x 3 − 6 x 2 + 5 x − 7 = 0Find 1 s 2 where r , s , t are roots of the equation.

Question

x    3    −  6      x    2    +  5  x  −  7  =  0Find       1          s      2       where   r  ,  s  ,  t are roots of the equation.

heilaritikermx · Accepted Answer

x    3    −  6      x    2    +  5  x  −  7  =  0Let   y  =      1    x    . Then      1          y      3        −      6          y      2        +      5    y    −  7  =  0    7      y    3    −  5      y    2    +  6  y  −  1  =  0.So, by Vieta&#039;s Formulae,      1          r      2        +      1          s      2        +      1          t      2          =      y    1    2    +      y    2    2    +      y    3    2      =  (      y    1    +      y    2    +      y    3        )    2    −  2  (      y    1        y    2    +      y    2        y    3    +      y    3        y    1    )    =            (              5        7            )        2    −  2      (          6      7        )      =  −      59    49    .

Landon Mckinney · Accepted Answer

You can also use this theorem:Let       α    1    ,      α    2    ,      α    3    ⋅  ⋅  ⋅      α    n   be the root of the polynomial equation   f  (  x  )  =  0 of the nth degree. Then the symmetric function:      S    r    =      α    1          −      r        +      α    2          −      r        +  ⋅  ⋅  ⋅  +      α    n          −      r      where r is a positive integer, is equal to the coefficient of       x          r      −      1       in the expansion of   −      f    ′    (  x  )      /    f  (  x  ) in the ascending powers of x, where       f    ′    (  x  ) is the first derived function.For your equation we have:      f    ′    (  x  )  =  3      x    2    −  12  x  +  5            −              f        ′            (      x      )              f      (      x      )        =            −      5      +      12      x      −      3              x        2                    −      7      +      5      x      −      6              x        2            +              x        3              =      5    7    −      59    47    x  +  ⋅  ⋅  ⋅You can find the expansion by direct division of       f    ′    (  x  ) by   f  (  x  ). So we have:      1          r      2        +      1          s      2        +      1          t      2        =  −      59    49

x 3 </msup> − 6 x 2 </msup> + 5 x −<!

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