- Oct 12, 2004
- 5,139
- 1
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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!
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!
