Square root of `13 - 4sqrt3` is ?

Options

A. `sqrt(12)+1`
B. `2sqrt3+7`
C. `sqrt(12)-1`
D. `2sqrt3-7`

Answer

To find the square root of \( 13 - 4\sqrt{3} \), we can express it in a simpler form. Let's denote \( x = \sqrt{13 - 4\sqrt{3}} \).

### Step 1: Assume the form of the square root
Assume that \( x \) can be expressed as \( \sqrt{a} - \sqrt{b} \). Then we have:
\[ x^2 = a + b - 2\sqrt{ab} \]

### Step 2: Set up the equation
From our original expression, we know:
\[ x^2 = 13 - 4\sqrt{3} \]
This gives us two equations:
1. \( a + b = 13 \)
2. \( -2\sqrt{ab} = -4\sqrt{3} \)

### Step 3: Solve for \( ab \)
From the second equation, we can simplify:
\[ 2\sqrt{ab} = 4\sqrt{3} \implies \sqrt{ab} = 2\sqrt{3} \implies ab = (2\sqrt{3})^2 = 4 \cdot 3 = 12 \]

### Step 4: Solve the system of equations
Now we have the system of equations:
1. \( a + b = 13 \)
2. \( ab = 12 \)

We can treat this as a quadratic equation where \( a \) and \( b \) are the roots:
\[ t^2 - (a+b)t + ab = 0 \implies t^2 - 13t + 12 = 0 \]

### Step 5: Find the roots of the quadratic equation
Using the quadratic formula:
\[ t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{13 \pm \sqrt{13^2 - 4 \cdot 1 \cdot 12}}{2 \cdot 1} \]
Calculating the discriminant:
\[ 13^2 - 48 = 169 - 48 = 121 \]
Thus,
\[ t = \frac{13 \pm \sqrt{121}}{2} = \frac{13 \pm 11}{2} \]
This gives us:
\[ t = \frac{24}{2} = 12 \quad \text{and} \quad t = \frac{2}{2} = 1 \]

### Step 6: Identify \( a \) and \( b \)
So, we have \( a = 12 \) and \( b = 1 \) (or vice versa).

### Step 7: Write the final expression
Thus, we can express \( x \) as:
\[ x = \sqrt{12} - \sqrt{1} = 2\sqrt{3} - 1 \]

### Final Answer
Therefore, the square root of \( 13 - 4\sqrt{3} \) is:
\[ \sqrt{13 - 4\sqrt{3}} = 2\sqrt{3} - 1 \]