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| 1. |
Find the zero of the quadratic polynomials 7yy-11/3y-2/3 |
| Answer» p(y) = 7y2 -\xa0{tex}\\frac{11}{3}{/tex}y -\xa0{tex}\\frac{2}{3}{/tex}\xa0=\xa0{tex}\\frac{1}{3}{/tex}\xa0(21y2 - 11y - 2)=\xa0{tex}\\frac{1}{3}{/tex}[(7y + 1)(3y - 2)]{tex}\\therefore{/tex}\xa0Zeroes are\xa0{tex}\\frac{2}{3}{/tex}, -\xa0{tex}\\frac{1}{7}{/tex}Sum of Zeroes =\xa0{tex}\\frac{2}{3}{/tex}\xa0-\xa0{tex}\\frac{1}{7}{/tex}\xa0=\xa0{tex}\\frac{11}{21}{/tex}{tex}\\frac{-b}{a}{/tex}\xa0=\xa0{tex}\\frac{11}{21}{/tex}\xa0{tex}\\therefore{/tex}\xa0sum of zeroes =\xa0{tex}\\frac{-b}{a}{/tex}Product of Zeroes = ({tex}\\frac{2}{3}{/tex})(-{tex}\\frac{-1}{7}{/tex}) = -{tex}\\frac{2}{21}{/tex}{tex}\\frac{c}{a}{/tex}\xa0= -\xa0{tex}\\frac{2}{3}{/tex}({tex}\\frac{1}{7}{/tex}) = -{tex}\\frac{2}{21}{/tex}{tex}\\therefore{/tex}\xa0Product =\xa0{tex}\\frac{c}{a}{/tex} | |