Square root of `(7-4sqrt3)` is equal to:

Options

A. `+-(2+sqrt3)`
B. `+-(2-sqrt3)`
C. `+-(sqrt2-3)`
D. `+-(sqrt2+3)`

Answer

To find the square root of \( 7 - 4\sqrt{3} \), we can express it in a different form that allows us to simplify it.

### Step-by-Step Solution:

1. **Assume the form of the square root**:
We can assume that \( \sqrt{7 - 4\sqrt{3}} \) can be expressed in the form \( \sqrt{a} - \sqrt{b} \).

2. **Square both sides**:
If \( \sqrt{7 - 4\sqrt{3}} = \sqrt{a} - \sqrt{b} \), then squaring both sides gives:
\[ 7 - 4\sqrt{3} = a + b - 2\sqrt{ab} \]

3. **Equate the rational and irrational parts**:
From the equation above, we can equate the rational parts and the irrational parts:
- Rational part: \( a + b = 7 \)
- Irrational part: \( -2\sqrt{ab} = -4\sqrt{3} \)

4. **Solve for \( ab \)**:
From the irrational part, we can simplify:
\[ 2\sqrt{ab} = 4\sqrt{3} \implies \sqrt{ab} = 2\sqrt{3} \implies ab = 4 \cdot 3 = 12 \]

5. **Set up a system of equations**:
Now we have a system of equations:
- \( a + b = 7 \)
- \( ab = 12 \)

6. **Use the quadratic formula**:
We can express \( a \) and \( b \) as the roots of the quadratic equation:
\[ x^2 - (a+b)x + ab = 0 \implies x^2 - 7x + 12 = 0 \]

7. **Solve the quadratic equation**:
Using the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
\[ x = \frac{7 \pm \sqrt{7^2 - 4 \cdot 1 \cdot 12}}{2 \cdot 1} = \frac{7 \pm \sqrt{49 - 48}}{2} = \frac{7 \pm 1}{2} \]
This gives us two solutions:
\[ x = \frac{8}{2} = 4 \quad \text{and} \quad x = \frac{6}{2} = 3 \]
Thus, \( a = 4 \) and \( b = 3 \) (or vice versa).

8. **Substitute back to find the square root**:
Therefore, we have:
\[ \sqrt{7 - 4\sqrt{3}} = \sqrt{4} - \sqrt{3} = 2 - \sqrt{3} \]

### Final Answer:
Thus, the square root of \( 7 - 4\sqrt{3} \) is:
\[ \sqrt{7 - 4\sqrt{3}} = 2 - \sqrt{3} \]