Value of `sqrt(5+2sqrt(6))-1/(sqrt(5+2sqrt(6)))` is

Options

A. `2sqrt2`
B. 0
C. `2sqrt3`
D. `sqrt5-1`

Answer

To find the value of the expression \( \sqrt{5 + 2\sqrt{6}} - \frac{1}{\sqrt{5 + 2\sqrt{6}}} \), we can follow these steps:

### Step 1: Simplify \( \sqrt{5 + 2\sqrt{6}} \)

We start by rewriting \( \sqrt{5 + 2\sqrt{6}} \). We can express \( 5 + 2\sqrt{6} \) in a different form. We want to find \( a \) and \( b \) such that:

\[ \sqrt{5 + 2\sqrt{6}} = \sqrt{a} + \sqrt{b} \]

Squaring both sides gives:

\[ 5 + 2\sqrt{6} = a + b + 2\sqrt{ab} \]

From this, we can equate the rational and irrational parts:

1. \( a + b = 5 \)
2. \( 2\sqrt{ab} = 2\sqrt{6} \) which simplifies to \( \sqrt{ab} = \sqrt{6} \) or \( ab = 6 \)

Now we have a system of equations:

\[ a + b = 5 \]
\[ ab = 6 \]

### Step 2: Solve for \( a \) and \( b \)

We can treat \( a \) and \( b \) as the roots of the quadratic equation:

\[ x^2 - (a+b)x + ab = 0 \implies x^2 - 5x + 6 = 0 \]

Factoring this gives:

\[ (x - 2)(x - 3) = 0 \]

Thus, \( a = 2 \) and \( b = 3 \) (or vice versa). Therefore:

\[ \sqrt{5 + 2\sqrt{6}} = \sqrt{2} + \sqrt{3} \]

### Step 3: Substitute back into the original expression

Now we substitute back into the expression:

\[ \sqrt{5 + 2\sqrt{6}} - \frac{1}{\sqrt{5 + 2\sqrt{6}}} = (\sqrt{2} + \sqrt{3}) - \frac{1}{\sqrt{2} + \sqrt{3}} \]

### Step 4: Rationalize the denominator

To simplify \( \frac{1}{\sqrt{2} + \sqrt{3}} \), we multiply the numerator and denominator by \( \sqrt{2} - \sqrt{3} \):

\[ \frac{1}{\sqrt{2} + \sqrt{3}} \cdot \frac{\sqrt{2} - \sqrt{3}}{\sqrt{2} - \sqrt{3}} = \frac{\sqrt{2} - \sqrt{3}}{(\sqrt{2})^2 - (\sqrt{3})^2} = \frac{\sqrt{2} - \sqrt{3}}{2 - 3} = -(\sqrt{2} - \sqrt{3}) = \sqrt{3} - \sqrt{2} \]

### Step 5: Combine the terms

Now we can combine the terms:

\[ \sqrt{2} + \sqrt{3} - (\sqrt{3} - \sqrt{2}) = \sqrt{2} + \sqrt{3} - \sqrt{3} + \sqrt{2} = 2\sqrt{2} \]

### Final Answer

Thus, the value of the expression \( \sqrt{5 + 2\sqrt{6}} - \frac{1}{\sqrt{5 + 2\sqrt{6}}} \) is:

\[ \boxed{2\sqrt{2}} \]