The Better Quadratic Formula You Won't Be Taught
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- Опубликовано: 28 сен 2024
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Corrections:
2:32 Make sure a=1 (divide by the leading term)
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Wait isn't it m +- sqrt(m^2 - p) instead of m +- sqrt(m^2 - c) ? that would make c = c/a
Hello can we use this formula for solving the equation consist of iota i in the coefficient of b in the equation?
@@That_One_Guy...Exactly what I know it as!
This is a great way to use the quadratic formula without "using" the quadratic formula. I've seen this method used on other channels.
It's true! It really sort of is the formula (with a bit of trickery)
It slow tho .original best
@@magicbaboon6333 but if you practice it is very much faster than quadractic formula
@@BriTheMathGuy but it doesn't work for some quadratic equations which dosent form a parabola in the graph. Example :
x²-350x+660.,,
.....(to be continued for eternities lol)
@@somerandomuserfromootooob It does form a parabola (plot it on desmos and zoom out), every quadratic does
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
Yeah that’s why it gives you the same answer, it is mathematically equivalent. The point is to show an intuitive way to “derive” the expression that isn’t completing the square
Thanks for explaining it😭 Quadratic makes my life hell
The jokes not funny if you explain it 🙄
Yes it’s basically x=-b/2a also graphically, x=-b/2a moves the vertex of the parabola to the y axis in the function f(x)=ax^2+bx+c which leaves us with only the constant and the square. This is kinda better than completing the square as moving special points of a graph is useful for simplifying any equation to ax^n+bx+c=0 (though there isn’t a simple formula way of solving this for n >=5
If you derive m=-b/2a, then you get (-b +- sqrt(b²- 4a²c))/2a. Why does this happen?
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.
good luck find imaginary roots with this
@@marius4363 I mean, it actually works just as well, and arguably the same. If U ends up as the square root of a negative number, congrats! You've just found the imaginary roots.
Yeah i would say, its about the same
@@marius4363 cykabled??
@@marius4363 this method IS the quadratic formula, so it's completely possible to find complex solutions, u2 will be negative and that's it
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.
We don't generally use this method because it requires external calculations. If you're going to use a canned formula, you might as well use one that has everything necessary in it when available. All this is to hide details that you still have to remember and deal with in an opaque variable.
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.
you cannot solve higher degree polynomials in any straight forward general way past degree 2, approximation methods are used past degree 2. The main reason we learn to solve polynomials is to solve linear differential equations and it turns out degree 2 is in fact the most important case, one way to think about why this is so is because the most of the important equations of Physics are degree 2. Hence the quadratic formula is in fact very important and not some afterthought.
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! :)
Can you (or someone) help me out? I tried this on a very simple problem that I can do in my head, and using this video's method my answer comes out with the wrong signs.
When I apply this method to:
x^2 - 5x -14 = 0
I get a mid point of 5/2 and a distance U of 9/2
My set of solutions for the zeros of the quadratic then become
5/2 - 9/2 = -2 and
5/2 + 9/2 = 7
But that's not correct. The correct solution is positive 2 and negative 7.
What am I doing wrong?
@@kevinkasp The set of solutions you got from the video are actually the correct solutions. So I'll do this both the way the video did it, and the way I did it in my comment. :)
We'll start with the videos approach:
x² - 5x - 14 = 0
M = -(-5)/(2(1)) = 5/2
c = -14
U = √(M² - c)
U = √((5/2)² - (-14)) = √(25/4 + 14) = √(25/4 + 56/4) = 9/2
x₁ = M - U
x₁ = 5/2 - 9/2 = -4/2 = -2
x₂ = M + U
x₂ = 5/2 + 9/2 = 14/2 = 7
Now if we do it the way shown in my comment:
x² - 5x - 14 = 0
-x² + 5x + 14 = 0
B = 5/2
C = 14
x = B ± √(B² + C)
x = 5/2 ± √((5/2)² + 14) = 5/2 ± √(25/4 + 56/4) = 5/2 ± 9/2
x₁ = 5/2 - 9/2 = -4/2 = -2
x₂ = 5/2 + 9/2 = 14/2 = 7
Either way you get to the correct solutions of x = -2 or x = 7. We can check this too and see that our answers are correct.
(-2)² - 5(-2) - 14 = 0
4 + 10 - 14 = 0
0 = 0
True!
(7)² - 5(7) - 14 = 0
49 - 35 - 14 = 0
0 = 0
True!
We can also check the other two solutions you came up with, and see they don't work.
(2)² - 5(2) - 14 = 0
4 - 10 - 14 = 0
-20 = 0
False!
(-7)² - 5(-7) - 14 = 0
49 + 35 - 14 = 0
70 = 0
False!
