SOLUTION: Can you help me rewrite this equation in quadratic form? (x-(2+i))(x-(2-i))= 0

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Question 395529: Can you help me rewrite this equation in quadratic form?
(x-(2+i))(x-(2-i))= 0

Found 2 solutions by stanbon, richard1234:
Answer by stanbon(75887) About Me  (Show Source):
You can put this solution on YOUR website!
(x-(2+i))(x-(2-i))= 0
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Rearrange:
((x-2)-i)((x-2)+i) = 0
-----
(x-2)^2-i^2 = 0
---
(x-2)^2+1 = 0
---
x^2-4x+4+1 = 0
---
x^2-4x+5 = 0
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Use Quadratic Formla:
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x = [4 +- sqrt(16-4*5)]/2
---
x = [4 +- sqrt(-4)]/2
---
x = [4+- 2i]/2
---
x = 2+i or x = 2-i
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Cheers,
Stan H.
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Answer by richard1234(7193) About Me  (Show Source):
You can put this solution on YOUR website!
We could simply multiply the whole thing out, rewriting as %28%28x-2%29-i%29%28%28x-2%29%2Bi%29 as the other tutor did (so that the difference of two squares can be carried out). This method uses a set of formulas called Vieta's formula:

The polynomial is in the form ax%5E2+%2B+bx+%2B+c. Right off the back we deduce a+=+1 since we can tell that the x^2 coefficient is 1 without even multiplying. Vieta's formulas say that:

The sum of the roots of the polynomial is -b/a = -b. Since the roots are 2+i, 2-i, their sum is 4, so b = -4.

The product of the roots is c/a = c. Since %282%2Bi%29%282-i%29+=+4+%2B+1+=+5 (applying difference of two squares) we get c = 5.

Therefore the polynomial is x%5E2+-+4x+%2B+5.

Here are a couple articles on Vieta's formulas, just in case (I rarely see these formulas taught in school, but they're pretty useful to know):
http://en.wikipedia.org/wiki/Fran%C3%A7ois_Vi%C3%A8te
http://www.artofproblemsolving.com/Wiki/index.php/Vieta's_Formulas