Question

if the zeros of the polynomial f(x) ax³+3bx²+3cx+d are alpha -beeta,alpha,alpha+beeta then prove that 2b³+3abc+a²d​

Answer 1

Answer:

Step-by-step explanation:

F(X) = ax³ + 3bx² + cx+d = 0

Let three roots be p,q and r

A/Q : roots in A.P : p+r = 2q

Now,

Sum of roots = p+q+r = -3b/a  ⇒ 2q+q = 3q = -3b/a

 ⇒ q = -b/a

Products of roots = pqr = - d/a ⇒ pr (-b/a) = -d/a

 ⇒ pr = d/b

Sum of products of two roots = pq + qr + rp = 3c/a

⇒ q(p+r) + rp = 3c/a

⇒ q(2q) + d/b = 3c/a

⇒ 2(-b/a)² + d/b = 3c/a

⇒ 2b²/a² + d/b = 3c/a

⇒ 2b³+ da² = 3cab

⇒2b³ - 3abc + a²d = 0

Hence proved.

Answer 2

Answer:

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