## the product of zeroes of P(x) = ax3 -6×2 +11x – 6 is 4 , then a = ​

Question

the product of zeroes of P(x) = ax3 -6×2

+11x – 6 is 4 , then a = ​

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1 month 2021-08-16T20:08:40+00:00 2 Answers 0 views 0

## Answers ( )

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 ///