Find the zeroes of the quadratic polynomial `3x^(2)-2` and verify the relationship between the zeroes and the coefficients .

Answer

To find the zeroes of the quadratic polynomial \(3x^2 - 2\) and verify the relationship between the zeroes and the coefficients, we can follow these steps:

### Step 1: Set the polynomial equal to zero
We start with the polynomial:
\[ 3x^2 - 2 = 0 \]

### Step 2: Rearrange the equation
Add \(2\) to both sides to isolate the \(x^2\) term:
\[ 3x^2 = 2 \]

### Step 3: Divide by the coefficient of \(x^2\)
Now, divide both sides by \(3\):
\[ x^2 = \frac{2}{3} \]

### Step 4: Take the square root of both sides
To find \(x\), take the square root of both sides:
\[ x = \pm \sqrt{\frac{2}{3}} \]

### Step 5: Simplify the square root
This can be simplified further:
\[ x = \pm \frac{\sqrt{2}}{\sqrt{3}} = \pm \frac{\sqrt{6}}{3} \]

Thus, the zeroes of the polynomial \(3x^2 - 2\) are:
\[ x_1 = \frac{\sqrt{6}}{3}, \quad x_2 = -\frac{\sqrt{6}}{3} \]

### Step 6: Verify the relationship between the zeroes and coefficients
The general form of a quadratic polynomial is given by:
\[ ax^2 + bx + c = 0 \]
For our polynomial \(3x^2 - 2\), we have:
- \(a = 3\)
- \(b = 0\)
- \(c = -2\)

#### Sum of the zeroes
The sum of the zeroes \(x_1 + x_2\) can be calculated using the formula:
\[ -\frac{b}{a} \]
Substituting the values:
\[ -\frac{0}{3} = 0 \]

Calculating the sum of the zeroes:
\[ x_1 + x_2 = \frac{\sqrt{6}}{3} + \left(-\frac{\sqrt{6}}{3}\right) = 0 \]

#### Product of the zeroes
The product of the zeroes \(x_1 \cdot x_2\) can be calculated using the formula:
\[ \frac{c}{a} \]
Substituting the values:
\[ \frac{-2}{3} \]

Calculating the product of the zeroes:
\[ x_1 \cdot x_2 = \left(\frac{\sqrt{6}}{3}\right) \cdot \left(-\frac{\sqrt{6}}{3}\right) = -\frac{6}{9} = -\frac{2}{3} \]

### Conclusion
Thus, we have verified the relationships:
- The sum of the zeroes is \(0\) which matches \(-\frac{b}{a}\).
- The product of the zeroes is \(-\frac{2}{3}\) which matches \(\frac{c}{a}\).