The zeroes of the polynomial `sqrt3x^2 - 8x+4sqrt3` are:

Options

A. `2sqrt3,2/sqrt3`
B. `2sqrt3,sqrt3/2`
C. `6sqrt2,3`
D. `3sqrt2,6`

Answer

To find the zeroes of the polynomial \( \sqrt{3}x^2 - 8x + 4\sqrt{3} \), we can use the quadratic formula. The quadratic formula states that for a quadratic equation of the form \( ax^2 + bx + c = 0 \), the roots can be found using:

\[ x = \frac{-b \pm \sqrt{D}}{2a} \]

where \( D \) is the discriminant given by \( D = b^2 - 4ac \).

### Step 1: Identify coefficients
From the polynomial \( \sqrt{3}x^2 - 8x + 4\sqrt{3} \), we identify:
- \( a = \sqrt{3} \)
- \( b = -8 \)
- \( c = 4\sqrt{3} \)

### Step 2: Calculate the discriminant \( D \)
Now, we calculate the discriminant \( D \):

\[ D = b^2 - 4ac \]
\[ D = (-8)^2 - 4(\sqrt{3})(4\sqrt{3}) \]
\[ D = 64 - 4(3)(4) \]
\[ D = 64 - 48 \]
\[ D = 16 \]

### Step 3: Apply the quadratic formula
Now that we have \( D = 16 \), we can find the roots using the quadratic formula:

\[ x = \frac{-(-8) \pm \sqrt{16}}{2\sqrt{3}} \]
\[ x = \frac{8 \pm 4}{2\sqrt{3}} \]

### Step 4: Calculate the two possible values for \( x \)
Calculating the two roots:

1. For the positive case:
\[ x_1 = \frac{8 + 4}{2\sqrt{3}} = \frac{12}{2\sqrt{3}} = \frac{6}{\sqrt{3}} = 2\sqrt{3} \]

2. For the negative case:
\[ x_2 = \frac{8 - 4}{2\sqrt{3}} = \frac{4}{2\sqrt{3}} = \frac{2}{\sqrt{3}} = \frac{2\sqrt{3}}{3} \]

### Conclusion
Thus, the zeroes of the polynomial \( \sqrt{3}x^2 - 8x + 4\sqrt{3} \) are:

\[ x_1 = 2\sqrt{3}, \quad x_2 = \frac{2\sqrt{3}}{3} \]