Recall that $(a + b)^2 = a^2 + 2ab + b^2$. Let $a = x^{-1/2}$ and $b = x^{1/2}$, and find $a^2$, $b^2$, and $ab$.

When you multiply $x^{-1/2}$ and $x^{1/2}$, add the exponents. Also, when you square $x^{-1/2}$ or $x^{1/2}$, multiply the exponents by 2.

After expanding $(x^{-1/2} + x^{1/2})^2$, remember to subtract 2 and combine like terms at the end.

In Desmos, type f(x) = (x^(-1/2) + x^(1/2))^2 - 2. Make sure your calculator is set to allow only $x>0$ (for example, by looking at positive $x$-values in the table or graph, since square roots require $x>0$).

Type the following as separate functions: g(x) = x - 1/x, h(x) = (x^(1/2) + x^(-1/2))^2, j(x) = x + 1/x, and k(x) = x^2 + 1/x^2.

Either:

• Use a table for each function (tap the function name and select "table") and plug in several positive values of $x$ (such as $0.25$, $1$, $4$), or
• Look at the graphs of $f(x)$ and each choice.

The correct answer is the one whose function matches $f(x)$ for all tested positive $x$-values (same outputs and overlapping graph).

The expression involves a squared binomial:

This suggests using the formula for squaring a sum:

Here, take $a = x^{-1/2}$ and $b = x^{1/2}$.

Apply $(a + b)^2 = a^2 + 2ab + b^2$:

• $a^2 = (x^{-1/2})^2 = x^{-1}$
• $b^2 = (x^{1/2})^2 = x$
• $ab = x^{-1/2} \cdot x^{1/2} = x^{(-1/2 + 1/2)} = x^0 = 1$

So the expansion is:

Now substitute the expanded form back into the original expression:

Combine like terms:

Since $x^{-1} = \dfrac{1}{x}$, the simplified form of the original expression is

which matches choice C.