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What is m if roots of Equation x^3-3x^2-6x+m=0 are arithmetic progression?

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mccee | Student, Undergraduate | (Level 1) eNoter

Posted July 9, 2012 at 3:10 PM via web

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What is m if roots of Equation x^3-3x^2-6x+m=0 are arithmetic progression?

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sciencesolve | Teacher | (Level 3) Educator Emeritus

Posted July 9, 2012 at 3:59 PM (Answer #1)

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You should identify the number of roots of equation, hence, since the equation is of third order, then it has three roots.

The problem provides the information that the roots are members of arithmetic progression such that:

`x_2 = (x_1 + x_3)/2 =gt x_1 + x_3 = 2x_2`

You should use the first Vieta's relations such that:

`x_1+x_2+x_3 = 3`

Substituting `2x_2`  for `x_1 + x_3`  yields:

`2x_2 + x_2 = 3 =gt 3x_2 = 3 =gt x_2 = 1`

You should use the third and second Vieta's relation such that:

`x_1*x_2*x_3 = -m`

Since `x_2 = 1 =gt x_1*x_3 = -m`

`x_1*x_2 + x_1*x_3 + x_2*x_3 = -6`

`x_1 + x_1*x_3 + x_3 = -6`

Using `x_1 + x_3 = 2x_2 =gtx_1 + x_3 = 2 ` yields:

`2 + x_1*x_3 =-6 =gt x_1*x_3 = -6-2 =gt x_1*x_3 = -8`

Notice that the third Vieta's relation states that `x_1*x_3 = -m` , hence`-m = -8 =gt m = 8`

Hence, evaluating m under given conditions yields m=8.

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