Perhaps you were confused when this is factored. Remember, that "x² - 5x - 14 = 0" can be factored as the following:
(x - 7)(x + 2) = 0
Remember though, x is still "-2" or "7". This can be seen by plugging in for x.
((-2) - 7)((-2) + 2) = 0
(-9)(0) = 0
0 = 0
True!
((7) - 7)((7) + 2) = 0
(0)(9) = 0
0 = 0
True!
Hope this helps! :)
@@superiontheknight963 Got it. Brain fart on my part. It's been so many years since I had to do problems like this I did in fact make the exact mistake you suggested. Meaning, I got the correct solution, but "checked" my answer by looking at the factored form of the equation and did the brain fart of thinking ( x + 2) shows +2 is a factor, instead of solving x + 2 = 0, which would have proved that I had the correct solution.
That's as bad as it gets. To solve a problem correctly and then discard the answer because you take the time to check it, but then do that incorrectly.
Thank you for taking the time to set me straight.
Also, your method is the way I would teach it to kids learning algebra. I would use the video's method of visually showing what is to be accomplished, and yours to do problems.
Thanks again.
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!
But beware, the coefficient A must be equal to 1
@@reio4641 yeah, I said that we can remove it by dividing. (By dividing I mean dividing the polynomial by a). Thanks for highlighting it for the readers of this comment.(if that was what you actually meant to do)
Actually it mixes the summ and product rule (factoring) in a way that you don't have to guess.
To me it is good because you can use summ and product trying to guess and, if you can't, you use this method as an expansion.
@@TamissonReis You don't have to guess with the quadratic formula (which this is just a poor imitation of) either.
@@programmingpi314 This method puts you more 'in touch' with what is happening graphically, imo. The quadratic formula just has you blithely punching numbers into a calculator without lending much understanding of what you're doing. Although if you learn the derivation of the quadratic formula from 'completing the square' you will gain some insight into what you're doing. (All this presuming the person doing the calculations actually cares about understanding, rather than just 'getting the right answer' for a test or something.)
while this formula looks shorter, when you sub -b/2a for M and do a little manipulation, you still get the same old quadratic formula, so I don’t know how effective it actually is in the real world.
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.
Well of course you're not going to find a ℝeal distance to the roots when the roots themselves aren't ℝeal. However I have actually seen a concept of an sphere with an imaginary radius in the past which could be thought of in a similar manner. The two roots of a quadratic can be thought of as a single 0-sphere with a centre of M and a radius of U. In this case U is imaginary.
@@angeldude101
Where did he say in words that he was restricting the solution to only the real numbers? He does only show examples where the solution are real. I'll stick to the "Clunky" formula, as its solution allow complex solutions also.
@@angeldude101
Of course, we could use the discriminant to determine if the solution has complex roots or not 🤣😂🤣😂.
@@davidbrisbane7206 Ultimately, this formula is algebraically equivalent to the other. M = -b/2a, U² = M² - c/a.
= M ± √(M² - c/a)
= -b/2a ± √((-b/2a)² - c/a)
= -b/2a ± √(b²/4a² - c/a)
= -b/2a ± √((b² - 4a²c/a)/4a²)
= -b/2a ± (√(b² - 4ac))/2a
= (-b ± √(b² - 4ac))/2a
Any result the traditional formula will give, this alternate formula will also give. The only difference is that the traditional formula is "simplified" into a single fraction.
@@angeldude101
Except for the graph 😀
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. "
I love that you're teaching so many people and getting them to reexamine the quadratic formulae ❤🙏👍 That being said, It's a prerequisite for High School level Math classes here in the Indian Sub-Continent. I'm back in Pakistan and yep, I make my students create their own sums. That's the best way to test their skills i.e. working backwards and forward. A good tip for teachers/tutors is to collect them and make a test using all the questions
@@asadmahmood2007 SO, if I understand you correctly, you are saying that as a professional teacher at High School level, your main task is to teach students how to meet the the pedagogic dictates of the examining authority. One wonders if there is any room in such a system to imbue students with a feeling of awe and wonderment regarding the nature and immensity of mathematics.
Speaking for myself I endured the pedagogic dictates of public exams up to the age of 22 when I graduated in engineering. It is only since then that I have developed a reverence for the greater part of the mathematical world, but with the reservation that I am not only incapable of understanding anything but a small part of that world, but there is much of it which holds no appeal at all.
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
In germany, we learn the pq formula, which is a bit simpler than the quadratic formula you showed at the beginning:
-p/2 ± √(p/2)^2 - q)
(p is the x¹ term and q the x⁰ term)
You need to normalize the quadratic equation by deciding by the factor of the x² term and then plug p and q into the equation.
I don't know if it makes it a lot easier to calculate, but I think it's more elegant than having the whole formula devided by 2a at the end
IIRC, we wouldn't use that in the US typically because the rational zeros theorem that that's based on isn't taught until later. It can be used, but there's little utility in using it on a 2nd degree polynomial when we've got both completing the square and the quadratic formula as well.
