Question 166538
Remember the quadratic formula is:



{{{x = (-b +- sqrt( b^2-4ac ))/(2a)}}} note: {{{a<>0}}}



Now let {{{w=sqrt( b^2-4ac )}}} (to simplify things a bit)



So the quadratic formula becomes



{{{x = (-b +- w)/(2a)}}} 



which really breaks down to



{{{x = (-b + w)/(2a)}}} or {{{x = (-b - w)/(2a)}}}



So the first root is {{{x[1] = (-b + w)/(2a)}}} and the second root is {{{x[2] = (-b - w)/(2a)}}} 



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<h4>Adding the Roots:</h4>


{{{x[1]+x[2]}}} Now let's add the roots



{{{(-b + w)/(2a)+(-b - w)/(2a)}}} Plug in {{{x[1] = (-b + w)/(2a)}}} and  {{{x[2] = (-b - w)/(2a)}}} 



{{{(-b + w-b - w)/(2a)}}} Combine the fractions.



{{{((-b-b) + (w-w))/(2a)}}} Group like terms.



{{{(-2b)/(2a)}}} Combine like terms. Notice how the "w" terms cancel each other out completely.



{{{-b/a}}} Reduce



So this shows us that the sum of the roots of {{{ax^2+bx+c}}} is {{{-b/a}}}