Find the zeroes of the following quadratic polynomials and verify the relationship between the zeroes and the coefficients.

To find the zeros of the polynomial let us first solve the polynomial by equating it to zero. Factorizing the given polynomial

x² – (√3 + 1)x + √3 = 0

To factorize the polynomial we have,

Sum of the value should be equal = -(√3 + 1)

Product should be equal to = 1 × √3

= √3

x² – [2(√3 + 1)x]/2 + √3 = 0

x² - 2(√3 + 1)x/2 + (√3 + 1)²/2² - (√3 + 1)²/2² + √3 = 0

[x - (√3 + 1)/2]² - (3 + 1 + 2√3)/4 + √3 = 0

[x - (√3 + 1)/2]² = (3 + 1 + 2√3 - 4√3)/4

[x - (√3 + 1)/2]² = (√3 - 1)²/2²

[x - (√3 + 1)/2] = ± (√3 - 1)/2

Solving for positive value,

x = (√3 - 1)/2 + (√3 + 1)/2

x = (√3 - 1 + √3 + 1)/2

x = 2√3/2 = √3

Solving for negative value,

x = -(√3 - 1)/2 + (√3 + 1)/2

x = (-√3 + 1 + √3 + 1)/2

x = 2/2 = 1

When we compare the above quadratic equation with the generalized one we get,

ax² + bx + c = 0

∴ a = 1, b = - (√3 + 1), c = √3

Sum of zeroes = -b / a

= (√3 + 1) / 1

= √3 + 1

Product of zeroes = c / a

= √3 / 1

= √3