Question 166121
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 any quadratic is {{{-b/a}}}




Note: once you have the sum of the two roots you can find the vertex (even if you do NOT know the two roots all by themselves).



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<h4>Multiplying the Roots:</h4>


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



{{{x[1]*x[2]}}} Now multiply 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)(2a))}}} Combine the fractions.



{{{((-b + w)(-b - w))/(4a^2)}}} Multiply {{{(2a)(2a)}}} to get {{{4a^2}}}



{{{(b^2+bw-bw-w^2)/(4a^2)}}} FOIL the numerator



{{{(b^2-w^2)/(4a^2)}}} Combine like terms. Notice how the "bw" terms cancel out.



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Now remember all the way back to the start of the problem. We let {{{w=sqrt( b^2-4ac )}}}. So this means that 


{{{w^2=(sqrt( b^2-4ac ))^2=b^2-4ac}}}



Or in other words: {{{w^2=b^2-4ac}}}


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{{{(b^2-(b^2-4ac))/(4a^2)}}} Replace {{{w^2}}} with {{{b^2-4ac}}}



{{{(b^2-b^2+4ac)/(4a^2)}}} Distribute the negative



{{{(4ac)/(4a^2)}}} Combine like terms. The {{{b^2}}} term is now gone.



{{{c/a}}} Reduce



So this shows us that the product of two roots of any quadratic is {{{c/a}}}