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

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    0
    2021-08-16T20:10:25+00:00

    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 ///

    0
    2021-08-16T20:10:34+00:00

    Answer:

    chal bosdk ans galt hai mader chodar tere me demag nhi h to ans mat dala kar sale hamare paper h tere baap ki bharat nhi kuch bhi lik rha h bhai eska ans galt hai koi mat likna ok

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