Question

If \(x=3+2\sqrt 2\) what will be the value of \((x^2+\frac {1}{x^2})\)?

• A. 35
• B. 32
• C. 36
• D. 34

Answer 1

The Correct Option is D

Given \(x = 3 + 2\sqrt{2}\), we are to find the value of \((x^2 + \frac{1}{x^2})\).

First, let's calculate \(x^2\): 

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

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

Using the identity \((a+b)^2 = a^2 + 2ab + b^2\), we have:

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

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

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

Next, let's calculate \(\frac{1}{x}\) and then \(\frac{1}{x^2}\):

To find \(\frac{1}{x}\), rationalize the denominator:

\(\frac{1}{x} = \frac{1}{3 + 2\sqrt{2}}\)

Multiply numerator and denominator by the conjugate:

\(\frac{1}{x} = \frac{1 \cdot (3 - 2\sqrt{2})}{(3 + 2\sqrt{2})(3 - 2\sqrt{2})}\)

\(= \frac{3 - 2\sqrt{2}}{9 - (2\sqrt{2})^2}\)

\(= \frac{3 - 2\sqrt{2}}{9 - 8}\)

\(= 3 - 2\sqrt{2}\)

Now calculate \(\frac{1}{x^2}\):

\(\frac{1}{x^2} = (3 - 2\sqrt{2})^2\)

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

\(= 17 - 12\sqrt{2}\)

Add \(x^2\) and \(\frac{1}{x^2}\):

\(x^2 + \frac{1}{x^2} = (17 + 12\sqrt{2}) + (17 - 12\sqrt{2})\)

\(= 17 + 17\)

\(= 34\)

The value of \((x^2 + \frac{1}{x^2})\) is \(\boxed{34}\).