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| 1. |
Zeros of quadratic polynomial f(x)=6x-3,verify relationship between zeros and coefficients |
| Answer» f(x)=6x2-3To find zeros6x2-3=06x2=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 =\xa0{tex} \\alpha + \\beta = \\frac { 1 } { \\sqrt { 2 } } + \\left( - \\frac { 1 } { \\sqrt { 2 } } \\right) = 0{/tex}\xa0and,\xa0{tex} - \\frac { \\text { Coefficient of } x } { \\text { Coefficient of } x ^ { 2 } } = - \\frac { 0 } { 6 } = 0{/tex}{tex}\\therefore{/tex}\xa0Sum of the zeros\xa0{tex} = - \\frac { \\text { Coefficient of } x } { \\text { coefficient of } x ^ { 2 } }{/tex}Also, Product of the zeroes =\xa0{tex} \\alpha \\beta = \\frac { 1 } { \\sqrt { 2 } } \\times \\frac { - 1 } { \\sqrt { 2 } } = \\frac { - 1 } { 2 } {/tex}, and\xa0{tex} \\frac { \\text { Constant term } } { \\text { Coefficient of } x ^ { 2 } }= \\frac { - 3 } { 6 } = \\frac { - 1 } { 2 }{/tex}{tex}\\therefore{/tex}\xa0Product of the zeros\xa0{tex}= \\frac { \\text { Constant term } } { \\text { Coefficient of } x ^ { 2 } }{/tex} | |