Zeros of quadratic polynomial f(x)=6x-3,verify relationship …

f(x)=6x²-3
To find zeros
6x²-3=0
6x²=3

{tex}x=\pm \frac {1}{\sqrt 2}{/tex}

Henc the zeros f(x) are:

{tex} \alpha = \frac { 1 } { \sqrt { 2 } } \text { and } \beta = - \frac { 1 } { \sqrt { 2 } }{/tex}
Sum of the zeroes = {tex} \alpha + \beta = \frac { 1 } { \sqrt { 2 } } + \left( - \frac { 1 } { \sqrt { 2 } } \right) = 0{/tex} and, {tex} - \frac { \text { Coefficient of } x } { \text { Coefficient of } x ^ { 2 } } = - \frac { 0 } { 6 } = 0{/tex}
{tex}\therefore{/tex} Sum of the zeros {tex} = - \frac { \text { Coefficient of } x } { \text { coefficient of } x ^ { 2 } }{/tex}
Also, Product of the zeroes = {tex} \alpha \beta = \frac { 1 } { \sqrt { 2 } } \times \frac { - 1 } { \sqrt { 2 } } = \frac { - 1 } { 2 } {/tex}, and {tex} \frac { \text { Constant term } } { \text { Coefficient of } x ^ { 2 } }= \frac { - 3 } { 6 } = \frac { - 1 } { 2 }{/tex}
{tex}\therefore{/tex} Product of the zeros {tex}= \frac { \text { Constant term } } { \text { Coefficient of } x ^ { 2 } }{/tex}