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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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Some Thoughts re: agree, but . . . Comment 3 of 3 |
(In reply to agree, but . . . by xdog)

x^3+ax+a=0; assume that a=-8; then x^3-8x-8=0 also solves for x=-2:(-8)+16-8=0.

Since this implies that (x+2)(x^2-2x-4)=0, the roots are: {-2, 1-5^(1/2), 1+5^(1/2)}

But in all combinations the stipulated summations total -8, not 8.


  Posted by broll on 2012-12-24 04:23:35
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