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If one zero of a polynomial 3x^-8x+2k+1 is seven time other find the value of k |
| Answer» Let\xa0{tex}\\alpha{/tex}\xa0and\xa0{tex}\\beta{/tex}\xa0be the zeroes of the polynomial,\xa0{tex}3x^2-8x+2k+1{/tex}Given,\xa0{tex}\\beta = 7 \\alpha{/tex}{tex}\\therefore \\quad \\alpha + 7 \\alpha = 8 \\alpha = - \\left( - \\frac { 8 } { 3 } \\right)=\\frac{8}{3}{/tex}So,\xa0{tex}\\alpha = \\frac { 1 } { 3 }{/tex}and\xa0{tex}\\alpha \\times 7 \\alpha = \\frac { 2 k + 1 } { 3 }{/tex}{tex}\\Rightarrow{/tex}\xa0{tex}7 \\alpha ^ { 2 } = \\frac { 2 k + 1 } { 3 }{/tex}{tex}\\Rightarrow{/tex}\xa0{tex}7 \\left( \\frac { 1 } { 3 } \\right) ^ { 2 } = \\frac { 2 k + 1 } { 3 }{/tex}{tex}\\Rightarrow{/tex}\xa0{tex}7 \\times \\frac { 1 } { 9 } = \\frac { 2 k + 1 } { 3 }{/tex}{tex}\\Rightarrow{/tex}\xa0{tex}\\frac { 7 } { 3 } - 1 = 2 k \\quad \\Rightarrow 2 k = \\frac { 4 } { 3 }{/tex}{tex}\\therefore k=\\frac{2}{3}{/tex} | |