Find the condition that zeros pf x^3-px^2+qx-r may be in A.P

1.	Find the condition that zeros pf x^3-px^2+qx-r may be in A.P
Answer» Let a - d, a and a + d be the zeros of the polynomial F(x). Then,Sum of the zeroes =\xa0{tex}- \\frac { \\text { Coefficient of } x ^ { 2 } } { \\text { Coefficient of } x ^ { 3 } }{/tex}{tex}\\Rightarrow (a-d)+a+(a+d)=-\\frac { ( - p ) } { 1 }{/tex}{tex}\\Rightarrow 3a=p{/tex}{tex}\\Rightarrow a=\\frac { p } { 3 }{/tex}Since {tex}a{/tex}\xa0is a zero of the polynomial {tex}f(x){/tex}. Therefore,{tex}f(a)=0{/tex}{tex}\\Rightarrow a^3-pa^2+qa-r=0{/tex}{tex}\\Rightarrow \\left( \\frac { p } { 3 } \\right) ^ { 3 } - p \\left( \\frac { p } { 3 } \\right) ^ { 2 } + q \\left( \\frac { p } { 3 } \\right) - r = 0{/tex}{tex}\\Rightarrow p^3-3p^3+9pq-27r=0{/tex}\xa0{tex}\\Rightarrow 2p^3-9pq+27r=0{/tex}hence,\xa0{tex}2p^3-9pq+27r=0{/tex}\xa0is the required condition

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