PLEASE HELP!!1

imported_Lucifer

Diamond Member
Oct 12, 2004
5,139
1
0
Ok I need to find 2 non real solutions. The original equation is x^2+2x+5. I used the quadratic formula to try to find the 2 non real solutions.

a=1
b=2
c=5

-(2)± v(2)^2-4(1)(5)
--------------------------
2(1)

The square root for (2)^2-4(1)(5) comes to -16
I know you cannot square root a - so you have to take the negative from the number like this:

v-1*v16 Square root of 16 is 4, while square root of the -1 is i.

Result:

-2± 4i
-------
2


Ok I am stuck there. What do I do to get the solutions? Do the 4 and the 2 cancel out, leaving me with one solution as -2+2i and the other as -2-2i? I am soo confused.

Thanks for any help! :)
 

ts3433

Platinum Member
Jun 29, 2004
2,731
0
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Simplify that fraction and then expand the part with the plus or minus sign so that you have two solutions (alternatively, leave the plus or minus sign in there because that sign means you have two solutions).
 

minendo

Elite Member
Aug 31, 2001
35,560
22
81
Originally posted by: Lucifer
Originally posted by: minendo
1-2i and 1+2i

So the negative 2 and the postive 2 divide with each other too?

Yes, you have to divide the whole numerator by denominator to simplify. Do not forget to include your - sign in from of the 1.
 

SaturnX

Diamond Member
Jul 16, 2000
3,415
0
76
Damnit man, it's not i, it's j!!! j I tell you! (well me and anyone else in EE)

--Mark