Question

Quadratics, Completing The Square, And Non-Linear Systems Explanations?

Answer 1

I'd like to better understand these subjects.

Answer 2

tis easier if you were to ask an specific in precise terms

Answer 3

and if you posted anew :), thus we can see it and help and also revise each other

Answer 4

Quadratic Formula would be a good place to start, I think.

Answer 5

yeah, like that as Loser66 just did, he seem to do that often =)

Answer 6

ok.. so anything confusing about the quadratic formula?

Answer 7

Just an explanation, really. Because I'm kind of confused about the whole thing, really.

Answer 8

ok... gimme a quadratic equation...

Answer 9

any... you can make up

Answer 10

Would 2x^2 - 8 + 4 work?

Answer 11

Or do you need something else?

Answer 12

Here is one:
\[(A + B)(A-B) = 0\]

Answer 13

I will explain you the concept of "Completing the Square" and rest I will leave for other fellows.

Answer 14

@FrostFelon :please tag me when you are back so I can explain the concept of "Completing the Square".

Answer 15

@imer Back.

Answer 16

did you understand the quadratic formula?

Answer 17

No.

Answer 18

ok let us do the one you posted => 2x^2 - 8 + 4

Answer 19

Alright.

Answer 20

tis quite simple really so \(\bf \textit{quadratic formula}\\
y={\color{blue}{ 2}}x^2{\color{red}{ -8}}x{\color{green}{ +4}}
\qquad \qquad
x= \cfrac{ - {\color{red}{ b}} \pm \sqrt { {\color{red}{ b}}^2 -4{\color{blue}{ a}}{\color{green}{ c}}}}{2{\color{blue}{ a}}}\)

and that's all there's to it :)

then you just plug in the values and simplify

Answer 21

I see. So it'd be

Answer 22

yeap

Answer 23

\(\bf \textit{quadratic formula}\\
y={\color{blue}{ 2}}x^2{\color{red}{ -8}}x{\color{green}{ +4}}
\qquad \qquad
x= \cfrac{ - {\color{red}{ (-8)}} \pm \sqrt { {\color{red}{ (-8)}}^2 -4{\color{blue}{ (2)}}{\color{green}{ (4)}}}}{2{\color{blue}{ (2)}}}
\\ \quad \\
x=\cfrac{8\pm \sqrt{64-32}}{4}\implies x=\cfrac{8\pm \sqrt{32}}{4}\implies x=\cfrac{8\pm \sqrt{4^2\cdot 2}}{4}
\\ \quad \\
x=\cfrac{\cancel{ 8 }\pm \cancel{ 4 }\sqrt{2}}{\cancel{ 4 }}\implies x=2\pm\sqrt{2}\)

Answer 24

Going into this:

Answer 25

yeap

Answer 26

Whoops. Seems you did it for me. Well, thank you for that.

Answer 27

yw

Answer 28

so you see, is quite simple, all you do is grab the "a", "b" and "c" and plug them in and simplify

Answer 29

so... .to understand what "completing the square" means

do you know what a "perfect square trinomial" is? or often called just "perfect square"

Answer 30

How did step five go to step six, though?

Answer 31

\(\bf x=\cfrac{8\pm \sqrt{4^2\cdot 2}}{4}
\\ \quad \\
x=\cfrac{\cancel{ 8 }\pm \cancel{ 4 }\sqrt{2}}{\cancel{ 4 }}\implies x=2\pm\sqrt{2} \quad?\)

Answer 32

you mean how we got "4" from the radicand?

Answer 33

Yes. And what about it being squared? there's no sixteen.

Answer 34

well true well \(\bf 36\to 2\cdot 2\cdot 2\cdot 2\cdot 2\to 4\cdot 4\cdot 2\to 4^2\cdot 2\)

Answer 35

woops 32 even \(\bf 32\to 2\cdot 2\cdot 2\cdot 2\cdot 2\to 4\cdot 4\cdot 2\to 4^2\cdot 2\)

Answer 36

Okay, that makes sense, but again, how did the four get out of the square, and without the exponent?

Answer 37

\(\Large \bf \pm\sqrt[{\color{red}{ n}}]{x^{\color{red}{ n}}} \iff \pm x\)

Answer 38

\(\large \bf \sqrt[2]{32}\to \sqrt[2]{4^2\cdot 2}\to \sqrt[{\color{red}{ 2}}]{4^{\color{red}{ 2}}}\cdot \sqrt[2]{2}\to 4\sqrt{2}\)

Answer 39

Oh...huh. So the four basically says, "later, numbers" to the twos and skips over the square-root fence?

Answer 40

yeap, pretty much, that is so for a FACTOR that has an exponent that MATCHES the root

Answer 41

Alright.

Answer 42

so... .to understand what "completing the square" means

do you know what a "perfect square trinomial" is? or often called just "perfect square"

Answer 43

Yes.

Answer 44

so... let us try say another quadratic

say for example \(\bf x^2+6x+20\)

Answer 45

Okay.

Answer 46

\( \bf x^2+6x+20\implies (x^2+6x)+20\implies (x^2+6x+{\color{red}{ \square ?}}^2)+20\)

what do you think we need there to get a "perfect square trinomial"

Answer 47

recall that \(\large \bf ({\color{blue}{ a}}+{\color{brown}{ b}})^2={\color{blue}{ a}}^2+2{\color{blue}{ a}}{\color{brown}{ b}}+{\color{brown}{ b}}^2\)