Question

If the zeros of the polynomial = 3 + 3 2 + 3 + are in AP then prove that 2 3 − 3 + 2 = 0.
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Answer 1

$\Large\mathtt\green{ }\huge\underline\mathtt\red{Answer : }$

Let the zeros of the given polynomial be p,q,r. As the roots are in A.P., then it can be assumed as p−k,p,p+k, where k is the common difference.

p−k+p+p+k=−a3b

p=−ab

And, (p−k)(p)(p+k)=−ad

p(p2−k2)=−ad

−ab(p2−k2)=−ad

p2−k2=bd

And, p(p−k)+p(p+k)+(p−k)(p+k)=a3c

2p2+(p2−k2)=a3c

a22b2+bd=a3c

a2b2b3+a2d=a3c

2b3−3abc+a2d=0.

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Answer 2

Let the zeros of the given polynomial be p,q,r. As the roots are in A.P., then it can be assumed as p−k,p,p+k, where k is the common difference.

⇒ p−k+p+p+k=−a3b

⇒ p=−ab

⇒ And, (p−k)(p)(p+k)=−ad

⇒ p(p2−k2)=−ad

⇒ −ab(p2−k2)=−ad

⇒ p2−k2=bd

⇒ And, p(p−k)+p(p+k)+(p−k)(p+k)=a3c

⇒ 2p2+(p2−k2)=a3c

⇒ a22b2+bd=a3c

⇒ a2b2b3+a2d=a3c

⇒ 2b3−3abc+a2d=0.