Question 1197055
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One solution of z3+ A*z^2 - z - 14 = 0 is z = -2-root(3)*i, where A is a real number. 
Find A and other two solutions
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            The solution in the post by math_tutor2020 may scare a reader - so complicated it is

            with tons of calculations.


            Meanwhile,  the problem can be solved mentally in a very simple way,  if to use  Vieta's theorem.



<pre>
Since the given equation is with real coefficients (given), its complex roots are conjugated complex numbers.

Since one such root is  {{{z[1]}}} = {{{-2-i*sqrt(3)}}}  (given), the other root is  {{{z[2]}}} = {{{-2+i*sqrt(3)}}}.


Vieta's theorem says that the product of three roots of the given equation equals to the constant term 
with the opposite sign. So

    {{{z[1]*z[2]*z[3]}}} = 14.


The product of the two complex conjugated numbers  {{{-2-i*sqrt(3)}}}  and  {{{-2+i*sqrt(3)}}}  is 

    {{{(-2)^2 + (sqrt(3))^2}}} = 4 + 3 = 7.


Therefore, the third root of the given equation is  {{{z[3]}}} = {{{(14)/7}}} = 2.


Now the coefficient A equals to the sum of the three roots with the opposite sign, due to the same Vieta's theorem

    A = {{{-(z[1] + z[2] + z[3])}}} = {{{-( (-2-i*sqrt(3)) + (-2+i*sqrt(3)) + 2)}}} = -(-2-2+2) = 2.


<U>ANSWER</U>.  A = 2.  The other two roots are {{{-2+i*sqrt(3)}}} and 2.
</pre>

Solved.


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Actually, &nbsp;the solution and the calculations are so simple that can be made mentally in the head.


The problems of this type are usually &nbsp;INTENDED, &nbsp;DESIGNED &nbsp;and &nbsp;EXPECTED &nbsp;to be solved in this way, &nbsp;using &nbsp;Vieta's theorem.


Then the solution is easy, elegant and provides fun for a student, &nbsp;showing a beauty of &nbsp;Math,

instead of making many tons of unnecessary complicated calculations.



&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;And only in this context it has an educational value.



On &nbsp;Vieta's theorem see, &nbsp;for example, &nbsp;this &nbsp;Wikipedia article

https://en.wikipedia.org/wiki/Vieta%27s_formulas



This problem is a typical of a high school Math circle level.