Posted by **KELLY** on Tuesday, December 21, 2010 at 5:33pm.

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

- trig -
**MathMate**, Tuesday, December 21, 2010 at 5:56pm
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=-(7-3i+7+3i)=-14

c=(7-3i)*(7+3i)=49+9=58

So the quadratic equation is

x²-14x+58=0

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

- trig -
**Henry**, Wednesday, December 22, 2010 at 6:39pm
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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