Question

If zeroes of the polynomial p(x)=ax³+3bx²+3cx+d are in A.P.,prove that 2b³-3abc+a²d=0

Answer 1

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:

$\sf \small \: \: \: let \: \: p(x) = a {x}^{3} + 3b {x}^{2} + 3cx + d \: \: and \: \: \alpha \: , \: \beta \\ \\ \sf \small \: \: \: , \: r \: their \: \: three \: \: zeroes \: \: but \: \: zero \: \: are \: \: in \: \: \\ \\ \sf \small \: \: \: \: AP$

______________

$\ \\ \quad \sf \: let \: \alpha = m - n, \: \beta = m \: , \: r = m + n \\ \\ \\ \quad \sf \: sum = \alpha + \beta + r = \frac{ - b}{a} \\ \\ \\ \quad \sf \: substitute \: \: this \: \: term, \: to \: \: get = m = \frac{ - b}{a}$

Now taking two zeroes as sum

$\\ \qquad \quad \sf \: \alpha \beta + \beta \: r + \alpha r = \frac{ - c}{a} \\ \\ \\ \rightarrow \sf \: (m - n)m + m(m + n) + (m + n) + (m - n) = \\ \\ \sf \: \frac{3c}{a}$

Now , solving this problem , we get

$\\ \sf \frac{3 {b}^{2} - 3ac }{ {a}^{2} } = {n}^{2} \\ \\ \\ \sf Product \: \: \alpha \beta \: r = \frac{ d}{a} \\ \\ \\ \sf(m - n)m \: \: (m + n) = \frac{ - d}{a} \\ \\ \\ \sf \: ( {m}^{2} - {n}^{2} )m = \frac{ - d}{a} \\ \\ \\ \sf \bigg[ \bigg( \frac{ - b}{a} \bigg) {}^{2} - \bigg( \frac{ - 3 {b}^{2} - 3ac }{ {a}^{2} } \bigg) {}^{2} \bigg] \bigg( \frac{ - b}{a} \bigg) = \frac{ - d}{a }$

Simplifying , we get

$\\ \\ \qquad \quad \sf \blue{2 {b}^{3} - 3abc + {a}^{2} d = 0}$