Accepted Answer:
$ \text{1. } x^2 - 2x - 8 $
$\text{The standard form of the quadratic equation is } ax^2 + bx + c = 0. \text{ For this polynomial: } a = 1, \, b = -2, \, c = -8.$
$\text{Using the quadratic formula: }$
$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$
$\text{Substitute the values of } a, b, c:$
$x = \frac{-(-2) \pm \sqrt{(-2)^2 - 4(1)(-8)}}{2(1)}$
$x = \frac{2 \pm \sqrt{4 + 32}}{2}$
$x = \frac{2 \pm \sqrt{36}}{2}$
$x = \frac{2 \pm 6}{2}$
$\text{So, the zeroes are: }$
$x = \frac{2 + 6}{2} = 4 \quad \text{or} \quad x = \frac{2 - 6}{2} = -2.$
$\text{The sum of the zeroes is: } \alpha + \beta = 4 + (-2) = 2 = -\frac{b}{a}, \quad \text{and}$
$\text{the product of the zeroes is: } \alpha \beta = 4 \times (-2) = -8 = \frac{c}{a}.$
$\text{Thus, the relationship between the zeroes and the coefficients is verified.}$
$ \text{2. } 4s^2 - 4s + 1 $
$\text{For this polynomial, } a = 4, \, b = -4, \, c = 1.$
$x = \frac{-(-4) \pm \sqrt{(-4)^2 - 4(4)(1)}}{2(4)}$
$x = \frac{4 \pm \sqrt{16 - 16}}{8}$
$x = \frac{4 \pm \sqrt{0}}{8}$
$x = \frac{4 \pm 0}{8}$
$x = \frac{4}{8} = \frac{1}{2}$
$\text{So, the only zero is: } x = \frac{1}{2}.$
$\text{The sum of the zeroes is: } \alpha + \beta = \frac{1}{2} = -\frac{b}{a}, \quad \text{and}$
$\text{the product of the zeroes is: } \alpha \beta = \frac{1}{2} = \frac{c}{a}.$
$ \text{3. } 6x^2 - 3 - 7x $
$\text{Rearrange the equation to the standard form: } 6x^2 - 7x - 3 = 0$
$\text{For this polynomial, } a = 6, \, b = -7, \, c = -3.$
$x = \frac{-(-7) \pm \sqrt{(-7)^2 - 4(6)(-3)}}{2(6)}$
$x = \frac{7 \pm \sqrt{49 + 72}}{12}$
$x = \frac{7 \pm \sqrt{121}}{12}$
$x = \frac{7 \pm 11}{12}$
$\text{So, the zeroes are: }$
$x = \frac{7 + 11}{12} = \frac{18}{12} = 1.5 \quad \text{or} \quad x = \frac{7 - 11}{12} = \frac{-4}{12} = -\frac{1}{3}.$
$\text{The sum of the zeroes is: } \alpha + \beta = 1.5 + (-\frac{1}{3}) = \frac{3}{2} - \frac{1}{3} = \frac{5}{3} = -\frac{b}{a}, \quad \text{and}$
$\text{the product of the zeroes is: } \alpha \beta = 1.5 \times (-\frac{1}{3}) = -\frac{1.5}{3} = -\frac{1}{2} = \frac{c}{a}.$
$ \text{4. } 4u^2 + 8u $
$\text{Rearrange the equation to the standard form: } 4u^2 + 8u = 0$
$\text{For this polynomial, } a = 4, \, b = 8, \, c = 0.$
$x = \frac{-8 \pm \sqrt{8^2 - 4(4)(0)}}{2(4)}$
$x = \frac{-8 \pm \sqrt{64}}{8}$
$x = \frac{-8 \pm 8}{8}$
$\text{So, the zeroes are: }$
$x = \frac{-8 + 8}{8} = 0 \quad \text{or} \quad x = \frac{-8 - 8}{8} = -2.$
$\text{The sum of the zeroes is: } \alpha + \beta = 0 + (-2) = -2 = -\frac{b}{a}, \quad \text{and}$
$\text{the product of the zeroes is: } \alpha \beta = 0 \times (-2) = 0 = \frac{c}{a}.$
$ \text{5. } t^2 - 15 $
$\text{Rearrange the equation to the standard form: } t^2 + 0t - 15 = 0$
$\text{For this polynomial, } a = 1, \, b = 0, \, c = -15.$
$x = \frac{-0 \pm \sqrt{0^2 - 4(1)(-15)}}{2(1)}$
$x = \frac{0 \pm \sqrt{60}}{2}$
$x = \frac{0 \pm \sqrt{4 \times 15}}{2}$
$x = \frac{0 \pm 2\sqrt{15}}{2}$
$x = \pm \sqrt{15}$
$\text{So, the zeroes are: } x = \sqrt{15} \quad \text{or} \quad x = -\sqrt{15}.$
$\text{The sum of the zeroes is: } \alpha + \beta = \sqrt{15} + (-\sqrt{15}) = 0 = -\frac{b}{a}, \quad \text{and}$
$\text{the product of the zeroes is: } \alpha \beta = \sqrt{15} \times (-\sqrt{15}) = -15 = \frac{c}{a}.$
$ \text{6. } 3x^2 - x - 4 $
$\text{For this polynomial, } a = 3, \, b = -1, \, c = -4.$
$x = \frac{-(-1) \pm \sqrt{(-1)^2 - 4(3)(-4)}}{2(3)}$
$x = \frac{1 \pm \sqrt{1 + 48}}{6}$
$x = \frac{1 \pm \sqrt{49}}{6}$
$x = \frac{1 \pm 7}{6}$
$\text{So, the zeroes are: }$
$x = \frac{1 + 7}{6} = \frac{8}{6} = \frac{4}{3} \quad \text{or} \quad x = \frac{1 - 7}{6} = \frac{-6}{6} = -1.$
$\text{The sum of the zeroes is: } \alpha + \beta = \frac{4}{3} + (-1) = \frac{4}{3} - \frac{3}{3} = \frac{1}{3} = -\frac{b}{a}, \quad \text{and}$
$\text{the product of the zeroes is: } \alpha \beta = \frac{4}{3} \times (-1) = -\frac{4}{3} = \frac{c}{a}.$