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

To solve the problem, we need to find the value of \(x^2 + \dfrac{1}{x^2}\) given that \(x = 3 + 2\sqrt{2}\).

1. First, let's find the value of \(x^2\).
2. Using the identity \((a + b)^2 = a^2 + 2ab + b^2\), let \(a = 3\) and \(b = 2\sqrt{2}\). Then:

\(x^2 = (3 + 2\sqrt{2})^2 = 3^2 + 2 \times 3 \times 2\sqrt{2} + (2\sqrt{2})^2\) 

1. Calculate each term:

\(= 9 + 12\sqrt{2} + 8\)

1. Combine the numbers:

\(x^2 = 17 + 12\sqrt{2}\)

1. Next, find the value of \(\dfrac{1}{x}\).

\(= 3 + 2\sqrt{2}\) is the conjugate of \(3 - 2\sqrt{2}\), then \(\dfrac{1}{x} = 3 - 2\sqrt{2}\\)

1. To find \(x^2 + \dfrac{1}{x^2}\), we use the identity \((a + b) + (a - b) = 2a\).

\(3 + 2\sqrt{2} + (3 - 2\sqrt{2}) = 2 \times 3 = 6\)

1. Using the identity \(x^2 + \dfrac{1}{x^2} = (x + \dfrac{1}{x})^2 - 2\), where \(x + \dfrac{1}{x} = 6\).

\((x^2 + \dfrac{1}{x^2}) = (6)^2 - 2 = 36 - 2 = 34\)

1. Thus, the solution yields \(x^2 + \dfrac{1}{x^2} = 34\).

Therefore, the correct answer is 34.

Answer 2

Step 1: Rationalize to find 1/x.
We have \(x = 3 + 2\sqrt{2}\). Multiply the numerator and denominator by the conjugate \(3 - 2\sqrt{2}\) to get \(1/x\).
\(\dfrac{1}{x} = \dfrac{1}{3+2\sqrt{2}} \times \dfrac{3-2\sqrt{2}}{3-2\sqrt{2}} = \dfrac{3-2\sqrt{2}}{9-8} = 3 - 2\sqrt{2}\).

Step 2: Add x and 1/x.
\(x + \dfrac{1}{x} = (3+2\sqrt{2}) + (3-2\sqrt{2}) = 6\). The surd terms cancel out, which is why this trick works so cleanly.

Step 3: Use the identity to get x^2 + 1/x^2.
We know \(\left(x+\dfrac{1}{x}\right)^2 = x^2 + \dfrac{1}{x^2} + 2\), so \(x^2+\dfrac{1}{x^2} = \left(x+\dfrac{1}{x}\right)^2 - 2\). Put in the value: \(x^2+\dfrac{1}{x^2} = 6^2 - 2 = 36 - 2 = 34\).

Final Answer:
The value of x^2 + 1/x^2 is 34.
\[ \boxed{34} \]