From question number 1 to 16, find zeroes of the given quadratic polynomials and verify the relation between zeroes and coefficients :
`u^(2)+2u`

Explanation

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

### Step 1: Write the polynomial in standard form
The given polynomial is:
\[ u^2 + 2u = 0 \]
This can be compared with the standard form of a quadratic equation:
\[ au^2 + bu + c = 0 \]
Here, we can identify:
- \( a = 1 \)
- \( b = 2 \)
- \( c = 0 \)

### Step 2: Find the zeroes of the polynomial
To find the zeroes, we can factor the polynomial:
\[ u^2 + 2u = u(u + 2) = 0 \]
Setting each factor to zero gives us:
1. \( u = 0 \)
2. \( u + 2 = 0 \) which simplifies to \( u = -2 \)

Thus, the zeroes of the polynomial are:
\[ u_1 = 0 \quad \text{and} \quad u_2 = -2 \]

### Step 3: Verify the relation between zeroes and coefficients
The sum and product of the zeroes can be calculated as follows:
- Sum of the zeroes \( \alpha + \beta \):
\[ \alpha + \beta = 0 + (-2) = -2 \]
- Product of the zeroes \( \alpha \cdot \beta \):
\[ \alpha \cdot \beta = 0 \cdot (-2) = 0 \]

According to the relations derived from the coefficients:
- The sum of the zeroes is given by \( -\frac{b}{a} \):
\[ -\frac{b}{a} = -\frac{2}{1} = -2 \]
- The product of the zeroes is given by \( \frac{c}{a} \):
\[ \frac{c}{a} = \frac{0}{1} = 0 \]

### Conclusion
Both relations are verified:
1. The sum of the zeroes \( (-2) \) matches \( -\frac{b}{a} \).
2. The product of the zeroes \( (0) \) matches \( \frac{c}{a} \).

Thus, the zeroes of the polynomial \( u^2 + 2u \) are \( 0 \) and \( -2 \), and the relations between zeroes and coefficients are verified.

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