Question

If the zeros of the polynomial f(x)=ax cube +3bx square +3cx+d are in A.P.,prove that 2b cube-3abc+a square d=0

Answer 1

roots of given polynomial is in AP
let P- r , P, and P+ r are the roots of

f( x) = ax³ + 3bx² + 3cx + d

sum of roots = -coefficient of x²/coefficient of x³

P -r + P + P+ r = -( 3b/a)

3P = -( 3b/a)

P = -b/a -----------(1)


sum of products of two roots =
( P -r)×P + P×(P+r) + (P-r)(P+r) = coefficient of x/coefficient of x³

P² -Pr + P² +Pr + P² -r² = 3c/a

3P² - r² = 3c/a ------------------(2)


again ,

products of all roots = -constant /co-efficient of x³

(P-r)×P× (P+r) = - d/a

P³ - Pr² = -d/a ----------(3)

now,

equation (1) put in eqn (2) ,

3(-b/a)² -r ² = (3c/a)

r² = 3(b²/a²) -3(c/a) ----------(4)

eqn (4) and (1) put in eqn(3)

(-b/a)³ -(-b/a)( 3b²/a² -3c/a ) = -d/a

-b³/a³ +b/a(3b²/a² -3c/a ) = -d/a

-b³/a³ +3b³/a³ -3bc/a² = - d/a

2b³/a³ -3bc/a² = -d/a

2b³ -3abc = -da²

2b³ -3abc + a²d = 0

hence, proved ///

Answer 2

here is your answer