Question

Find the solutions of g(x)=x^2+6x+1

Answer 1

explain each step.

Answer 2

I assume you mean the zeros of g(x)?

Answer 3

or the solution to \(x^2+6x+1=0\)

Answer 4

yes but idk what to do from that poing @zzr0ck3r

Answer 5

have you seen this \(\frac{-b\pm\sqrt{b^2-4ac}}{2a}\) before ?

Answer 6

yes but I never really got what it was used for

Answer 7

its used for this question

Answer 8

oh okay

Answer 9

it will give the zeros of the polynomial \(ax^2+bx+c\) for you \(a=1, b=6,\) and \(c=1\)

Answer 10

so then you plug the number in and whatever you get is the answer?

Answer 11

yeah

Answer 12

\(\frac{-6\pm\sqrt{6^2-4*1*1}}{2*1}\)

Answer 13

you will get two answers

Answer 14

how?

Answer 15

the \(\pm\) will lead to two answers

Answer 16

what that reall is, is \(\frac{-6+\sqrt{6^2-4*1*1}}{2*1}\) and \(\frac{-6-\sqrt{6^2-4*1*1}}{2*1}\)

Answer 17

really*

Answer 18

ooooh okay. so how would you solve this part \[-6-\sqrt{6^{2}-4*1*1}\] I don't do good with radicals

Answer 19

what is 6^2?

Answer 20

6 squared

Answer 21

\(\frac{-6+\sqrt{6^2-4*1*1}}{2*1}=\frac{-6+\sqrt{36-4}}{2}=\frac{-6+\sqrt{32}}{2*1}=\frac{-6+\sqrt{16*2}}{2}=\\\frac{-6+\sqrt{16}\sqrt{2}}{2}=\frac{-6+4\sqrt{2}}{2}=-3+2\sqrt{2}\)

Answer 22

its the same thing with the other equation too?