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Question
If zeros of the polynomial f(x) = x3 − 3px2 + qx − r are in A.P., then
Options
2p3 = pq − r
2p3 = pq + r
p3 = pq − r
None of these
Solution
Let `a-d,a,a+d` be the zeros of the polynomial f(x) = x3 − 3px2 + qx − r then
`\text{sum of zero }= - (text{coefficient of x})/(text{coefficient of } x^2)`
`(a - d) + a(a +b)= -(-3p)/1`
`a - cancel(d)+a+a+cancel(d)= 3p`
`3a = 3p`
`a = 3/3p`
`a = p`
Since a is a zero of the polynomial `f(x)`
Therefore,
`f(a)= 0`
`a ^3 - 3pa^2 + qa - r=0`
Substituting `a=p`.we get
`p^3 - 3p(p)^2 + q xxp -r =0`
`p^3 - 3p^3 + qp-r =0`
`-2p^3 + qp - r =0`
` qp-r = 2p^3`
Hence, the correct choice is (a).
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