This is why you should learn the reasoning behind formulas. I do get mostly all A's in mathematics but when asked to define how i did it i just say i used stuff written in textbooks and it actually has worked this far
Well explained but I think you left some things like saying a has to be different from 0 when you divided everything by a. You also could have explained why the number of solutions depends on the sign of the V(b^2)-4ac.
@@demianpryima1015well, it'd be very in line with the rest of mathematics for it to be a "degenerate quadratic" and for the formula to still apply. it's a wonder of algebra that equations of different orders behave so differently. (also, video creator, you have a typo in the parenthesis in the thumbnail)
It’s not a problem at all, just write a =/= 0. But it’s a quadratic, so that fact is already assumed anyways. Moreover, he’s just deriving the formula and not explaining it. I still think it’s relatively obvious - there are always 2 solutions including multiplicity
@@jkid1134very interesting, u could take the limit a->0 of the quadratic formula and try to show that it tends to -c/b, the expression for the linear root (this limit does exist for certain conditions on b,c, and this is indeed the case!). If q(a,b,c) is the quadratic formula for the + root, then one can show (by multiplying numerator and denominator by b + sqrt(b^2 - 4ac) that: lim a->0 q(a,b,c) = -2c/(b + |b|) Thus if b > 0, lim a->0 q(a,b,c) = -c/b , the linear root!!!! However u may notice that this only holds for b > 0, so what about b < 0? The limit does not converge any longer, can u see why? It’s because there are two roots to the quadratic, and we were using the + root before. If r(a,b,c) was the quadratic formula with the - root, this would converge to the linear root -c/b as a->0, which is very reassuring! So the way the convergence works is, one root tends to the linear setting, whereas the other to infinity. As for the case b = 0, ax^2 + c = 0 and so x = +- sqrt(-c/a); as a->0, this limit is actually undefined as a->0, rather than giving some sort of convergence to c like we might have expected. Can you see why?
Way to go. The math teaching sounds like telling a fascinating story if you would like to inviting your listeners exploring the math realm. You are almost there...
The derivation becomes a bit simpler in this way: at 3:33, multiply with 4a², then take the square root, use 2a (x + b/2a) = 2ax + b on the left hand side, subtract the b and divide by 2a. So you don't have to combine the fractions at 3:35 and again at 5:55.
Bro I've been trying to prove that formula because [√(b2 - 4ac)] / 2a is the distance between the centre and one of the root but I can't find it. I did not think of the perfect square root technique. You're a cool guy!
We know that |a|≠0, but can a, b, and/or c be imaginary or complex as well? Write the quadratic equation as x^2+(b/a)x+(c/a)=0 and assume the roots are u and v. Then (x-u)(x-v)=x^2-(u+v)x+uv=x^2+(b/a)x+(c/a)=0. Proof follows if (u+v)=-(b/a) and uv=c/a. No other restrictions on a, b, or c other than |a|≠0? (u+v)=[-b+√(b^2-4ac)]/(2a)+[-b-√(b^2-4ac)]/(2a)=-b/(2a)-b/(2a)=-b/a. OK uv={[-b+√(b^2-4ac)]/(2a)}{[-b-√(b^2-4ac)]/(2a)}=b^2/(4a^2)-(b^2-4ac)/(4a^2)=4ac/(4a^2)=c/a. OK
Another great question. At first we did include it where we had a positive and negative of the enite fraction (maybe I didn't make that clear in my video) but then when we include the other fraction we move it up to the numerator. Lets look at this in an easier fraction with just a negative. if you have (7/4)-(6/4) if you combine those two fractions you just make it (7-6)/4 and you don't consider the denominator to be negative. This is the same for ±. Hopefully that clears it up for you.
@@NoahBugbee sorry you had to hear such question one more time. People probably asked you this a thousand times. I had the curiosity, but never asked, because of that. You do a great job.
Yeah. Completing the square proves it. Also, you understood what he meant? No point in nit-picking like an asshole. Derivation and proving in this context means the same thing.
@@zadiczane7618 Not exactly, if you just want to prove the quadratic formula you only need to substitute the formula into the general quadratic equation and get zero. Thats why i said it was the wrong word because he did more than just proving it.
