If $x = 3 + \sqrt{8}$, then $x^2 + \frac{1}{x^2}$ is equal to

### Concept & Formula This problem is solved using the algebraic identity for the sum of squares and the properties of rationalizing surds. The core identity used is: $$ (x + \frac{1}{x})^2 = x^2 + \frac{1}{x^2} + 2 $$ ### Step-by-Step Solution * **Given:** $x = 3 + \sqrt{8}$ * First, find the value of the reciprocal, $\frac{1}{x}$. * $\frac{1}{x} = \frac{1}{3 + \sqrt{8}}$ * Rationalize the denominator by multiplying both numerator and denominator by the conjugate $(3 - \sqrt{8})$: * $\frac{1}{x} = \frac{3 - \sqrt{8}}{(3 + \sqrt{8})(3 - \sqrt{8})}$ * Apply the difference of squares formula $(a+b)(a-b) = a^2 - b^2$ to the denominator: * $\frac{1}{x} = \frac{3 - \sqrt{8}}{3^2 - (\sqrt{8})^2} = \frac{3 - \sqrt{8}}{9 - 8} = 3 - \sqrt{8}$ * Now, find the sum $x + \frac{1}{x}$: * $x + \frac{1}{x} = (3 + \sqrt{8}) + (3 - \sqrt{8}) = 6$ * Square both sides of the equation: * $(x + \frac{1}{x})^2 = 6^2$ * $x^2 + \frac{1}{x^2} + 2(x)(\frac{1}{x}) = 36$ * $x^2 + \frac{1}{x^2} + 2 = 36$ * $x^2 + \frac{1}{x^2} = 34$ ### Exam Strategy & Shortcut When given $x = a + \sqrt{b}$ and the difference of their squares is 1 (i.e., $a^2 - b = 1$), the reciprocal $\frac{1}{x}$ is simply the conjugate $a - \sqrt{b}$. Here, $3^2 - 8 = 9 - 8 = 1$. So, $\frac{1}{x} = 3 - \sqrt{8}$ instantly. The sum $x + \frac{1}{x}$ is always twice the rational term ($2a$), which is $2(3) = 6$. The required value $x^2 + \frac{1}{x^2}$ is simply $(2a)^2 - 2 = 36 - 2 = 34$. This mental check eliminates the need for written calculation. ### Common Pitfall A frequent mistake is attempting to directly square the initial value $x = 3 + \sqrt{8}$ to find $x^2$, and then doing the same for the fraction. This creates messy, error-prone arithmetic. Always use the $x + \frac{1}{x}$ identity for these patterns. ### Final Answer **Therefore, the correct answer is 34.**