Question

please help! :))
2. Solve for y: y^2 + y + 1 = 0
using the quadratic formula
I don't know how to do this so bare with me:)

Answer 1

Do you know the formula?

Answer 2

do you allow complex solutions? because the answer will not be real

Answer 3

y = [-b ± √(b^2 - 4ac)] / 2a

Answer 4

yup
use \(a=1,b=1,c=1\)

Answer 5

and @satellite73 I have no clue

Answer 6

oh

Answer 7

the number under the radical will be \(1-4=-3\) so the solutions are complex

Answer 8

write them directly in to your formula
get
\[\frac{-1\pm\sqrt{1-4}}{2}\] as a first step

Answer 9

then since \(1-4=-3\) you get
\[\frac{-1\pm\sqrt{-3}}{2}\]

Answer 10

okay so you basically cleaned it up

Answer 11

if you are working with complex numbers you can rewrite this as
\[\frac{-1\pm\sqrt{3}i}{2}\] and if you are not, you say "no solution"

Answer 12

yeah the only clean up is \(1-4=-3\) that is really all the computation necessary

Answer 13

\[Ax ^{2}+Bx+C\]

Quadratic formula: \[x= \frac{-B+-\sqrt{B ^{2}-4AC} }{ 2A }\]

Answer 14

but i'm a tad bit confused?

Answer 15

i'm trying to find y

Answer 16

yeah, that is \(y\)

Answer 17

actually you get two answer, the plus and the minus
i.e. either
\[y=\frac{-1+\sqrt{-3}}{2}\] or \[y=\frac{-1-\sqrt{-3}}{2}\] those are your two solutions

Answer 18

got it?

Answer 19

okay so i basically understand except how you got the number under the radical

Answer 20

the number under the radical is \(b^2-4ac\)
in your case \(a=1\) and \(b=1\) and even \(c=1\) so literally you get \(1^2-4\times 1\times 1\) under the radical

Answer 21

but this is really \(1-4=-3\)

Answer 22

if you got another one we can try that and see how it works in a different example

Answer 23

I'm sorta getting the concept. I'm gonna write this down and look at it more to understand it and that was the only quadratic problem so I don't have another example.

Answer 24

oh ok

Answer 25

thank you so much though:)

Answer 26

yw