Question

g(x) = x2 +6x + 1
Find the solutions of g(x). Show each step.
Can someone please help?

Answer 1

this needs the general quadratic equation

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

in your question a= 1, b = 6 and c = 1

substitute the values then simplify

Answer 2

okay, one sec

Answer 3

x=-6(sqrt)32/2

Answer 4

where do i go from here?

Answer 5

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

so make the substitution and then divide by 2

Answer 6

I'm really confused on this

Answer 7

ok... so using some laws for radicals
\[\sqrt{32} = \sqrt{16 \times 2} = \sqrt{16} \times \sqrt{2} = 4\sqrt{2}\]

so the problem is

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

so divide -6 by 2 and then 4 by 2

and you get

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

Answer 8

okay, thank you. i wasn't sure if i divided 4 by 2 as well

Answer 9

where do i from here?

Answer 10

thats the answer

the 2 values of x are

\[x = - 3 + \sqrt{2}...and.... x = -3 - \sqrt{2}\]

Answer 11

oh! okay thank you so much!

Answer 12

oops it I missed a bit

\[x = -3 + 2\sqrt{2} .... and ..... x = -3 - 2\sqrt{2}\]

sorry about that