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

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

$ \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 $

$\text{For this polynomial, } a = 3, \, b = -1, \, c = -4.$

$x = \frac{-(-1) \pm \sqrt{(-1)^2 - 4(3)(-4)}}{2(3)}$
$x = \frac{1 \pm \sqrt{1 + 48}}{6}$
$x = \frac{1 \pm \sqrt{49}}{6}$
$x = \frac{1 \pm 7}{6}$

$\text{So, the zeroes are: }$
$x = \frac{1 + 7}{6} = \frac{8}{6} = \frac{4}{3} \quad \text{or} \quad x = \frac{1 - 7}{6} = \frac{-6}{6} = -1.$

$\text{The sum of the zeroes is: } \alpha + \beta = \frac{4}{3} + (-1) = \frac{4}{3} - \frac{3}{3} = \frac{1}{3} = -\frac{b}{a}, \quad \text{and}$
$\text{the product of the zeroes is: } \alpha \beta = \frac{4}{3} \times (-1) = -\frac{4}{3} = \frac{c}{a}.$

Thanks

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