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Rooting for Three Roots (Posted on 2012-12-23) Difficulty: 3 of 5
The cubic equation x3 + ax+ a = 0 has three roots x1, x2 and x3 with x1 ≤ x2 ≤ x3, where a is real and non zero, such that:

x12/ x2 + x22/ x3 + x32/ x1 = 8

Determine x1, x2 and x3

No Solution Yet Submitted by K Sengupta    
Rating: 4.0000 (1 votes)

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Hints/Tips solving for a (spoiler) | Comment 1 of 3

Lets say the three roots are b, c, and d (just to avoid subscripts).

Then (x-b)(x-c)(x-d) = 0  Multiplying and comparing to the 1st equation gives us:    b+c+d = 0    bc+bd+cd = a    bcd = -a      The 2nd equation becomes b^2/c + c^2/d + d^2/b = 8    Multiplying by bcd gives b^3*d + c^3*b + d^3*c = 8bcd  but we can substitute b^3 = -ab -a                        c^3 = -ac -a                        d^3 = -ad -a  and after rearranging terms,    -a(bc + bd + cd) -a(b+c+d) = 8bcd      and (using the identities above) this becomes      -a(a) -a(0) = -8a  or   a^2 = 8a    This has two solutions, a = 0 and a = 8  But a = 0 implies b = c = d = 0, which doesn't work    so, a = 8    

And, if I haven't made mistakes, then all you need 

to do is solve

x^3 + 8x + 8 = 0
which is a known formula.

Edited on April 25, 2014, 4:33 pm
  Posted by Steve Herman on 2012-12-23 15:52:11

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