Proof of quadratic formula | Polynomial and rational functions | Algebra II | Khan Academy

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Proof of Quadratic Formula

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I've always hated how most schools don't teach how stuff like this is derived. Without knowing the proof, it feels meaningless and empty, but now it feels like it has meaning and is backed by logic instead of some wizard creating it in a puff of smoke.

littledude
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Man i love this guy, he has made maths so accessible, one of the most influential person i knw !

transes
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You are definitely better at this than your average modern HS math teacher. It's no wonder the schools recommend you, cause you ACTUALLY TEACH THE MATERIAL,

lilahdog
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My highschool failed to explain this to me. They failed to explain many things, actually. So here I am, re-educating myself.


Thank you Sal

teli
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I'm in freaking college and no one ever taught me this; in fact I learned to complete squares a while a go. Don't judge me I'm a freshman and my college is in fact very good.

SebastianLopez-nhrr
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Hello! do someone knows why the +/- sign is not maintained for the denominator? when Sal, at 7:28
, takes the square root of 4a^2 que does not matains the +/-2a, and only puts 2a. On the other side, the +/- sign is maintained for the numerator.

jordanr
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my jaw was dropped the whole time...Im an electrical engineer major and I've never seen such a good proof of the quad eq, Im very impressed.

luketechco
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Here's an alternative way to prove the formula which may be more appealing since it does not involve any fancy footwork.
The two roots of a quadratic we can take as R1 and R2, with the assumption that R2>R1 without loss of generality.
This results in the assertion that (x - R1)*(x - R2)=0. Expanding the LHS we have x^2 - (R1+R2)*x + R1*R2 = 0.
Referring back to the standard for a*x^2 + b^x + c = 0 we can see that b/a = (R1+R2) and c/a = R1*R2. 
We can put R1 = m - d and R2 = m + d, so 2*m = b/a and R1*R2 = (m - d)*(m + d) = m^2 - d^2
from which it follows that m = b/(2*a) and d^2 = m^2 - c/a. 
Finally that R1 = m - d = b/(2*a) - sqrt(m^2 - c/a) and R2 = m + d = b/(2*a) + sqrt(m^2 - c/a) where m^2 = b^2/(4*a^2). - which is the same as the better known form discussed in your piece, even though that not might be immediately obvious.

crustyoldfart
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Thanks for your clear and detailed video which is better than my teacher's explanations.

shuminchen
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4:51 i don't understand why you are suddenly multiplying -c/a by 4a

edit: nevermind lol

cozierelf
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htank you so much เป็นวีดีโอที่ชอบที่สุดด

chonnaphaubonpap
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this is the most amazing math video. I understand every single thing.

JokoGuerra-txfr
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why could you remove the 2a denominator from -b, but leave it for the rest of the right side?

oraz.
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I like the way you explain ( perfect ) except for the writing.

clumsygirl
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bro, u just taught me something my teacher just zoomed through. I couldn't even understand what he taught lol. Thanks a Bunch!

superss
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I tried to prove f(x)= a(x-h)^2+k but I keep getting a(x+h)^2+k

mrhnm
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You know it's a math video when it has 250k viewers but only 550 likes.

akivaweil
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How do you know that the square root of 4a^2 is NOT -2a. Since (-2a)*(-2a) = 4a^2?

marccepeci
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i gave up try to get it after 6 hours of trying i see i did achievement because i got the half way to get it while my grade 9 now its time to go flex in school to my math teacher that i got it without help lol hehe (jk)

mt
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Very nice.
I had been searching for an explanation to the formula, and this is GREAT!!!

lucastapia