Deriving is more impressive than proving in my eyes because proving a statement often involves someone else already deriving it. Even tho this guy didn't derive it first, him doing it shows he understands it which is hard to do with just a proof
This is why you should learn the reasoning behind formulas. I do get mostly all A's in mathematics but when asked to define how i did it i just say i used stuff written in textbooks and it actually has worked this far
This guy is on to something, no one turn a blind eye.
Lol
people, share some love to this guy please he deserves it
you are Euler 2.0
Keep it up and never stop uploading math vids. btw you earned a sub
Nicely explained
nice one mate. keep making more :)
This was very interesting and well explained!
Great explanation!
Good job bro keep going at it!
Very well paced
Well your passion for math definitely shines through. It will take you far.
Thank you great video!
Very cool, keep it up!
Nice one bro keep it up
Well explained but I think you left some things like saying a has to be different from 0 when you divided everything by a. You also could have explained why the number of solutions depends on the sign of the V(b^2)-4ac.
When a is zero it’s not a quadratic anymore
@@demianpryima1015well, it'd be very in line with the rest of mathematics for it to be a "degenerate quadratic" and for the formula to still apply. it's a wonder of algebra that equations of different orders behave so differently.
(also, video creator, you have a typo in the parenthesis in the thumbnail)
The discriminant is just b^2 I 4ac. There is no \/ in front of it.
It’s not a problem at all, just write a =/= 0. But it’s a quadratic, so that fact is already assumed anyways. Moreover, he’s just deriving the formula and not explaining it. I still think it’s relatively obvious - there are always 2 solutions including multiplicity
@@jkid1134very interesting, u could take the limit a->0 of the quadratic formula and try to show that it tends to -c/b, the expression for the linear root (this limit does exist for certain conditions on b,c, and this is indeed the case!). If q(a,b,c) is the quadratic formula for the + root, then one can show (by multiplying numerator and denominator by b + sqrt(b^2 - 4ac) that:
lim a->0 q(a,b,c) = -2c/(b + |b|)
Thus if b > 0, lim a->0 q(a,b,c) = -c/b , the linear root!!!!
However u may notice that this only holds for b > 0, so what about b < 0? The limit does not converge any longer, can u see why? It’s because there are two roots to the quadratic, and we were using the + root before. If r(a,b,c) was the quadratic formula with the - root, this would converge to the linear root -c/b as a->0, which is very reassuring! So the way the convergence works is, one root tends to the linear setting, whereas the other to infinity.
As for the case b = 0, ax^2 + c = 0 and so x = +- sqrt(-c/a); as a->0, this limit is actually undefined as a->0, rather than giving some sort of convergence to c like we might have expected. Can you see why?
another great video!
Way to go. The math teaching sounds like telling a fascinating story if you would like to inviting your listeners exploring the math realm. You are almost there...
Nicely done 👍
Well explained 🎉 ❤
very helpful!
The derivation becomes a bit simpler in this way: at 3:33, multiply with 4a², then take the square root, use 2a (x + b/2a) = 2ax + b on the left hand side, subtract the b and divide by 2a. So you don't have to combine the fractions at 3:35 and again at 5:55.
my algebra teacher once let my class do this exact problem for extra credit
Bro I've been trying to prove that formula because [√(b2 - 4ac)] / 2a is the distance between the centre and one of the root but I can't find it.
I did not think of the perfect square root technique.
You're a cool guy!
Thanks for explaining the formula. I've never understood it before.
Thanks for it❤❤🎉 bro
my algebra teacher showed this way of deriving the quadratic formula on the last day of school and everyone was in awe
go on my bro
I don't know why, but he is something special.
Here before you go famous 🔥
thank u , that was the easiest way I've seen
came for the proof, stayed for the crooked writing and then continuing on the left side when you ran out of room. subbed.
Great!
Well done.
when (x+b/2a)²= b²-4ac/2a is the equation, after you take the sqrt on both sides, shouldn't |x+b/2a| be the result on the left hand side?
