Find the zeroes of the following quadratic polynomials and verify the relationship

Between the zeroes and the coefficients.

• (i) x² - 2x - 8
• (ii) 4s² - 4s + 1
• (iii) 6x² - 7x - 3
• (iv) 4u² + 8u
• (v) t² - 15
• (vi) 3x² - x - 4

To find the zeroes of the given quadratic polynomials, we can use the quadratic formula or factorization. The relationship between the zeroes and the coefficients can be verified using Vieta's formulas, which state that for a quadratic equation ax² + bx + c = 0, the sum of the zeroes (α + β) is equal to -b/a and the product of the zeroes (αβ) is equal to c/a.

1. Zeroes of x² - 2x - 8

Using factorization:

• (x - 4)(x + 2) = 0

The zeroes are x = 4 and x = -2.

Sum: 4 + (-2) = 2 = -(-2)/1

Product: 4 * (-2) = -8 = -8/1

2. Zeroes of 4s² - 4s + 1

This can be factored as:

• (2s - 1)² = 0

The zero is s = 1/2 (a repeated root).

Sum: 1/2 + 1/2 = 1 = -(-4)/4

Product: (1/2) * (1/2) = 1/4 = 1/4

3. Zeroes of 6x² - 7x - 3

Using the quadratic formula:

• x = [7 ± √(49 + 72)] / 12
• x = [7 ± 11] / 12

The zeroes are x = 3/2 and x = -1/3.

Sum: 3/2 + (-1/3) = 7/6 = -(-7)/6

Product: (3/2) * (-1/3) = -1/2 = -3/6

4. Zeroes of 4u² + 8u

Factoring gives:

• 4u(u + 2) = 0

The zeroes are u = 0 and u = -2.

Sum: 0 + (-2) = -2 = -8/4

Product: 0 * (-2) = 0 = 0/4

5. Zeroes of t² - 15

This can be expressed as:

• (t - √15)(t + √15) = 0

The zeroes are t = √15 and t = -√15.

Sum: √15 + (-√15) = 0 = -0/1

Product: (√15)(-√15) = -15 = -15/1

6. Zeroes of 3x² - x - 4

Using the quadratic formula:

• x = [1 ± √(1 + 48)] / 6
• x = [1 ± 7] / 6

The zeroes are x = 4/3 and x = -1.

Sum: 4/3 + (-1) = 1/3 = -(-1)/3

Product: (4/3)(-1) = -4/3 = -(-12)/3

In each case, the relationships between the zeroes and the coefficients hold true, confirming Vieta's formulas.