Find the zeroes of the quadratic …

x² - 6
Let p(x) = x² - 6
For zeroes of p(x), p(x) = 0
{tex}\Rightarrow x ^ { 2 } - 6 = 0 \Rightarrow ( x ) ^ { 2 } - ( \sqrt { 6 } ) ^ { 2 } = 0{/tex}
{tex}\Rightarrow ( x - \sqrt { 6 } ) ( x + \sqrt { 6 } ) = 0{/tex}
Using the identity a² - b² = (a - b) (a + b)
{tex}\Rightarrow x - \sqrt { 6 } = 0 \text { or } x + \sqrt { 6 } = 0{/tex}
{tex}\Rightarrow x = \sqrt { 6 } \text { or } x = - \sqrt { 6 } \Rightarrow x = \sqrt { 6 } , - \sqrt { 6 }{/tex}
So, the zeroes of x² - 6 are {tex}\sqrt 6 {/tex} and {tex} - \sqrt 6 {/tex}
Sum of zeroes
{tex}= ( \sqrt { 6 } ) + ( - \sqrt { 6 } ) = 0 = \frac { - 0 } { 1 } = \frac { - \text { Coefficient of } x } { \text { Coefficient of } x ^ { 2 } }{/tex}
Product of zeroes
{tex}= ( \sqrt { 6 } ) \times ( - \sqrt { 6 } ) = - 6 = \frac { - 6 } { 1 } = \frac { \text { Constant term } } { \text { Coefficient of } \mathrm { x } ^ { 2 } }{/tex}

Hence the relation between zeroes and coefficient is verified.