Finding Value of $\sqrt{x} + \frac{1}{\sqrt{x}}$

We are given the expression $x = 7 + 4\sqrt{3}$ and asked to find the value of $\sqrt{x} + \frac{1}{\sqrt{x}}$.

Simplify $\sqrt{x}$

First, we simplify $\sqrt{x}$. We need to express $7 + 4\sqrt{3}$ as a perfect square, possibly in the form $(a+b)^2 = a^2 + b^2 + 2ab$.

• We can rewrite $4\sqrt{3}$ as $2 \times 2 \times \sqrt{3}$.
• Let $a=2$ and $b=\sqrt{3}$. Then $a^2 = 2^2 = 4$ and $b^2 = (\sqrt{3})^2 = 3$.
• Checking the sum of squares: $a^2 + b^2 = 4 + 3 = 7$. This matches the constant term in the expression for $x$.
• Thus, $x = 7 + 4\sqrt{3}$ can be written as $(2)^2 + (\sqrt{3})^2 + 2(2)(\sqrt{3}) = (2 + \sqrt{3})^2$.
• Therefore, $\sqrt{x} = \sqrt{(2 + \sqrt{3})^2}$. Since $2 + \sqrt{3}$ is positive, $\sqrt{x} = 2 + \sqrt{3}$.

Calculate $\frac{1}{\sqrt{x}}$

Next, we find the value of $\frac{1}{\sqrt{x}}$.

• Substitute the simplified value of $\sqrt{x}$: $\frac{1}{\sqrt{x}} = \frac{1}{2 + \sqrt{3}}$.
• To simplify this fraction, we rationalize the denominator by multiplying the numerator and denominator by the conjugate of $2 + \sqrt{3}$, which is $2 - \sqrt{3}$.
• Calculation: \[ \frac{1}{2 + \sqrt{3}} \times \frac{2 - \sqrt{3}}{2 - \sqrt{3}} = \frac{2 - \sqrt{3}}{(2)^2 - (\sqrt{3})^2} = \frac{2 - \sqrt{3}}{4 - 3} = \frac{2 - \sqrt{3}}{1} = 2 - \sqrt{3} \]
• So, $\frac{1}{\sqrt{x}} = 2 - \sqrt{3}$.

Find the Sum

Finally, we add the values of $\sqrt{x}$ and $\frac{1}{\sqrt{x}}$.

• $\sqrt{x} + \frac{1}{\sqrt{x}} = (2 + \sqrt{3}) + (2 - \sqrt{3})$.
• Combine the terms: $2 + \sqrt{3} + 2 - \sqrt{3} = 4$.

The value of $\sqrt{x} + \frac{1}{\sqrt{x}}$ is 4.