The absolute value on the left side is the same as the plus or minus on the right side
Thanks 🙏🏿
Glad you liked it.
I was looking for this lol
Glad you found it then!
thanks
Good
Nice job kid! Going for a career in math or engineering? Physics?
Probably math but physics interests me as well.
We know that |a|≠0, but can a, b, and/or c be imaginary or complex as well? Write the quadratic equation as x^2+(b/a)x+(c/a)=0 and assume the roots are u and v. Then (x-u)(x-v)=x^2-(u+v)x+uv=x^2+(b/a)x+(c/a)=0. Proof follows if (u+v)=-(b/a) and uv=c/a. No other restrictions on a, b, or c other than |a|≠0?
(u+v)=[-b+√(b^2-4ac)]/(2a)+[-b-√(b^2-4ac)]/(2a)=-b/(2a)-b/(2a)=-b/a. OK
uv={[-b+√(b^2-4ac)]/(2a)}{[-b-√(b^2-4ac)]/(2a)}=b^2/(4a^2)-(b^2-4ac)/(4a^2)=4ac/(4a^2)=c/a. OK
Yeah, of course. Also, you can, and should, write "a≠0" instead of "|a|≠0". Simply because a=0 if, and only if, |a|=0.
Wouldn't it be possible to just plug the formula for X back into quadratic? It should simplify to 0=0, no?
Yes, correct. If you just want to prove it you only need to substitute it into the general quadratic and get that 0=0.
That's not a complete proof, because you still need to prove that there is no other solution to the equation.
at 5:30 i don’t get why you don’t consider the positive and negative of the sqrt(4a^2)
i am pretty sure it doesn't matter, as in it would give the same results and just be more complicated
Another great question. At first we did include it where we had a positive and negative of the enite fraction (maybe I didn't make that clear in my video) but then when we include the other fraction we move it up to the numerator. Lets look at this in an easier fraction with just a negative. if you have (7/4)-(6/4) if you combine those two fractions you just make it (7-6)/4 and you don't consider the denominator to be negative. This is the same for ±. Hopefully that clears it up for you.
Because he doesn't need to.
(±a)/(±b) = ±a/b
i remember proving this when i was 15, me and my friends had a contest to see who could do it first, i wont lol.
Thsnk you, i hate how my teacher just shows and tells conceps without actually explaining them
Now derive the cubic formula
what happened to your eye bro?
Umm...Will this be on the test?
ive literally done this by accident before
😉
You should fix your title. You are not proving or deriving the quadratic equation. You are deriving the quadratic formula.
You should skip the parentheses in the title of the video.
Type
What happened to your eye?
I had brain surgery when I was much younger.
@@NoahBugbee sorry you had to hear such question one more time. People probably asked you this a thousand times. I had the curiosity, but never asked, because of that.
You do a great job.
@@samueldeandrade8535 Yeah same here ahah
Popeye?
k
You just complete the square. Also prove is the wrong word. You derive it.
Yeah. Completing the square proves it. Also, you understood what he meant? No point in nit-picking like an asshole. Derivation and proving in this context means the same thing.
@@zadiczane7618 Not exactly, if you just want to prove the quadratic formula you only need to substitute the formula into the general quadratic equation and get zero. Thats why i said it was the wrong word because he did more than just proving it.
L@@antenym8947
Deriving is more impressive than proving in my eyes because proving a statement often involves someone else already deriving it. Even tho this guy didn't derive it first, him doing it shows he understands it which is hard to do with just a proof
Aye Aye Matey 🏴☠️
p
I don't need your proof I can already prove it by transposing constant term making perfect squares taking square roots a\amd done
Right…
why is one of ur eyes closed?
Oh wow nice, also your handwriting is rather wonky
You write like an infant. C-
Yeah, im trying to work on that lol.
@@NoahBugbee Nothing wrong with writing like a infant. The best engineers have the worst handwriting
wtf??? his writing is immaculate dude. What do you want from this guy? Incredibly readable, definitely cleaner than mine.
@@jiggelnaut3907where did you get that ☠️
@@judomaster8629Experience I guess. It really is like that