Question

I NEED HELP PLEASE! find the solutions for g(x) = x^2 +6x + 1

Answer 1

need help?

Answer 2

yes please!

Answer 3

Have you learned about the quadratic formula at all?

Answer 4

yeah i just forget how to get the solutions of this formula i know i need an a, b and c after that i kind forget

Answer 5

x^2 +6x + 1

really is

1x^2 +6x + 1

Answer 6

1x^2 +6x + 1

is in the form

ax^2 + bx + c

where in this case

a = 1
b = 6
c = 1

Answer 7

yeah i got that far but after that I'm confused on what to do next

Answer 8

Then you would use the quadratic formula

\[\Large x=\frac{-b \pm \sqrt{b^2-4ac}}{2a}\]

Answer 9

Plug in a = 1, b = 6, c = 1

\[\Large x=\frac{-b \pm \sqrt{b^2-4ac}}{2a}\]

\[\Large x=\frac{-6 \pm \sqrt{6^2-4*1*1}}{2*1}\]

\[\Large x=\frac{-6 \pm \sqrt{36-4*1*1}}{2*1}\]

\[\Large x=\frac{-6 \pm \sqrt{36-4}}{2}\]

I'll let you finish

Answer 10

do i have to simplify the sqrt or 32?

Answer 11

yes if you want to fully simplify the roots

Answer 12

it would be -6 - 2sqrt4 over 2 and -6 + 2sqrt4 over 2

Answer 13

close

Answer 14

\[\Large \sqrt{32} = \sqrt{16*2}\]

\[\Large \sqrt{32} = \sqrt{16}*\sqrt{2}\]

\[\Large \sqrt{32} = 4\sqrt{2}\]

Answer 15

So here is the full step by step picture.

\[\Large x=\frac{-b \pm \sqrt{b^2-4ac}}{2a}\]

\[\Large x=\frac{-6 \pm \sqrt{6^2-4*1*1}}{2*1}\]

\[\Large x=\frac{-6 \pm \sqrt{36-4*1*1}}{2*1}\]

\[\Large x=\frac{-6 \pm \sqrt{36-4}}{2}\]

\[\Large x=\frac{-6 \pm \sqrt{32}}{2}\]

\[\Large x=\frac{-6 \pm \sqrt{16*2}}{2}\]

\[\Large x=\frac{-6 \pm \sqrt{16}*\sqrt{2}}{2}\]

\[\Large x=\frac{-6 \pm 4\sqrt{2}}{2}\]

\[\Large x=\frac{2(-3 \pm 2\sqrt{2})}{2}\]

\[\Large x=\frac{\cancel{2}(-3 \pm 2\sqrt{2})}{\cancel{2}}\]

\[\Large x=-3 \pm 2\sqrt{2}\]

Answer 16

thank you so much!

Answer 17

Also, the plus minus symbol \(\Large \pm\) tells us to break the solutions down

so we really have these two solutions

\[\Large x=-3 + 2\sqrt{2} \text{ or } x=-3 - 2\sqrt{2}\]

but I find it more convenient and compact to write the two solutions as

\[\Large x=-3 \pm 2\sqrt{2}\]

Answer 18

you're welcome