1.

FInd the condition that the zeroes of the polynomial `f(x) =x^3+3px^2+3qx+r` may be in A.P

Answer» Let `alpha-d,alpha and alpha+d` are the zeroes of the given polynomial.
Then, sum of zeroes ` = alpha-d+alpha+alpha+d = 3alpha`
`=> 3alpha = -3p => alpha = -p->(1)`
Product of zeroes taking two at a time ` = alpha(alpha-d)+ alpha(alpha+d)+(alpha-d)(alpha+d)`
`=alpha^2-alphad+alpha^2+alphad+alpha^2-d^2`
`=3alpha^2-d^2`
`=> 3alpha^2-d^2 = 3q->(2)`
Now, product of zeroes ` = alpha(alpha^2-d^2)`
`=> alpha(alpha^2-d^2) = -r `
`=>(alpha^2-d^2) = -r/alpha`
`=>(alpha^2-d^2) = (-r)/(-p)`
`=>(alpha^2-d^2) = (r)/(p)->(3)`
Now, from (2),
`2alpha^2+alpha^2-d^2 = 3q`
`=>2p^2+(r/p) = 3q`
`=>2p^3+r = 3pq`, which is the required condition.


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