SOLUTION: Can you help me rewrite this equation in quadratic form?
(x-(2+i))(x-(2-i))= 0
Algebra.Com
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) (Show Source): You can put this solution on YOUR website!
(x-(2+i))(x-(2-i))= 0
----
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
----
Use Quadratic Formla:
----
x = [4 +- sqrt(16-4*5)]/2
---
x = [4 +- sqrt(-4)]/2
---
x = [4+- 2i]/2
---
x = 2+i or x = 2-i
========================
Cheers,
Stan H.
==========
Answer by richard1234(7193) (Show Source): You can put this solution on YOUR website!
We could simply multiply the whole thing out, rewriting as 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 . Right off the back we deduce 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 (applying difference of two squares) we get c = 5.
Therefore the polynomial is .
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
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