Question

Derive a formula for the sum of solutions of the quadratic equation ax^2 + bx + c in terms of a and b

Answer 1

every quadratic equation will factor to \[(x-r _{1})(x-r _{2})=0\]and thus the complex roots will be \[r _{1} ;\ \ r _{2}\]if we take\[ax^2+bx+c=0\]and divide it by a (assuming a does not equal 0) we get\[x^2+\frac{b}{a}x+\frac{c}{a}=0\]now FOIL the first equation\[(x-r _{1})(x-r _{2})=x^2-(r _{2}+r_{1})x+(r_{1}r_{2})^2\]equating the second and third terms of the last two equations we see that\[\frac{c}{a}=r _{1}*r _{2}\]and the negative of the roots must sum to b/a:\[\frac{b}{a}=-(r _{1}+r _{2})\]or, as it is often written\[-\frac{b}{a}=r_{1}+r_{2}\]

Answer 2

Thanks I get it now

Answer 3

ur welcome