Question

hey guys, I have a little question....

Answer 1

\(\large\color{black}{ \displaystyle \color{#009900}{\rm a}x^2+\color{#009900}{\rm b}x= -\color{#009900}{\rm c} }\)

\(\large\color{black}{ \displaystyle \color{#009900}{\rm a}\left(x^2+\color{#009900}{\rm \frac{b}{a}}x\right)= -\color{#009900}{\rm c} }\)

\(\large\color{black}{ \displaystyle \color{#009900}{\rm a}\left(x^2+\color{#009900}{\rm \frac{b}{a}}x+\left(\color{#009900}{\rm \frac{b}{2a}}\right)^2\color{black}{-}\left(\color{#009900}{\rm \frac{b}{2a}}\right)^2\right)= -\color{#009900}{\rm c}}\)

\(\large\color{black}{ \displaystyle \color{#009900}{\rm a}\left(x^2+\color{#009900}{\rm \frac{b}{a}}x+\color{#009900}{\rm \frac{b^2}{4a^2}}\color{black}{-}\color{#009900}{\rm \frac{b^2}{4a^2}}\right)= -\color{#009900}{\rm c}}\)

\(\large\color{black}{ \displaystyle \color{#009900}{\rm a}\left(x^2+\color{#009900}{\rm \frac{b}{a}}x+\color{#009900}{\rm \frac{b^2}{4a^2}}\right)\color{black}{-}\color{#009900}{\rm a\frac{b^2}{4a^2}}= -\color{#009900}{\rm c}}\)

\(\large\color{black}{ \displaystyle \color{#009900}{\rm a}\left(x^2+\color{#009900}{\rm \frac{b}{a}}x+\color{#009900}{\rm \frac{b^2}{4a^2}}\right)\color{black}{-}\color{#009900}{\rm \frac{b^2}{4a}}= -\color{#009900}{\rm c}}\)

\(\large\color{black}{ \displaystyle \color{#009900}{\rm a}\left(x+\color{#009900}{\rm \frac{b}{2a}}\right)^2= -\color{#009900}{\rm c}+\color{#009900}{\rm \frac{b^2}{4a}}}\)

\(\large\color{black}{ \displaystyle \left(x+\color{#009900}{\rm \frac{b}{2a}}\right)^2= -\color{#009900}{\rm \frac{c}{a}}+\color{#009900}{\rm \frac{b^2}{4a^2}}}\)

\(\large\color{black}{ \displaystyle \left(x+\color{#009900}{\rm \frac{b}{2a}}\right)= \pm \sqrt{-\color{#009900}{\rm \frac{c}{a}}+\color{#009900}{\rm \frac{b^2}{4a^2}}}}\)

\(\large\color{black}{ \displaystyle x= -\color{#009900}{\rm \frac{b}{2a}}\pm\sqrt{-\color{#009900}{\rm \frac{c}{a}}+\color{#009900}{\rm \frac{b^2}{4a^2}}}}\)

\(\large\color{black}{ \displaystyle x= -\color{#009900}{\rm \frac{b}{2a}}\pm\sqrt{-\color{#009900}{\rm \frac{4ac}{4a^2}}+\color{#009900}{\rm \frac{b^2}{4a^2}}}}\)

\(\large\color{black}{ \displaystyle x= -\color{#009900}{\rm \frac{b}{2a}}\pm\sqrt{\color{#009900}{\rm \frac{b^2-4ac}{4a^2}}}}\)

\(\large\color{black}{ \displaystyle x= -\color{#009900}{\rm \frac{b}{2a}}\pm\sqrt{\color{#009900}{\rm \frac{b^2-4ac}{(2a)^2}}}}\)

\(\large\color{black}{ \displaystyle x= -\color{#009900}{\rm \frac{b}{2a}}\pm\color{#009900}{\rm \frac{\sqrt{b^2-4ac}}{\sqrt{(2a)^2}}}}\)

\(\large\color{black}{ \displaystyle x= -\color{#009900}{\rm \frac{b}{2a}}\pm\color{#009900}{\rm \frac{\sqrt{b^2-4ac}}{|2a|}}}\) \(\LARGE \color{red}{\Longleftarrow}\)

Answer 2

The last row is due to the following definition of the absolute value:

\(\large\color{blue}{\displaystyle \sqrt{{\rm Q}^2}=\left|\rm Q\right| }\)

Answer 3

so we will say that the quadratic formula shouldn't work when the "a" is negative?

Answer 4

let me think

Answer 5

it does look strange

Answer 6

Absolutely, in standard form a > 0. This has been the case since I first recall.

Answer 7

I didn't know that rule, but I once again generated it using completing the square, I guess.
Thanks for telling me that this is in case part of the definition :)

Answer 8

does negative value affect in any way

Answer 9

Well, if "a" is negative (lets say -3), then according to what I said in my last line (with absolute value of 2a), we get 6 in the denominator below the sqrt.

If you plug in the value of "a" into a regular quadratic formula, you get a -6 in the denominator below the sqrt.

Answer 10

This is the defferentiation when a<0, and therefore, using the proof via completing the square, must be that a>0 whenever you come to use the standard version (the one you would usually see) of the squadratic formula.

Answer 11

another proofs, with no modules

Answer 12

This is worth noting that if u want the equation for a negative 'a', u have to derive it likewise from the beginning so that -ax^2 + bx = -c

Answer 13

Why in Russian? I understand English too

Answer 14

i just didn't found on english
it was too long to type it that's why i scrt it.

