Question

Algebra 2 help! Medal :)

Answer 1

g(x)= x^2+6x+1

Answer 2

@ganeshie8

Answer 3

@SolomonZelman

Answer 4

The attachment doesn't work for me, what are you asked to do, to find the zeros ?

Answer 5

no to find the solutions of g(x)

Answer 6

\(\large\color{blue}{ \rm g(x)~=~ x^2~+~6x~+~1 }\)

\(\large\color{blue}{ \rm x^2~+~6x~+~1~=~0 }\)

\(\large\color{blue}{ \rm x^2~+~6x~+~1~+8=~0+8 }\)

\(\large\color{blue}{ \rm x^2~+~6x~+~9=~+8 }\)

\(\large\color{blue}{ \rm (x~+~3)^2=~+8 }\)

\(\large\color{blue}{ \rm x~+~3~=~±~ \sqrt{8} }\)

\(\large\color{blue}{ \rm x~+~3~=~±~ 2\sqrt{2} }\)

\(\large\color{blue}{ \rm x~=~-~3±~ 2\sqrt{2} }\)

Answer 7

These are the zeros of the function. The y intercept is (0,1)

( As, ax^2+bx+ ` c ` = 0, and in this case c is 1 )

Answer 8

The solution is any point on the line, plug in any x value to solve for y.

\(\large\color{darkgreen}{ \rm g(x)= x^2+6x+1 }\)
\(\large\color{darkgreen}{ \rm g(x)= (-2)^2+6(-2)+1 }\)
\(\large\color{darkgreen}{ \rm g(x)= 4+(-12)+1 }\)
\(\large\color{darkgreen}{ \rm g(x)= -7 }\)

hence, ` ( -2, -7 ) `

Answer 9

` (-2,-7) ` is just an example .

Answer 10

so what are the solutions? im confused @SolomonZelman

Answer 11

@ganeshie8

Answer 12

http://prntscr.com/3r54v5

Answer 13

the solutions are :
\(\large\color{blue}{ \rm x~=~-~3±~ 2\sqrt{2} } \)

Answer 14

that's it?

Answer 15

Those are the zeros. The solutions are all the points on the line. (∞ number of points )

Answer 16

work is there in that link ^

Answer 17

they mean solutions of g(x) = 0 i think @SolomonZelman - they're bit careless :)

Answer 18

Yes, I also thought that they mean that. Otherwise, the question is easier. Just plugging in numbers for x is easier than solving the quadratic.
(For people that don't solve quadratics and plug in numbers instantly )

Answer 19

true lol

Answer 20

thank you guys @SolomonZelman @ganeshie8

Answer 21

`Y` `W` `!`

Answer 22

wait one question to that... @SolomonZelman how did you get 8 in the second step?

Answer 23

I added 8 to both sides.

Answer 24

but whered you get 8 from?

Answer 25

I was thinking like this.

What should I add to both sides to make the left side a specter square.

Answer 26

I mean a perfect square, not a specter square.

Answer 27

oh haha ok I was like that wouldn't make sense

Answer 28

Alright, so now you see what I did, right ?