Question

-3x^2 - 4x - 4 = 0

Answer 1

if it was me, i would start with
\[3x^2+4x+4=0\] and try to factor

Answer 2

it does not factor, and the zeros are complex
is that ok?

Answer 3

yes the possible answers are
X= 2 plus or minus 4i square root of 2 over 3

X= 2plus or minus 2i square root of 2 over 3

X= -2plus or minus 2i square root of 2 over 3

X= -2 plus or minus 4i square root of 2 over 3

Answer 4

Use what is commonly called as the `box method`.

Multiply the entire equation by negative 1 across so you get \(3x^2+4x+4=0\)

Answer 5

Then multiply the leading coefficient 3 by the constant term 4 so your equation represents that of standard quadratic. \[x^2 +4x+12=0\]

Answer 6

Now find two numbers that add to give you 4, but multiply to give you 12. what would those be?

Answer 7

Thank you

Answer 8

6 and -2

Answer 9

what??

Answer 10

Nope, that wouldnt since \((x-2)(x+6) \ne x^2+4x+12\)

Answer 11

And since you cannot find any factors, we abandon this method and move onto completing the square.

Answer 12

either this is something that i do not know, or it is straight up wrong
how do you turn
\[3x^2+4x+4\] in to
\[x^2+4x+12\]?

Answer 13

the zeros are complex, use the quadratic formula, it always works

Answer 14

multiplying \(\color{red}{3}x^2+4x\color{red}{+4} \rightarrow x^2+4x+12\)

It works (sometimes) to help us factor out the original equation to which you divide both factors by the coefficient of the leading variable, and obtain you answer. But in this case, it did not work as we could not simplify this equation to produce factors.

Answer 15

Therefore, I resort to my favorite method, completing the square.

Answer 16

still lost
you multiplied in one place, divided in the other, left the middle term alone

Answer 17

Yeah. Pretty much. http://www.purplemath.com/modules/factquad2.htm

Answer 18

wow another factoring technique that i do not get, but i believe you

Answer 19

Im still so confused

Answer 20

@esvey u know quadratic formula?

Answer 21

use the quadratic formula, i guarantee you will get it

Answer 22

\[-3x^2 - 4x - 4 = 0\]\[3x^2+4x+4=0\]\[3x^2+4x=-4\]\[3\left(x^2+\frac{4}{3}x\right)=-4\]\[~~~~~~ c= \left(\frac{4}{\dfrac{3}{2}}\right)^2 = \left(\color{blue}{\frac{2}{3}}\right)^2 = \frac{4}{9}\]\[3\left(x^2+\frac{4}{3}x\color{red}{+\frac{4}{9}}\right)=-4\color{red}{+\frac{4}{3}}\]\[3\left(x+\color{blue}{\frac{2}{3}}\right)^2 = -4+\frac{4}{3}\]

Answer 23

this is how I would go about solving it because I dislike solving it by the quadratic formula.

Answer 24

Just on a side note: When you have a number multiplying the quadratic (stuff inside the parenthesis) and you're completing the square, you always multiply your `c` value to that number that is multiplying the stuff within the parenthesis

Answer 25

So would that be the last answer ?

Answer 26

That's how I obtained the \(\dfrac{4}{3}\) on the RHS.

Answer 27

No, you've got to simplify it from the step I left you with

Answer 28

So what is \(-4 +\dfrac{4}{3}\)

Answer 29

2 and 2/3

Answer 30

Why doesnt anything look like the answer choices

Answer 31

X= 2 plus or minus 4i square root of 2 over 3

X= 2plus or minus 2i square root of 2 over 3

X= -2plus or minus 2i square root of 2 over 3

X= -2 plus or minus 4i square root of 2 over 3

Answer 32

\[3\left(x+\color{blue}{\frac{2}{3}}\right)^2 = -4+\frac{4}{3}\]\[3\left(x+\frac{2}{3}\right)^2 = -\frac{8}{3}\]\[\left(x+\frac{2}{3}\right)^2 = -\frac{8}{\dfrac{3}{3}}\]\[\left(x+\frac{2}{3}\right)^2 = -\frac{8}{9}\]\[x+\frac{2}{3} = \pm \left(\sqrt{\frac{-8}{9}}\right)\]\[x+\frac{2}{3} = \pm \frac{\color{red}{\sqrt{-1}}\cdot 2\sqrt{2}}{3}~,~~~ i = \sqrt{-1} \]\[x=-\frac{2}{3}\pm \frac{2i\sqrt{2}}{3}\]\[\boxed{x=\dfrac{-2\pm 2i\sqrt{2}}{3}} \]