Solve `x^(2)-(sqrt3+1)x+sqrt3=0`.

Answer

To solve the quadratic equation \( x^2 - (\sqrt{3} + 1)x + \sqrt{3} = 0 \), we will follow these steps:

### Step 1: Identify the coefficients
The given quadratic equation is in the standard form \( ax^2 + bx + c = 0 \). Here, we identify:
- \( a = 1 \)
- \( b = -(\sqrt{3} + 1) \)
- \( c = \sqrt{3} \)

### Step 2: Use the quadratic formula
The quadratic formula to find the roots of the equation \( ax^2 + bx + c = 0 \) is given by:
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
Substituting the values of \( a \), \( b \), and \( c \):
\[ x = \frac{-(-(\sqrt{3} + 1)) \pm \sqrt{(-(\sqrt{3} + 1))^2 - 4 \cdot 1 \cdot \sqrt{3}}}{2 \cdot 1} \]

### Step 3: Simplify the expression
First, simplify \( -b \):
\[ -b = \sqrt{3} + 1 \]
Now calculate \( b^2 - 4ac \):
\[ b^2 = (-(\sqrt{3} + 1))^2 = (\sqrt{3} + 1)^2 = 3 + 2\sqrt{3} + 1 = 4 + 2\sqrt{3} \]
Now calculate \( 4ac \):
\[ 4ac = 4 \cdot 1 \cdot \sqrt{3} = 4\sqrt{3} \]
Now substitute these into the discriminant:
\[ b^2 - 4ac = (4 + 2\sqrt{3}) - 4\sqrt{3} = 4 - 2\sqrt{3} \]

### Step 4: Substitute back into the quadratic formula
Now we have:
\[ x = \frac{\sqrt{3} + 1 \pm \sqrt{4 - 2\sqrt{3}}}{2} \]

### Step 5: Simplify the square root
To simplify \( \sqrt{4 - 2\sqrt{3}} \), we can express it in a simpler form. Let’s assume:
\[ \sqrt{4 - 2\sqrt{3}} = \sqrt{a} - \sqrt{b} \]
Squaring both sides gives:
\[ 4 - 2\sqrt{3} = a + b - 2\sqrt{ab} \]
From this, we can equate:
1. \( a + b = 4 \)
2. \( -2\sqrt{ab} = -2\sqrt{3} \Rightarrow \sqrt{ab} = \sqrt{3} \Rightarrow ab = 3 \)

Solving these two equations:
Let \( a \) and \( b \) be the roots of the equation \( t^2 - 4t + 3 = 0 \):
\[ t = \frac{4 \pm \sqrt{(4)^2 - 4 \cdot 1 \cdot 3}}{2 \cdot 1} = \frac{4 \pm \sqrt{4}}{2} = \frac{4 \pm 2}{2} \]
This gives us \( t = 3 \) or \( t = 1 \). Therefore, \( a = 3 \) and \( b = 1 \).

Thus, we have:
\[ \sqrt{4 - 2\sqrt{3}} = \sqrt{3} - 1 \]

### Step 6: Substitute back into the quadratic formula
Now substitute back:
\[ x = \frac{\sqrt{3} + 1 \pm (\sqrt{3} - 1)}{2} \]
This gives us two cases:
1. \( x = \frac{(\sqrt{3} + 1) + (\sqrt{3} - 1)}{2} = \frac{2\sqrt{3}}{2} = \sqrt{3} \)
2. \( x = \frac{(\sqrt{3} + 1) - (\sqrt{3} - 1)}{2} = \frac{2}{2} = 1 \)

### Final Solution
The solutions to the equation \( x^2 - (\sqrt{3} + 1)x + \sqrt{3} = 0 \) are:
\[ x = \sqrt{3} \quad \text{and} \quad x = 1 \]