Question

Please help me write x^2+6x+8=0 in a vortex form ill give medal!!!!

Answer 1

x^2+6x+8=0
written in a vortex will be

x
^2
+
6x
+8
=
0
- --
v
vortex here
sucking up the
equation

Answer 2

do you know what a "perfect square trinomial" is?

Answer 3

Yes I believe so but if you could just re explain that to me that would be great

Answer 4

\(\bf (a\pm b)^2\implies a^2\pm {\color{red}{ 2}}ab+b^2\)

notice the middle term, is just 2 * "a" and "b" square rooted

Answer 5

oh ok

Answer 6

so let's grab yours and do some grouping first

\(\bf x^2+6x+8=0\implies (x^2+6x)+8=0\implies (x^2+6x+{\color{red}{ \square }}^2)+8=0\) so what do you think we need there to get a "perfect square trinomial" ?

Answer 7

b?

Answer 8

let's see that \(\large \begin{array}{llll}
(x^2+&6x+{\color{red}{ b }}^2)\\
&\uparrow \\
&2\cdot x\cdot b\ne 6x
\end{array}\)

Answer 9

so is not "b"... so what do you think might be?

Answer 10

is it a letter or number that's supposed to go there?

Answer 11

well, the middle term is \(\bf 2\cdot \sqrt{\textit{term on its left}}\cdot \sqrt{\textit{term on its right}}\) whatever those tems might be
I just happen to use "a" and "b", but they could be anything

Answer 12

so it would be "a" your saying

Answer 13

if you look at your equation, we can tell from the grouping that the middle term is 6x
so 2 * "term on the left" * "term on the right" has to equal 6x

Answer 14

yes, "a" and "b" just means, the left-term and right-term

Answer 15

sooooo....... that was what you wanted right, a, because honestly I don't know what it is you want from me now

Answer 16

we need to find a value for the right-side-term

Answer 17

What is the value supposed to be?

Answer 18

I don't understand:(

Answer 19

hmm the system is getting lagged

Answer 20

it is actually because your user want showing

Answer 21

lemme rewrite it a bit

\(\bf 2\cdot x\cdot {\color{red}{ \square }}=6x\) if you solve for \(\bf {\color{red}{ \square }}\) what would you get?

Answer 22

8?

Answer 23

let's try 8

2 * x * 8 = 6 * x
16x \(\ne\) 6x

Answer 24

so its wrong?

Answer 25

well, 16x is not equal 6x..... so is not 8

have you covered simplifying linear equations yet?

Answer 26

yes

Answer 27

its 3

Answer 28

look at it this way

2 * a * b = 6ab <---- what would you get for "b"?

Answer 29

right is 3

that means 2 * x * 3 = 6x <--- that IS your middle term in the trinomial

\(\bf 2\cdot \sqrt{\textit{term on its left}}\cdot \sqrt{\textit{term on its right}}\)

Answer 30

then the other side is 2

Answer 31

so now we know what is the number we need in order to complete the perfect square trinomial

now keep in mind that all we're doing is borrowing from "0", zero
that is, if we ADD \(3^2\) we have to also SUBTRACT \(3^2\)

so let us use 3 \(\bf x^2+6x+8=0\implies (x^2+6x)+8=0
\\ \quad \\
\implies (x^2+6x+{\color{red}{ 3 }}^2)+8-{\color{red}{ 3 }}^2=0\implies (x+3)^2+8-{\color{red}{ 3 }}^2=0\)

Answer 32

8 - 9 = -1 so \(\bf (x+3)^2-9=0\) <--- would be the vertex form
\(\bf (x+{\color{red}{ 3}})^2-{\color{blue}{ 9}}=0\implies (x-{\color{red}{ -3}})^2-{\color{blue}{ 9}}=0\qquad vertex\implies ({\color{red}{ -3}},{\color{blue}{ 9}})\)

Answer 33

wooops, lemme fix that... darn =(

Answer 34

8 - 9 = -1 \(\bf (x+{\color{red}{ 3}})^2-{\color{blue}{ 1}}=0\implies (x-{\color{red}{ -3}})^2-{\color{blue}{ 1}}=0\qquad vertex\implies ({\color{red}{ -3}},{\color{blue}{ 1}})\)

Answer 35

hmm shoud be -1 shoot anyhow \(\bf (x+{\color{red}{ 3}})^2-{\color{blue}{ 1}}=0\implies (x-{\color{red}{ -3}})^2-{\color{blue}{ 1}}=0\qquad vertex\implies ({\color{red}{ -3}},{\color{blue}{ -1}})\)

Answer 36

so the vertex form is (x+3)^2 -1=0

Answer 37

yeap

Answer 38

http://www.mathwarehouse.com/geometry/parabola/images/standard-vertex-forms.png

Answer 39

Thank u my friend and just as promised...............

Answer 40

yw