Answer 15

Wait, do you also speak Russian? (just asking, btw)

Answer 16

why also? i'm Russian lol

Answer 17

@ganeshie8

Answer 18

Oh, me too

Answer 19

@SolomonZelman like I said if u introduce the sign from beginning there won't be a problem
7th reply from this one, but upwards

Answer 20

what? lol. привет

Answer 21

I moved to the US several years ago, and now I attend college here;)
((I am not using Russian, because I have no Russian on my keyboard))

Answer 22

lol

Answer 23

order in which the quadratic formula spits out the rotos doesn't matter, so we have :
\[\pm |a|=\pm a\]

Answer 24

Yes, but for the bottom denominator

Answer 25

since the order in which we take the roots from quadratic formula doesn't matter, we can write :

\[\large\color{black}{ \displaystyle x= -\color{#009900}{\rm \frac{b}{2a}}\pm\color{#009900}{\rm \frac{\sqrt{b^2-4ac}}{|2a|}}=-\color{#009900}{\rm \frac{b}{2a}}\pm\color{#009900}{\rm \frac{\sqrt{b^2-4ac}}{2a}}} \]

Answer 26

"little question"

PFFT

my eyes hurt

Answer 27

Oh, so "a" does not have to be positive?

Answer 28

Would you agree that saying \(x \in\{2,3\} \) is same as \(x\in \{3,2\}\) ?

Answer 29

I suppose so

Answer 30

Yes, to my knowledge, quadratic formula does not require \(a\) to be positive...

Answer 31

I don't see how it should matter..

Answer 32

\(\color{#0cbb34}{\text{Originally Posted by}}\) @tkhunny
Absolutely, in standard form a > 0. This has been the case since I first recall.
\(\color{#0cbb34}{\text{End of Quote}}\)

Answer 33

To make \(a\) positive, you just have to multiply the function by \(-1\).
Observe that the quadratic funcitons \(f(x)\) and \(-f(x)\) both have same roots. When you multiply \(-1\) by the entire funciton, the graph of the function flips over the \(x\) axis. So, the points on the \(x\) axis won't change. The points on the \(x\) axis are the roots given by quadratic formula.

As you can see, it doesn't hurt to let \(a \gt 0\) for definiteness.

Answer 34

yes, I was thinking that you can simply multiply both sides times -1. Just needed to clarify how precisely does it matter for a to be negative, and clarified successfully, THANK YOU!

Answer 35

Do you mind another question?
Perhaps something in the past when I wasn't born.

Answer 36

lol

Answer 37

Btw, that is a very good observation... I had never bothered about that absolute value in the quadratic formula derivation before...

Answer 38

I had a debate with my grandfather about a very simple matter, whether \(x^0\) is 1 (as I thought) or undefined (as my grandfather claimed).

Answer 39

it is 1

Answer 40

x^0=x^(1-1)=x^1/x^1=x/x=1

Answer 41

proof is so easy

Answer 42

So I was once told that it was used to be undefined, but the mathematicians had "AGREED UPON" \(x^0\color{red}{=1}\) to make the rules work.

Answer 43

Yes, Alexander, that was the first thing that I cited in our conversation.

Answer 44

^

Answer 45

Well, not that I can think of a real example of \(x^0\) situation... Nor an explanation of what the "zero-power" really means.

So I can still see that there is a place for arguing that \(x^0\) is indeed undefined.

Answer 46

you wrote same thing i wrote @Nnesha , but i do ignore it lol

Answer 47

nvm I just saw that.

Answer 48

\(x^0\) could be interpreted as x multiplied by 1 zero times and \(x^1\) could be interpreted as x multiplied by 1 a single time, and then \(x^n\) is x multiplied by 1 a total of n times.

Also, for the first question, \(a<0?\) This is no problem, check this out:

\[ax^2+bx+c=0\]
\[-ax^2+(-b)x+(-c)=0\]
Now we can just plug in -a (which is positive) and use -b and -c instead of b and c and get the same answer.

\[x= \frac{-(-b)\pm \sqrt{(-b)^2-4(-a)(-c)} }{2(-a)} = \frac{-b \mp \sqrt{b^2-4ac} }{2a} \]

The - sign in the denominator cancels out with the extra - sign on b, and just flips the sign \(\pm\) to \(\mp\) but really this doesn't matter, I just wanted to show I was distributing it. And of course the two negatives inside will pretty clearly cancel themselves out too, so it doesn't matter what a is.

Answer 49

(I totally tl;dr everything lol )

Answer 50

Very clever to define
\(2^3=1\cdot2\cdot2\cdot2\)

Answer 51

Yes, with a<1, everything indeed does work...

Answer 52

Exponent rules are pretty cool I think making anyone memorize exponent rules is a sin against math lol.

Answer 53

And sin against their person.
Why raping a memory to remember something that does follow logically anytime you think about it?

Answer 54

Anyway, I think I have all of my concerns addressed.

Answer 55

well it was fun while it lasted :D

Answer 56

Yeah,perhaps and probably, but a movement is as well needed.

Answer 57

Thank y'all !!!!!!!!!!!!!!!

Answer 58

Again, we have Domain problems. \(x^{0}=1\) ONLY under appropriate conditions.