If r and s are the roots , then
(x-r)(x-s) = 0
x^2 - rx - sx + rs = 0
x^2 + x(-r-s) + rs = 0
compare with
ax^2 + bx + c = 0
or
x^2 + (b/a)x + c/a = 0
then -r-s = b/x ---> r+s = -b/a
and c/a = rs
so r+s = -b/a and rs = c/a
Assume r and s are the roots of the quadratic equation ax^2 + bx + c = 0, show that r + s = -b/a and rs = c/a.
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