Question 1104988
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0.  Let the original equation be  {{{ax^2 + bx + c}}} = 0.     (1)


1.  After the 1-st student incorrectly copied the coefficient at  {{{x^2}}},  the equation took the form

    {{{dx^2 + bx + c}}} = 0.     (2)


    Since its roots are 2 and 3, we have this decomposition

    {{{dx^2 + bx + c}}} = d*(x-2)*(x-3),     or

    {{{dx^2 + bx + c}}} = {{{dx^2 - 5dx + 6d}}}.

    So, the original equation (1) has the two lowest degree terms -5dx + 6d:

        {{{ax^2 + bx + c}}} = {{{ax^2 - 5dx + 6d}}}    (3)

    with some unknown coefficients  "a"  and  "d".



2.   After the 2-nd student incorrectly copied the constant term,  the equation took the form

    {{{ax^2 - 5dx + e}}} = 0.     (4)


    Since its roots are 4 and 6, we have this decomposition

    {{{ax^2 - 5dx + e}}} = a*(x-4)*(x-6),     or

    {{{ax^2 - 5dx + e}}} = {{{ax^2 - 10ax +24a}}}.

    It implies  -5d = -10a,  which in turn  implies  d = 2a.

    Now from (3) we conclude that the original equation (polynomial) is/was

        {{{ax^2 - 5dx + 6d}}} = {{{ax^2 - 10ax + 12a}}}.

    Its roots are the same as for equation  {{{x^2 - 10x + 12}}} = 0.

    And they are  {{{x[1,2]}}} = {{{(10 +- sqrt (10^2 -4*12))/2}}} = {{{5 +- sqrt(13)}}}.


<U>Answer</U>.  The roots of the original equation are  {{{x[1]}}} = {{{5 + sqrt(13)}}}  and  {{{x[2]}}} = {{{5 - sqrt(13)}}}.
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