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: Compute \(X=2+\sqrt{3}\approx2+1.732=3.732\).

Step 2: Then \(\sqrt{X}\approx\sqrt{3.732}\approx1.9319\), and \(\dfrac{1}{\sqrt{X}}\approx\dfrac{1}{1.9319}\approx0.5176\).

Step 3: Adding these, \(\sqrt{X}+\dfrac{1}{\sqrt{X}}\approx1.9319+0.5176=2.4495\), which matches \(\sqrt{6}\approx2.4495\) to four decimal places. \[ \boxed{\sqrt{6}} \]