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| 1. |
Find the zeros of polynomial x²+1/6x-2 |
| Answer» Let f(x) = x2 + {tex} \\frac 16{/tex}x - 2.Then, f(x) = {tex} \\frac 16{/tex}(6x2+ x -12) = {tex} \\frac 16{/tex}(6x2 + 9x - 8x - 12){tex} \\Rightarrow{/tex}\xa0f(x) = {tex} \\frac 16{/tex}{(6x2 + 9x) - (8x + 12)}= {tex} \\frac 16{/tex}{3x (2x + 3) - 4 (2x + 3)}= {tex} \\frac 16{/tex}(2x + 3) (3x - 4)Now f(x)=0 if{tex}\\style{font-family:Arial}{\\style{font-size:12px}{\\begin{array}{l}i.e.\\;x=-\\frac32\\;or\\;x=\\frac43\\\\\\end{array}}}{/tex}Hence, {tex} \u200b\u200b\\alpha = \\frac { - 3 } { 2 } \\text { and } \\beta = \\frac { 4 } { 3 }{/tex}\xa0are the zeros of the given polynomial.{tex}\\style{font-family:Arial}{\\style{font-size:12px}{\\begin{array}{l}Now\\;\\alpha+\\beta=\\frac{-3}2+\\frac43=\\frac{-9+8}6=-\\frac16\\\\\\end{array}}}{/tex}and\xa0{tex}\\style{font-family:Arial}{\\style{font-size:12px}{\\begin{array}{l}\\alpha\\beta=\\frac{-3}2\\times\\frac43=-2\\\\\\end{array}}}{/tex}The given polynomial is f(x) =x2 + {tex} \\frac 16{/tex}x - 2.so a=1 b={tex}\\frac16{/tex}\xa0c=-2{tex}\\style{font-family:Arial}{\\style{font-size:12px}{\\begin{array}{l}\\alpha+\\beta=\\frac ba=\\frac{-{\\displaystyle\\frac16}}1=-\\frac16\\\\\\end{array}}}{/tex}\xa0and {tex}\\style{font-family:Arial}{\\style{font-size:12px}{\\begin{array}{l}\\alpha\\beta=\\frac ca=\\frac{-2}1=-2\\\\\\end{array}}}{/tex}Hence, the relation between the coefficients and zeros is verified. | |