# trig

How do you write the quadratic equation with integer coefficients that have the roots 7-3i and 7+3i?

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ax²+bx+c=0
b/a = -(sum of the two roots)
c/a = product of the two roots.

Assume a=1, then
b=-(7-3i+7+3i)=-14
c=(7-3i)*(7+3i)=49+9=58
x²-14x+58=0

Check by solving the equation and you should get back 7±3i as the roots.

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2. 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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