Understanding the Problem: Finding \(x^2 + \frac{1}{x^2}\)

We are given the value of \(x\) as the sum of two square roots: \(x = \sqrt3 + \sqrt2\). Our goal is to find the numerical value of the expression \(x^2 + \frac{1}{x^2}\).

To solve this problem, we need to first calculate the values of \(x^2\) and \(\frac{1}{x^2}\) and then add them together. There are two main ways to approach this: either calculate \(x^2\) and \(\frac{1}{x^2}\) directly and sum them, or calculate \(x + \frac{1}{x}\) or \(x - \frac{1}{x}\) first and then use an algebraic identity.

Step-by-Step Calculation of \(x^2 + \frac{1}{x^2}\)

Method 1: Direct Calculation of \(x^2\) and \(\frac{1}{x^2}\)

Step 1: Calculate \(x^2\)

Given \(x = \sqrt3 + \sqrt2\), we square \(x\) to find \(x^2\):

Using the algebraic identity for squaring a sum, \((a+b)^2 = a^2 + 2ab + b^2\), where \(a = \sqrt3\) and \(b = \sqrt2\):

Thus, \(x^2 = 5 + 2\sqrt6\).

Step 2: Calculate \(\frac{1}{x}\) and then \(\frac{1}{x^2}\)

First, let's find the reciprocal of \(x\), which is \(\frac{1}{x}\):

To simplify this expression and remove the square roots from the denominator, we perform rationalization. We multiply the numerator and the denominator by the conjugate of the denominator. The conjugate of \(\sqrt3 + \sqrt2\) is \(\sqrt3 - \sqrt2\).

The denominator is in the form \((a+b)(a-b)\), which simplifies to \(a^2 - b^2\). Here \(a = \sqrt3\) and \(b = \sqrt2\):

Now that we have \(\frac{1}{x}\), we can find \(\frac{1}{x^2}\) by squaring the value of \(\frac{1}{x}\):

Using the algebraic identity for squaring a difference, \((a-b)^2 = a^2 - 2ab + b^2\), where \(a = \sqrt3\) and \(b = \sqrt2\):

Thus, \(\frac{1}{x^2} = 5 - 2\sqrt6\).

Step 3: Calculate \(x^2 + \frac{1}{x^2}\)

Now, we add the calculated values of \(x^2\) and \(\frac{1}{x^2}\):

Remove the parentheses and combine like terms:

Method 2: Using Algebraic Identities

We know that \(x = \sqrt3 + \sqrt2\) and we found that \(\frac{1}{x} = \sqrt3 - \sqrt2\).

Consider the sum \(x + \frac{1}{x}\):

We use the algebraic identity \((a+b)^2 = a^2 + b^2 + 2ab\). If we let \(a=x\) and \(b=\frac{1}{x}\), we get:

Rearranging this identity to solve for \(x^2 + \frac{1}{x^2}\):

Substitute the value of \(x + \frac{1}{x}\) we found:

Both methods yield the same result.

Final Result for \(x^2 + \frac{1}{x^2}\)

When \(x = \sqrt3 + \sqrt2\), the value of \(x^2 + \frac{1}{x^2}\) is 10.

Revision Table: Algebra Concepts for Square Roots

Additional Information: Properties of Square Roots and Expressions

Understanding the basic properties of square roots is crucial for solving problems like this:

• \(\sqrt a \times \sqrt b = \sqrt{ab}\)
• \((\sqrt a)^2 = a\) for \(a \ge 0\)
• \(\sqrt a + \sqrt a = 2\sqrt a\), but \(\sqrt a + \sqrt b\) cannot be simplified further if \(a\) and \(b\) are different and not perfect squares that lead to like terms.

The structure of the expression \(x = \sqrt3 + \sqrt2\) is quite common in problems that require finding \(x^2 + \frac{1}{x^2}\) or similar forms. This is because the reciprocal \(\frac{1}{x}\) simplifies nicely to \(\sqrt3 - \sqrt2\) after rationalization, due to the difference between the numbers under the square roots being 1 (\(3-2=1\)). This leads to the cancellation of the radical terms (\(\pm 2\sqrt6\)) when computing \(x^2 + \frac{1}{x^2}\), resulting in a simple integer.

The problem tests the ability to manipulate algebraic expressions involving square roots, including squaring binomials and rationalizing denominators, as well as recognizing useful algebraic identities.