SolomonZelman
  • SolomonZelman
hey guys, I have a little question....
Mathematics
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schrodinger
  • schrodinger
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
SolomonZelman
  • SolomonZelman
\(\large\color{black}{ \displaystyle \color{#009900}{\rm a}x^2+\color{#009900}{\rm b}x= -\color{#009900}{\rm c} }\) \(\large\color{black}{ \displaystyle \color{#009900}{\rm a}\left(x^2+\color{#009900}{\rm \frac{b}{a}}x\right)= -\color{#009900}{\rm c} }\) \(\large\color{black}{ \displaystyle \color{#009900}{\rm a}\left(x^2+\color{#009900}{\rm \frac{b}{a}}x+\left(\color{#009900}{\rm \frac{b}{2a}}\right)^2\color{black}{-}\left(\color{#009900}{\rm \frac{b}{2a}}\right)^2\right)= -\color{#009900}{\rm c}}\) \(\large\color{black}{ \displaystyle \color{#009900}{\rm a}\left(x^2+\color{#009900}{\rm \frac{b}{a}}x+\color{#009900}{\rm \frac{b^2}{4a^2}}\color{black}{-}\color{#009900}{\rm \frac{b^2}{4a^2}}\right)= -\color{#009900}{\rm c}}\) \(\large\color{black}{ \displaystyle \color{#009900}{\rm a}\left(x^2+\color{#009900}{\rm \frac{b}{a}}x+\color{#009900}{\rm \frac{b^2}{4a^2}}\right)\color{black}{-}\color{#009900}{\rm a\frac{b^2}{4a^2}}= -\color{#009900}{\rm c}}\) \(\large\color{black}{ \displaystyle \color{#009900}{\rm a}\left(x^2+\color{#009900}{\rm \frac{b}{a}}x+\color{#009900}{\rm \frac{b^2}{4a^2}}\right)\color{black}{-}\color{#009900}{\rm \frac{b^2}{4a}}= -\color{#009900}{\rm c}}\) \(\large\color{black}{ \displaystyle \color{#009900}{\rm a}\left(x+\color{#009900}{\rm \frac{b}{2a}}\right)^2= -\color{#009900}{\rm c}+\color{#009900}{\rm \frac{b^2}{4a}}}\) \(\large\color{black}{ \displaystyle \left(x+\color{#009900}{\rm \frac{b}{2a}}\right)^2= -\color{#009900}{\rm \frac{c}{a}}+\color{#009900}{\rm \frac{b^2}{4a^2}}}\) \(\large\color{black}{ \displaystyle \left(x+\color{#009900}{\rm \frac{b}{2a}}\right)= \pm \sqrt{-\color{#009900}{\rm \frac{c}{a}}+\color{#009900}{\rm \frac{b^2}{4a^2}}}}\) \(\large\color{black}{ \displaystyle x= -\color{#009900}{\rm \frac{b}{2a}}\pm\sqrt{-\color{#009900}{\rm \frac{c}{a}}+\color{#009900}{\rm \frac{b^2}{4a^2}}}}\) \(\large\color{black}{ \displaystyle x= -\color{#009900}{\rm \frac{b}{2a}}\pm\sqrt{-\color{#009900}{\rm \frac{4ac}{4a^2}}+\color{#009900}{\rm \frac{b^2}{4a^2}}}}\) \(\large\color{black}{ \displaystyle x= -\color{#009900}{\rm \frac{b}{2a}}\pm\sqrt{\color{#009900}{\rm \frac{b^2-4ac}{4a^2}}}}\) \(\large\color{black}{ \displaystyle x= -\color{#009900}{\rm \frac{b}{2a}}\pm\sqrt{\color{#009900}{\rm \frac{b^2-4ac}{(2a)^2}}}}\) \(\large\color{black}{ \displaystyle x= -\color{#009900}{\rm \frac{b}{2a}}\pm\color{#009900}{\rm \frac{\sqrt{b^2-4ac}}{\sqrt{(2a)^2}}}}\) \(\large\color{black}{ \displaystyle x= -\color{#009900}{\rm \frac{b}{2a}}\pm\color{#009900}{\rm \frac{\sqrt{b^2-4ac}}{|2a|}}}\) \(\LARGE \color{red}{\Longleftarrow}\)
SolomonZelman
  • SolomonZelman
The last row is due to the following definition of the absolute value: \(\large\color{blue}{\displaystyle \sqrt{{\rm Q}^2}=\left|\rm Q\right| }\)
SolomonZelman
  • SolomonZelman
so we will say that the quadratic formula shouldn't work when the "a" is negative?

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AlexandervonHumboldt2
  • AlexandervonHumboldt2
let me think
AlexandervonHumboldt2
  • AlexandervonHumboldt2
it does look strange
tkhunny
  • tkhunny
Absolutely, in standard form a > 0. This has been the case since I first recall.
SolomonZelman
  • SolomonZelman
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 :)
anonymous
  • anonymous
does negative value affect in any way
SolomonZelman
  • SolomonZelman
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.
SolomonZelman
  • SolomonZelman
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.
