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If the zeros of the polynomial f(x)=2xcube -15xsquare +37x -30 are in A.P.,find them. |
| Answer» Given\xa0{tex}f(x)=2{ x }^{ 3 }-15{ x }^{ 2 }+37x-30{/tex}Let the zeroes of the polynomial in A.P be a-d,a,a+dUsing the relationship between the zeroes and coefficients of a cubic polynomial ,we getSum of the roots = {tex}a-d+a+a+d=\\frac { -\\left( -15 \\right) }{ 2 } {/tex}\xa0{tex}\\Rightarrow 3a=\\frac { 15 }{ 2 } \\\\ \\Rightarrow a=\\frac { 5 }{ 2 } {/tex}Now the product of the roots=\xa0{tex}\\left( a-d \\right) a\\left( a+d \\right) =\\frac { -\\left( -30 \\right) }{ 2 } {/tex}{tex}\\Rightarrow a\\left( { a }^{ 2 }-{ d }^{ 2 } \\right) =\\frac { 30 }{ 2 } \\\\ \\Rightarrow \\frac { 5 }{ 2 } \\left( \\frac { 25 }{ 4 } -{ d }^{ 2 } \\right) =15\\\\ \\Rightarrow \\left( \\frac { 25 }{ 4 } -{ d }^{ 2 } \\right) =15\\times \\frac { 2 }{ 5 } =6\\\\ \\Rightarrow { d }^{ 2 }=\\frac { 25 }{ 4 } -6=\\frac { 25-24 }{ 4 } =\\frac { 1 }{ 4 } \\\\ \\Rightarrow { d }=\\pm \\frac { 1 }{ 2 } {/tex}Hence the roots are\xa0{tex}\\frac { 5 }{ 2 } -\\frac { 1 }{ 2 } ,\\frac { 5 }{ 2 } ,\\frac { 5 }{ 2 } +\\frac { 1 }{ 2 } =2,2.5,3{/tex} | |