SOLUTION: Given that 2 + 3i is a solution of a quadratic equation with real coefficients.Find the other solution. I need the steps.

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Question 1207477: Given that 2 + 3i is a solution of a quadratic equation with real coefficients.Find the other solution.

I need the steps.
Found 3 solutions by josgarithmetic, ikleyn, math_tutor2020:
Answer by josgarithmetic(39617) About Me  (Show Source):
You can put this solution on YOUR website!
One of the factors is %28x-%282%2B3i%29%29 and the other factor is %28x-%282-3i%29%29.

Equation may start as %28x-%282%2B3i%29%29%28x-%282-3i%29%29=0.

A little bit of algebra, %28x-2-3i%29%28x-2%2B3i%29=0.
%28%28x-2%29-3i%29%28%28x-2%29%2B3i%29=0
%28x-2%29%5E2-%283i%29%5E2=0
Now, you take this the rest of the way.

Much of the steps for this example are not necessary, since complex solutions for quadratic equations come in conjugate pairs.

Answer by ikleyn(52788) About Me  (Show Source):
You can put this solution on YOUR website!
.

The general theorem of algebra states that if a polynomial with real coefficients
has a complex number root  a+bi,  b=/= 0,  then it has another complex number root  a-bi,  too.

According to this theorem,  if  2+3i  is the root of your quadratic equation with real coefficients,
then it has the root  2-3i,  too.   It is your other root.



Answer by math_tutor2020(3817) About Me  (Show Source):
You can put this solution on YOUR website!

Answer: 2-3i

Reason: If a+bi is one solution then a-bi represents its mirrored counterpart.
This is known as the complex conjugate.
It only applies when all coefficients are real numbers.

For a more in-depth look at a similar problem, check out this link
https://www.algebra.com/algebra/homework/complex/Complex_Numbers.faq.question.1207475.html