The Better Quadratic Formula You Won't Be Taught

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Disclaimer: This video is for entertainment purposes only and should not be considered academic. Though all information is provided in good faith, no warranty of any kind, expressed or implied, is made with regards to the accuracy, validity, reliability, consistency, adequacy, or completeness of this information.

Corrections:
2:32 Make sure a=1 (divide by the leading term)

#math #brithemathguy #quadraticequation

[BriTheMathGuy]

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[SuperMtheory]

This is a great way to use the quadratic formula without "using" the quadratic formula. I've seen this method used on other channels.

[johnbutler]

I don't know if "better" is the right word, but I think this is "as good." It highlights a different facet of quadratic equations and functions in a really cool and enlightening way. But the quadratic formula we were taught in school also highlights some important relationships, namely, that of the vertex and the discriminant.

[yoyoezzijr]

m ± √(m² - c) is because we let m = -b/2, and so it is -b/2 ± √(b²/4 - c) which becomes -b/2 ± √(b² - 4c) / 2 which is the quadratic equation in the case of a = 1. And we can always ÷ the equation to get a = 1

[vice]

The formula we learn here in Germany is x1, x2=-p/2+-sqrt((p/2)^2-q) where in a quadratic ax^2+bx+c=0 p=b/a, q=c/a has the nice name of pq-Formel

[kingsleys.]

In Germany in school we learned to solve quadratic equations by setting a to 1, i.e. to bring it into the form of x^2 + px + q = 0. To solve you simply use x = - p/2 +- sqr ( (p/2)^2 - q). I personally think thats easier to memorize than the other formula

[fireflyslight]

I think the reason we are taught to solve these problems the way we are is because students quickly move on to equations that aren't quadratic. They need more robust strategies to deal with more advanced polynomials. While this is a great trick and something that I can definitely see coming in handy or saving some time, learning to factor through some method is truly much more helpful in my experience. Past more introductory levels of algebra, solving quadratics was never a consistent part of any of my courses.

[acidishot]

It’s interesting. This approach is more similar to how financial professionals approach pricing securities. One can use Ito’s lemma to price non-linear products. The key result is that the one can deduce the width of the quadratic curve using:

Sqrt(2*Θ/Γ)

Where Θ = the absolute value of the minimum of the curve, or F(F’(X)=0). This is given by:

B^2 * (1-2a)/4a^2 + c

Where Γ = F’’(X), which for a quadratic is just a constant.

Defining M as b/2a (from ax^2 + bx + c). The roots become:

-M +- sqrt(abs(b^2 * (1-2a)/4a^3 + c/a))

That all simplifies to:

-M +- sqrt(abs((M^2 -bM + c)/a)))

[joelpaddock]

I think possibly the reason we get the 'clunky' form of the quadratic formula is just to follow certain conventions: "always combine fractions" and "always bring common factors out of radical expressions." So often you can see recent textbook presentations of the quadratic formula will actually derive this more intuitive expression, and then they apply the rules to get everything in one big fraction. I sort of wonder if older books went a different way because typesetting for math was fairly limited, and these rules had a real benefit in cases like that.

[ianmathwiz]

This is essentially equivalent to dividing the entire equation by a, and replacing the linear term bx with 2Mx.

[eagle]

I always noticed that the difference between x1 and x2 was always the discriminant and due to almost always working with a = 1, I always felt that the value between those two was somewhat special. Well, now I at least know I was onto something!

[anudeepaghosh]

We learnt this extensively in school.. but in college the 'usual' formula was always used and i had forgotten the trick, ie, which was addition, which was multiplication. Thanks for helping in recalling it.

[mcalkis]

I had never heard of this before. Thanks Brian!

[superiontheknight]

I saw the Poh-shen-loh method before! It's really cool! It got me to experiment and see how much I could simplify it(like this video). I found a few fellas in a comment section to some other video(can't remember which, it was a while ago unfortunately) mention a few ideas that got really close to what I was looking for, but I still thought it could be slightly neater. So I played with it a little and found the following:

Starting from a typical trinomial:
ax² + bx + c = 0
-x² - (b/a)x - c/a = 0
B = -b/(2a)
C = -c/a
-x² + 2Bx + C = 0

x² - 2Bx + B² = B² + C
(x - B)² = B² + C

x = B ± √(B² + C)

So what all that means if we ignore the two steps before the last, is that we can divide everything by "-a", and jump directly from this:
-x² + 2Bx + C = 0

...to this:
x = B ± √(B² + C)

Basically, this is just yet another way of looking at what was just shown in the video. Really awesome however you look at it, and I'm glad to see someone put out a video to make this more known! :)

[Anmol_Sinha]

So, basically we just manually remove the coefficient 'a' by dividing and substitute m=-b/2 to make it more simpler in looks but actually harder. While I will not call it the 'better quadratic formula', I did like the new perspective of the derivation of the formula brought in the video!

[lukeerikblue]

The Po Shen Lo Method! I usually teach my kids the formula x = (-B/2) +/- sqrt( (B/2)^2 - C) after we've divided out a first. Very similar to the related formula x = M +/- sqrt( M^2 - P). Granted I show them the actual formula too. Both is good.

[davidbrisbane]

The midpoint approach fails geometrically when the roots of the polynomial are *complex.*

Consider x² + x + 1 = 0.
M = -1/2 and we discover that U² = -1/2.
If U is interpreted as a distance and it's positive, then how do we interpret U = √(-3/2) as a distance between the midpoint and the roots, as U isn't a real number?

Sure, the midpoint approach works algebraicly, but not geometrically when the roots are complex.

[peg]

this method was explained only in special cases where the a = 1. but incase you want to know them for the general cases, for the sum of x1 + x2 = -b/a, and for the product of x1 . x2 = c/a

[islandsociete]

This was in our high school textbook when learning the quadratic formula. This shows the intuition of the quadratic formula. It is much easier to just remember the one quadratic formula, than remember and figure out these steps in the video. Less room for mistakes too. (I did university engineering and economics btw).

[crustyoldfart]

Bravo ! Your approach is what I myself have been advocating for years. That is, it's based on the simple fact you point out - that the two roots are equidistant from the vertex value of the parabola..
A proof might run as follows :
Let R1 and R2 be the roots of the equation so
( x - R1 ) * ( x - R2 ) = 0 ; Arbitrarily R2 > R1.
x^2 - ( R1 + R2 ) * x + R1 * R2 = 0 which is equivalent to x^2 - b/a*x + c/a = 0
but m [ mean ] is ( R1 + R2 ) / 2 which is obvious from geometry.
So m = - b / ( 2 * a ).
We can define d as the difference between the value m and R1, R2 such that
R1= m - d and R2 = m + d .
But R1 * R2 = c / a -> ( m - d ) * ( m + d ) -> m^2 - d^2
From which we can assert that d = sqrt ( m^2 - c / a )
In summary then the two roots are :
( m - d ) and ( m + d ) where m = - b / ( 2* a ) and d = sqrt ( m^2 - c/a )
A quick check that if m^2 < c/a then the roots are imaginary.

Many teachers will deplore this approach, since they feel that the various ways of solving the quadratic [ completing the square, factorization ...etc ] are important pedagogic tools.
Schooldays behind them, those who have to solve quadratics in real life want the roots quickly and your method is logical and fast, and based on simple geometry.
" When I was a child, I thought as a child ; now I am a man, and have put aside childish things. "