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Thread: quadratic problem

  1. #1
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    quadratic problem

    if x^2-2ra[r]x+r=0,r=1,2,3 are three quadratic equations of which each pair has exactly one root common then the number of solutions of the triplet (a1,a2,a3) is
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  2. #2
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    Your notation is a bit strange. Do you mean x^2- 2a_1x+ 1= 0, x^2- 4a_2x+ 2= 0, and x^2- 6a_3x+ 3= 0?


    By the quadratic formula, the solution to the first is x= a_1\pm\sqrt{a_1^2- 1}, to the second x= 2a_2\pm\sqrt{4a_2^2- 2}, and to the third x= 3a_3\pm\sqrt{9a_3- 3}. Compare those in pairs so see how you can get one common root.
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  3. #3
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    how do we compare these two in pairs
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