Question

If \(x = 3 + 2\sqrt{2}\), find the value of \(x^2 + \dfrac{1}{x^2}\).

• A. 35
• B. 32
• C. 36
• D. 34

Hint:

Rationalize to find 1/x, add it to x, then use (x + 1/x)^2 - 2.

Answer 1

The Correct Option is D

Step 1: Square x directly instead of using the sum trick.
$x = 3+2\sqrt{2}$, so $x^2 = (3+2\sqrt{2})^2 = 9 + 12\sqrt{2} + 8 = 17+12\sqrt{2}$.

Step 2: Find 1/x and square it too.
Rationalizing gives $1/x = 3-2\sqrt{2}$. Squaring this: $1/x^2 = (3-2\sqrt{2})^2 = 9-12\sqrt{2}+8 = 17-12\sqrt{2}$.

Step 3: Add the two squared terms.
$x^2+1/x^2 = (17+12\sqrt{2})+(17-12\sqrt{2}) = 34$. The $\sqrt{2}$ parts cancel out here too, so we do not even need a calculator.

Final Answer:
Both the direct squaring method and the sum-of-reciprocals method agree on 34.
\[ \boxed{34} \]