Question 253132
To find any quadratic equation of the form {{{ax^2 + bx + c = 0}}}, we have to realize that:


a = 1 (always)
b = - (sum of roots)
c = product of roots


Since a is always 1, and the sum of the roots = 12, then b = - (12) = -12


We now have: {{{x^2 - 12x + c = 0}}} 


To find c, we need to 1st determine the roots and multiply them


Let root 1 be {{{r[1]}}}, and root 2, {{{r[2]}}}


Since the sum of the roots = 12, then ----- {{{r[1] + r[2] = 12}}} --------- eq (i)
Also, since the roots’ difference = 4i, then {{{r[1] - r[2] = 4i}}} --------- eq (ii)


Adding equations (i) & (ii), we get: {{{2r[1] = 12 + 4i}}} 


{{{r[1] = (12 + 4i)/2}}}  or {{{r[1] = 6 + 2i}}} 


Substituting 6 + 2i for {{{r[1]}}} in eq (i), we get: {{{6 + 2i + r[2] = 12}}}


{{{r[2] = 12 - 6 - 2i}}}


{{{r[2] = 6 - 2i}}} 


Since we now have both roots, {{{r[1] = 6 + 2i}}} and {{{r[2] = 6 - 2i}}}, we multiply these two roots to get “c.”


Therefore, “c” = (6 + 2i)(6 – 2i) = {{{36 - 4i^2}}}, or, 36 – 4(-1) = 40

With “a” being 1, “b” being – 12, and “c” being 40, the quadratic equation in the form {{{ax^2 + bx + c = 0}}} = {{{highlight_green(x^2 - 12x + 40 = 0)}}}