What is the value of
`sqrt(37+20sqrt(3))-sqrt(61+288sqrt(3))` ?

Options

A. `-1`
B. 1
C. 2
D. `-2`

Answer

To solve the expression \( \sqrt{37 + 20\sqrt{3}} - \sqrt{61 + 28\sqrt{3}} \), we will simplify each square root separately.

### Step 1: Simplifying \( \sqrt{37 + 20\sqrt{3}} \)

We want to express \( 37 + 20\sqrt{3} \) in the form of \( (a + b\sqrt{3})^2 \).

1. **Expand \( (a + b\sqrt{3})^2 \)**:
\[ (a + b\sqrt{3})^2 = a^2 + 2ab\sqrt{3} + 3b^2 \]
We want this to equal \( 37 + 20\sqrt{3} \).

2. **Set up equations**:
- From \( a^2 + 3b^2 = 37 \)
- From \( 2ab = 20 \)

3. **Solve for \( ab \)**:
\[ ab = 10 \implies b = \frac{10}{a} \]

4. **Substituting \( b \) into the first equation**:
\[ a^2 + 3\left(\frac{10}{a}\right)^2 = 37 \]
\[ a^2 + \frac{300}{a^2} = 37 \]
Multiply through by \( a^2 \):
\[ a^4 - 37a^2 + 300 = 0 \]
Let \( x = a^2 \):
\[ x^2 - 37x + 300 = 0 \]

5. **Using the quadratic formula**:
\[ x = \frac{37 \pm \sqrt{37^2 - 4 \cdot 300}}{2} \]
\[ x = \frac{37 \pm \sqrt{1369 - 1200}}{2} \]
\[ x = \frac{37 \pm \sqrt{169}}{2} \]
\[ x = \frac{37 \pm 13}{2} \]
This gives \( x = 25 \) or \( x = 12 \).

6. **Finding \( a \) and \( b \)**:
- If \( x = 25 \): \( a = 5 \), \( b = 2 \) (since \( ab = 10 \)).
- If \( x = 12 \): \( a = 2\sqrt{3} \), \( b = 5 \) (not applicable here).

Thus, we have:
\[ \sqrt{37 + 20\sqrt{3}} = 5 + 2\sqrt{3} \]

### Step 2: Simplifying \( \sqrt{61 + 28\sqrt{3}} \)

1. **Set up similar equations**:
- We want \( 61 + 28\sqrt{3} = (c + d\sqrt{3})^2 \).
- From \( c^2 + 3d^2 = 61 \)
- From \( 2cd = 28 \) implies \( cd = 14 \).

2. **Solving for \( d \)**:
\[ d = \frac{14}{c} \]
Substitute into the first equation:
\[ c^2 + 3\left(\frac{14}{c}\right)^2 = 61 \]
\[ c^2 + \frac{588}{c^2} = 61 \]
Multiply through by \( c^2 \):
\[ c^4 - 61c^2 + 588 = 0 \]
Let \( y = c^2 \):
\[ y^2 - 61y + 588 = 0 \]

3. **Using the quadratic formula**:
\[ y = \frac{61 \pm \sqrt{61^2 - 4 \cdot 588}}{2} \]
\[ y = \frac{61 \pm \sqrt{3721 - 2352}}{2} \]
\[ y = \frac{61 \pm \sqrt{1369}}{2} \]
\[ y = \frac{61 \pm 37}{2} \]
This gives \( y = 49 \) or \( y = 12 \).

4. **Finding \( c \) and \( d \)**:
- If \( y = 49 \): \( c = 7 \), \( d = 2 \) (since \( cd = 14 \)).
- If \( y = 12 \): \( c = 2\sqrt{3} \), \( d = 5 \) (not applicable here).

Thus, we have:
\[ \sqrt{61 + 28\sqrt{3}} = 7 + 2\sqrt{3} \]

### Step 3: Final Calculation

Now, we can substitute back into the original expression:
\[ \sqrt{37 + 20\sqrt{3}} - \sqrt{61 + 28\sqrt{3}} = (5 + 2\sqrt{3}) - (7 + 2\sqrt{3}) \]
\[ = 5 + 2\sqrt{3} - 7 - 2\sqrt{3} = -2 \]

### Final Answer
The value of \( \sqrt{37 + 20\sqrt{3}} - \sqrt{61 + 28\sqrt{3}} \) is \( -2 \).

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