trig
posted by KELLY .
How do you write the quadratic equation with integer coefficients that have the roots 73i and 7+3i?

In the quadratic equation
ax²+bx+c=0
b/a = (sum of the two roots)
c/a = product of the two roots.
Assume a=1, then
b=(73i+7+3i)=14
c=(73i)*(7+3i)=49+9=58
So the quadratic equation is
x²14x+58=0
Check by solving the equation and you should get back 7±3i as the roots. 
x = 7 + 3i, and X = 7  3i.
We can use either of the 2 values of X:
X = 7 + 3i,
(X  7) = 3i,
Square both sides:
X^2 14X + 49 = 9(1),
X^2  14X + 49 +9 = 0,
X^2  14X + 58 = 0.
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