$.

Answer 1

The Correct Option is C

Step 1: Square the Equation:
Let $A = \sqrt{x + 6\sqrt{2}}$ and $B = \sqrt{x - 6\sqrt{2}}$. Given $A - B = 2\sqrt{2}$. Square both sides: $(A - (b)^2 = (2\sqrt{2})^2 = 8$. $A^2 + B^2 - 2AB = 8$. $(x + 6\sqrt{2}) + (x - 6\sqrt{2}) - 2\sqrt{(x+6\sqrt{2})(x-6\sqrt{2})} = 8$ $2x - 2\sqrt{x^2 - (6\sqrt{2})^2} = 8$ $2x - 2\sqrt{x^2 - 72} = 8$.
Step 2: Solve for the Square Root:
Divide by 2: $x - \sqrt{x^2 - 72} = 4$. $\sqrt{x^2 - 72} = x - 4$.
Step 3: Square Again:
Square both sides: $x^2 - 72 = (x - 4)^2 = x^2 - 8x + 16$. Cancel $x^2$: $-72 = -8x + 16$. $8x = 16 + 72 = 88 \implies x = 11$.
Step 4: Verify:
Check $x=11$: LHS = $\sqrt{11+6\sqrt{2}} - \sqrt{11-6\sqrt{2}}$. Note $11+6\sqrt{2} = (3+\sqrt{2})^2$ and $11-6\sqrt{2} = (3-\sqrt{2})^2$. LHS = $(3+\sqrt{2}) - (3-\sqrt{2}) = 2\sqrt{2}$, which matches RHS.
Step 5: Final Answer:
$x = 11$.