Question

I need some help please Fan&&Medal

Can someone help me solve this using the Completing the Square method? I always get confused.

h(x)= x^2 +6x +12

Answer 1

Sure :)

\[\large x^2 + 6x + 12 = 0\]

Answer 2

Start by moving the constant to the other side of the equation

subtract 12 from both sides...

\[\large x^2 + 6x = -12\]

Answer 3

How...you take the coefficient of the 'x' term...which is 6 here...

Divide it by 2....so we now have 3

And the square the result.....so 3^2 = 9

We add this 9 to both sides of the equation

\[\large x^2 + 6x + 9 = -12 + 9\]

Answer 4

Simplify that right side a bit

\[\large x^2 + 6x + 9 = -3\]

Answer 5

now we just write the left side as the sum of 2 squares

\[\large (x + 3)^2 = -3\]

Make sense there? we need 2 numbers that add to 6 but also multiply to make 9...so 3 and 3

instead of writing

\[\large (x + 3)(x + 3) = -3\]

we just write

\[\large (x + 3)^2= -3\]

Answer 6

ok, that's what I got.. I just have a bad habit of putting the wrong sign in front of stuff and all that, lol :)) Thank you SO much!!!

Answer 7

No problem :)

and if you keep going you will get an imaginary number answer

Fun stuff :D lol :P

Answer 8

ughhh, lol I don't understand why you need to know about imaginary numbers... apparently they're not even real :P

Answer 9

....*ba dum *ching* .... :P

lol :D

Answer 10

but you do know how to proceed right?

Square root both sides

\[\large (x + 3) = \pm \sqrt{-3}\]

Subtract 3 from both sides and break it into the 2 equations

\(\large x = i\sqrt{3} - 3\) and \(\large x = -i\sqrt{3} - 3\)

Answer 11

Thanks :D