Question

can someone help me find the solution of •g(x) = x2 +6x + 1

Answer 1

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

I believe this method involves completing the square. Are you familiar with that?

Answer 2

yeah, I know it's irrational though because it has 2 real roots

Answer 3

Mmhmm, so let's begin by the completing the square method.

Answer 4

can you walk me through it please :3 I'm not exactly familiar with that that's why I asked :l it's my last question of the problem though

Answer 5

What you want to do first, is set your equation = 0. \[g(x) = x^2 +6x+1=0\]

Answer 6

Now subtract both sides of the equation by -1 to isolate your x values. \[x^2+6x=-1\]

Answer 7

so -1 =x^2+6x?

Answer 8

Now we need to complete the square. You can do this by taking your value of b, dividing by 2, and squaring it. This will produce your `c` value, which puts the left hand side of your equation into quadratic form, `ax+by+c`

Answer 9

Yes, you're on track :)

Answer 10

so c would = 9?

Answer 11

So now we will have \[c=\left(\frac{b}{2}\right)^2 = \left(\frac{6}{2}\right)^2= (3)^2 =9\]\[x^2+6x\color{blue}{+9} = -1\color{blue}{+9}\]

Answer 12

Yep, C = 9.

Answer 13

sweet thank you so much :)

Answer 14

Now we see that the left hand side of the equation is a perfect square, therfore we can reduce it to \((x+c)^2 \implies (x+3)^2\)\[(x+3)^2 = 8\]

Answer 15

You're following, yes? :)

Answer 16

yes :)

Answer 17

Awesome. To solve for x, we take the square root of both sides. \[\sqrt{(x+3)^2} = \pm \sqrt{8}\]

Answer 18

This will help us undo the square that is bounding the x, and help us solve for it. \[x+3=\pm \sqrt{8}\]

Answer 19

Now just subtract `-3` from both sides of your equation and you will have solved for x :D \[x= -3 \pm \sqrt{8}\]

Answer 20

take the square root of 8, then subtract for 3?

Answer 21

Precisely.

Answer 22

ouch spoke to fast, so you don't take the square root of 8?

Answer 23

depends how simplified you want it to be. \[\sqrt{8} \iff 2\sqrt{2}\]

Answer 24

So you would then have \[x=-3\pm 2\sqrt{2}\]

Answer 25

this makes a lot of sense, thanks :)

Answer 26

Awesome :) No problem.