AlexandervonHumboldt2
  • AlexandervonHumboldt2
1 Attachment
AlexandervonHumboldt2
  • AlexandervonHumboldt2
another proofs, with no modules
anonymous
  • anonymous
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
SolomonZelman
  • SolomonZelman
Why in Russian? I understand English too
AlexandervonHumboldt2
  • AlexandervonHumboldt2
i just didn't found on english it was too long to type it that's why i scrt it.
SolomonZelman
  • SolomonZelman
Wait, do you also speak Russian? (just asking, btw)
AlexandervonHumboldt2
  • AlexandervonHumboldt2
why also? i'm Russian lol
AlexandervonHumboldt2
  • AlexandervonHumboldt2
@ganeshie8
SolomonZelman
  • SolomonZelman
Oh, me too
anonymous
  • anonymous
@SolomonZelman like I said if u introduce the sign from beginning there won't be a problem 7th reply from this one, but upwards
AlexandervonHumboldt2
  • AlexandervonHumboldt2
what? lol. привет
SolomonZelman
  • SolomonZelman
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))
AlexandervonHumboldt2
  • AlexandervonHumboldt2
lol
ganeshie8
  • ganeshie8
order in which the quadratic formula spits out the rotos doesn't matter, so we have : \[\pm |a|=\pm a\]
SolomonZelman
  • SolomonZelman
Yes, but for the bottom denominator
ganeshie8
  • ganeshie8
since the order in which we take the roots from quadratic formula doesn't matter, we can write : \[\large\color{black}{ \displaystyle x= -\color{#009900}{\rm \frac{b}{2a}}\pm\color{#009900}{\rm \frac{\sqrt{b^2-4ac}}{|2a|}}=-\color{#009900}{\rm \frac{b}{2a}}\pm\color{#009900}{\rm \frac{\sqrt{b^2-4ac}}{2a}}} \]
ShadowLegendX
  • ShadowLegendX
"little question" PFFT my eyes hurt
SolomonZelman
  • SolomonZelman
Oh, so "a" does not have to be positive?
ganeshie8
  • ganeshie8
Would you agree that saying \(x \in\{2,3\} \) is same as \(x\in \{3,2\}\) ?
SolomonZelman
  • SolomonZelman
I suppose so
ganeshie8
  • ganeshie8
Yes, to my knowledge, quadratic formula does not require \(a\) to be positive...
ganeshie8
  • ganeshie8
I don't see how it should matter..
SolomonZelman
  • SolomonZelman
\(\color{#0cbb34}{\text{Originally Posted by}}\) @tkhunny Absolutely, in standard form a > 0. This has been the case since I first recall. \(\color{#0cbb34}{\text{End of Quote}}\)
ganeshie8
  • ganeshie8
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.
SolomonZelman
  • SolomonZelman
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!
SolomonZelman
  • SolomonZelman
Do you mind another question? Perhaps something in the past when I wasn't born.
AlexandervonHumboldt2
  • AlexandervonHumboldt2
lol
ganeshie8
  • ganeshie8
Btw, that is a very good observation... I had never bothered about that absolute value in the quadratic formula derivation before...
SolomonZelman
  • SolomonZelman
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).
AlexandervonHumboldt2
  • AlexandervonHumboldt2
it is 1
AlexandervonHumboldt2
  • AlexandervonHumboldt2
x^0=x^(1-1)=x^1/x^1=x/x=1
AlexandervonHumboldt2
  • AlexandervonHumboldt2
proof is so easy
SolomonZelman
  • SolomonZelman
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.
SolomonZelman
  • SolomonZelman
Yes, Alexander, that was the first thing that I cited in our conversation.
AlexandervonHumboldt2
  • AlexandervonHumboldt2
^
SolomonZelman
  • SolomonZelman
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.
AlexandervonHumboldt2
  • AlexandervonHumboldt2
you wrote same thing i wrote @Nnesha , but i do ignore it lol
Nnesha
  • Nnesha
nvm I just saw that.
Empty
  • Empty
\(x^0\) 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.
Empty
  • Empty
(I totally tl;dr everything lol )
SolomonZelman
  • SolomonZelman
Very clever to define \(2^3=1\cdot2\cdot2\cdot2\)
SolomonZelman
  • SolomonZelman
Yes, with a<1, everything indeed does work...
Empty
  • Empty
Exponent rules are pretty cool I think making anyone memorize exponent rules is a sin against math lol.
SolomonZelman
  • SolomonZelman
And sin against their person. Why raping a memory to remember something that does follow logically anytime you think about it?
SolomonZelman
  • SolomonZelman
Anyway, I think I have all of my concerns addressed.
Empty
  • Empty
well it was fun while it lasted :D
SolomonZelman
  • SolomonZelman
Yeah,perhaps and probably, but a movement is as well needed.
SolomonZelman
  • SolomonZelman
Thank y'all !!!!!!!!!!!!!!!
tkhunny
  • tkhunny
Again, we have Domain problems. \(x^{0}=1\) ONLY under appropriate conditions.

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