Question

If \(X = 2 + \sqrt{3}\), then the value of \( \sqrt{X} + \dfrac{1}{\sqrt{X}} \) is:

• A. \(\sqrt{3}\)
• B. \(\sqrt{6}\)
• C. \(2\sqrt{6}\)
• D. \(6\)

Hint:

When dealing with expressions like \(\sqrt{X} + \frac{1}{\sqrt{X}}\), squaring helps eliminate radicals and makes use of the identity \(y^2 = X + \frac{1}{X} + 2\).

Answer 1

The Correct Option is B

Step 1: Expression to evaluate
We need: \[ \sqrt{X} + \frac{1}{\sqrt{X}}. \] Step 2: Rationalize and simplify
Let \[ y = \sqrt{X} + \frac{1}{\sqrt{X}}. \] Then \[ y^2 = X + \frac{1}{X} + 2. \] Step 3: Simplify \(X + \tfrac{1}{X}\)
Given \(X = 2 + \sqrt{3}\).
\[ \frac{1}{X} = \frac{1}{2+\sqrt{3}} \times \frac{2-\sqrt{3}}{2-\sqrt{3}} = 2 - \sqrt{3}. \] So \[ X + \frac{1}{X} = (2 + \sqrt{3}) + (2 - \sqrt{3}) = 4. \] Step 4: Substitute into \(y^2\)
\[ y^2 = X + \frac{1}{X} + 2 = 4 + 2 = 6. \] \[ y = \sqrt{6}. \] \[ \boxed{\sqrt{6}} \]