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Arithmetic Roots (Posted on 2016-03-02)

For what values k does the polynomial f(x) = x^4 + 7x^3 + 9x^2 - 7x + k have three roots in arithmetic progression?

See The Solution Submitted by Brian Smith

Rating: 4.0000 (1 votes)

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Analytic solution (outline)

Comment 2 of 2 |

Outline Solution

Let the roots be r, r + t, r – t, s, so that they satisfy the equation

            (x – r)(x – r – t)(x – r + t)(x – s) = 0

Equating coefficients with those in x⁴ + 7x³ + 9x² – 7x + k = 0

gives:                - 3r – s = 7                                (1)

                        3r² + 3rs – t² = 9                       (2)

                        - r³ - 3r²s + rt² + st² = -7            (3)

                        r³s – rst² = k                              (4)

From (1) and (2), s = -3r – 7      and       t² = 3r² + 3rs – 9    (5)

Thus (3) gives - r³ – 3r²(-3r – 7) + (-2r – 7)( 3r² + 3rs – 9) = -7

which simplifies to          20r³ + 105r² + 165r + 70 = 0

                                    (r + 2)(20r² + 65r + 35) = 0

with roots                      r = - 2, (-13 +/- sqrt(57))/8

Using (4),          k = rs(r² – t²)      which expands, using (5),

to                     k = -(21r⁴ + 112r³ + 174r² + 63r)

The three values found above for r can now be substituted to
give
            k = -10, (-553 – 159(-13 +/– sqrt(57))/8)/64

i.e.        k = -10, -2.258938..., -6.948093…

which agree with Charlie’s posting.

Posted by Harry on 2016-03-02 12:19:05

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