Find zeroes of the given quadratic polynomials and verify the relation between zeroes and coefficients :
`x^(2)-2`

Explanation

To find the zeroes of the quadratic polynomial \( p(x) = x^2 - 2 \) and verify the relation between the zeroes and coefficients, we will follow these steps:

### Step 1: Identify the coefficients
The general form of a quadratic polynomial is given by:
\[ p(x) = ax^2 + bx + c \]
For the polynomial \( p(x) = x^2 - 2 \):
- \( a = 1 \) (coefficient of \( x^2 \))
- \( b = 0 \) (coefficient of \( x \))
- \( c = -2 \) (constant term)

### Step 2: Set the polynomial equal to zero
To find the zeroes, we set the polynomial equal to zero:
\[ x^2 - 2 = 0 \]

### Step 3: Rearrange the equation
Rearranging the equation gives:
\[ x^2 = 2 \]

### Step 4: Take the square root of both sides
Taking the square root of both sides, we get:
\[ x = \sqrt{2} \quad \text{or} \quad x = -\sqrt{2} \]

### Step 5: Identify the zeroes
Thus, the zeroes of the polynomial \( p(x) = x^2 - 2 \) are:
\[ \alpha = \sqrt{2} \quad \text{and} \quad \beta = -\sqrt{2} \]

### Step 6: Verify the relation between zeroes and coefficients
1. **Sum of the zeroes**:
The sum of the zeroes \( \alpha + \beta \) is given by:
\[ \alpha + \beta = \sqrt{2} + (-\sqrt{2}) = 0 \]
According to the relation, the sum of the zeroes is given by:
\[ -\frac{b}{a} = -\frac{0}{1} = 0 \]
Thus, the sum of the zeroes is verified.

2. **Product of the zeroes**:
The product of the zeroes \( \alpha \cdot \beta \) is given by:
\[ \alpha \cdot \beta = \sqrt{2} \cdot (-\sqrt{2}) = -2 \]
According to the relation, the product of the zeroes is given by:
\[ \frac{c}{a} = \frac{-2}{1} = -2 \]
Thus, the product of the zeroes is also verified.

### Final Answer
The zeroes of the polynomial \( x^2 - 2 \) are \( \sqrt{2} \) and \( -\sqrt{2} \). The relations between the zeroes and coefficients are verified as follows:
- Sum of zeroes: \( 0 \) (verified)
- Product of zeroes: \( -2 \) (verified)

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