A quadratic equation with imaginary numbers

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Question from Alissa, a student:

I am solving a quadratic equation and I got this far;
(x-4+i)(x-4-i)=0
but how do I add the imaginary numbers
i know you multiply x by x and then add -4 + -4 but what do you do with the i's?
Thanks for your help.

Alissa,

It looks like you factored x² - 8x + 17 and got (x - 4 + i)(x - 4 - i) and hence your quadratic equation x² - 8x + 17 became (x - 4 + i)(x - 4 - i) = 0. At this point notice that you have two numbers (x - 4 + i) and (x - 4 - i), and their product is zero. Thus one of the two numbers must be zero. That is either

x - 4 + i = 0

x - 4 - i = 0.

If the first is true then add 4 - i to each side to obtain x = 4 - i. Similarly solve the second equation.

I hope this helps,
Penny

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