I' m having trouble to prove Viete' s relation, which is the following: X1+X2+X3= -b/a With a quadratic equation I can work out two of the X' s so: [(-b+sqrt(D))/2a]+[(-b-sqrt(D))/2a] +X3= -b/a If I work this out I' m left with: (-2b/2a)+ X3 = -b/a or -b/a +X3 = -b/a So either I' m terrible wrong or either I have to proof that X3 equals 0.I would be very grateful if anyone could help me on this

Expert Answers

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Viete's formulas exist for all algebraic equations. `x_1+x_2+x_3=-b/a` is one of the formulas for 3rd degree polynomial equation (for quadratic equation equivalent formula is `x_1+x_2=-b/a,` that's the reason you get 0 - there are only two solutions to quadratic equation).

Now to prove the formula.

First we write general cubic equation.


Now because of the fundamental theorem of algebra this is equivalent to


Multiply terms in first two parethesis.


Now multiply terms in remaining two parentesis.


Group the terms with the same power of `x` and multiply everything by `a.`


Since this equation is equivalent to the first equation it follows that



You can similarly prove other Viete's formulas.

` ` `x_1x_2+x_1x_3+x_2x_3=c/a`



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