Question 1199134
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Method 1


Use the quadratic formula to see the two roots are {{{x = (-b+sqrt(b^2-4ac))/(2a))}}} and {{{x = (-b-sqrt(b^2-4ac))/(2a))}}}


We can simplify things down a bit to say the two roots are {{{x = (-b+S)/(2a))}}} and {{{x = (-b-S)/(2a))}}} where {{{S = sqrt(b^2-4ac)}}}
When adding those roots together, the "S" terms cancel (since we have a positive S added to a negative S).


We'll then have {{{(-b+(-b))/(2a) = (-2b)/(2a) = -b/a}}} as the sum of the two roots, where 'a' is nonzero.


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Method 2


The vertex is located at the point (h,k)
The formula for h is
{{{h = -b/(2a)}}}
where 'a' is nonzero.
This formula is useful for completing the square to get the equation {{{y = ax^2+bx+c}}} into vertex form {{{y = a(x-h)^2+k}}}


The equation {{{x = h}}} represents the vertical line through the vertex.
This vertical line is known as the axis of symmetry.


Let the two roots be p and q.
If we knew the roots, we can average them to determine the value of h.
We'll use this fact to isolate {{{p+q}}}.


{{{h = (p+q)/2}}}    ....    h is the midpoint of p and q


{{{2*h = p+q}}}


{{{p+q = 2*h}}}


{{{p+q = 2*(-b/(2a))}}}     ....   plug in h = -b/(2a)


{{{p+q = -b/a}}}


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Method 3


p and q are the roots of {{{y = ax^2+bx+c}}}


Meaning:
{{{ap^2+bp+c = 0}}}
and
{{{aq^2+bq+c = 0}}}
Each input (x = p and x = q) lead to an output of y = 0.
i.e. p and q are the x-intercepts on the parabola.


Subtract the equations straight down
{{{(ap^2+bp+c) - (aq^2+bq+c) = 0}}}


{{{ap^2+bp+c - aq^2-bq-c = 0}}}


{{{(ap^2-aq^2) + (bp-bq) + (c-c) = 0}}}


{{{a(p^2-q^2)+b(p-q) = 0}}}


{{{a(p-q)(p+q)+b(p-q) = 0}}}


{{{(p-q)(  a(p+q)+b  ) = 0}}}


From that, either
{{{p-q = 0}}}, or
{{{a(p+q)+b = 0}}}


If the first scenario is the case, then {{{p-q = 0}}} leads to {{{p = q}}}. 
This isn't particularly useful.


So we move onto the second scenario.
{{{a(p+q)+b = 0}}}


{{{a(p+q) = -b}}}


{{{p+q = -b/a}}}
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