We are taught this in Sweden too. I feel like it is less clunky compared to the other formula.
I just checked it. If you puzzle the steps he did in the video together, this exact formula is the result
In India we do this at the very beginning of the quadratic equation chapter
Yes, this is what i do with the po-shen loh style. The most simple if you use the -b/2a , but a=1, so it becomes -b/2 only. Vertex formula to find the “h” value of the vertex: (h,k). That is actually the AOS or Axis of Symmetry. Po-Shen Loh is hiding this one.
This is what I've always been taught in school, I was always confused why others use the weird one that I still don't understand
Factoring, completing the square, the quadratic formula are all easy to learn and use. Learning how to derive these is the key to enjoying this level of maths and sets the groundwork for harder things to come.
It makes so much more sense... I've discovered it myself the first time I tried to write a computer program to solve this. Later I've found out that this is the method taught in high schools in some countries (Germany? China?)
To find an analytical solution for ax^2+bx+c = 0 is easy. However, what is the analytical solution for ax^(2+e)+bx+c=0 with ‘e’ being a real number? The solutions are
x1=(b/(az))Wq(((-c/b)^z)(a/b)z)^(1/z), where z = (1+e) and q = 1-1/z.
x2 = (-y(a/b)Wq((-1/y)(b/a)((-c/a)^(-1/y))))^(-1/(1+e)) where y = (2+e)/(1+e) and q = 1+y
Wq is the Lambert-Tsallis Wq function (a generalization of the Lambert function). Sometimes the correct solution is x1, in other cases the correct one is x2 and there are cases where x1 = x2, depending on the values of a, b and c. For example the solution of x^(2.5)+x-1 = 0 is x1 = x2 = 0.6540 (up to 4 decimals).
for basic quadratic equation i was taught in school to do: -b/a = Sum of both x and c/a = Product of both x so for instance in the example you showed -b/a= -(-4)/1=4 and c/a= 3 that means the x1 and x2 are equivalent to 1 and 3
That’s for factoring, but yea
What I do is x^2 - 4x + 3=> x(x-4)=-3
And then I take -3’s multipliers (which is 1,-3; -1 and 3) and just try them to see which one is 4 smaller than the other.
This is such a cool way of visualizing how (a + b)(a - b) = a^2 - b^2 with two points b distance away from a
I actually had derived this on my own to save time 😭😭😭
they taught us that solution in school along with the classic formula (-b ± √(b^2-4ac)/2a
lets say the formula is x^2 + 3x + 2
first you find the factors of 2 (try the easiest ones first)
its 1 and 2
second you find the factors of the x^2
its x and x
then you multiplie the factors of x and factors of 2 together, if it makes the "b" which is 3, then they are the true zeros for the equation
x^2 + 3x + 2
x1 2
x2 1
first x factor times second number factor, second x factor times first number factor, its a diagonal multiplication
1 . x1 + 2 . x2 = x + 2x = 3x
it worked, so we can do the addition with the first factors of x and number, and second factors of x and number
(x1 + 2)(x2 + 1) = (x+2)(x+1)
x is -2 or -1
it doesnt matter how you multiplie if the x^2 multiplier is 1, but if its like 2 or 3 or something else, you should be careful
for example
2x^2 + 5x + 2
2x 2
x 1
lets multiply diagonally and try to get the "b" value
1.2x + 2.x = 2x + 2x = 4x
b is 5 but we got 4 so lets change the order of x factors (x on the top, 2x on the down instead)
2x^2 + 5x + 2
x 2
2x 1
1.x + 2.2x = x+4x = 5x
b is 5 and we got 5 so its true, so lets do the addition now, but not diagonally, addition is straight
2x^2 + 5x + 2
x -------> 2
2x -------> 1
(x+2)(2x+1)=0
x is -2 or -1/2
it looks hard and so long to do but its because of explanation, you just have to factor out the number here and multiply it diagonally with the x factors, if it gives the "bx" value, then add them straightly. putting them in the (ax - x1)(bx - x2) equation, and factoring it successfully. thats what they taught us in my country
sometimes it has no factors or hard-to-find factors, then you should do the regular formula
hope that helped :)
Since M= -a/(2b), you'll often be stuck with simplifying the square root of a fraction, even when the coefficients are integers. The reason why the quadratic formula is written the way it is, is to eliminate the simplification necessary every time. Notice that the presenter only chooses equations with integral values of M...
There is no need for any form of quadratic formula when you can just complete the square.
if i cannot factor the equation i will usually just manually derive the quadratic formula by completing the square, shifting c-(b²/2a)² over to the right and then solving for x - it just feels nicer that way
bro why is my brain just not braining rn
We learned both and called it abc formula and pq formula
The midpoint M, a.k.a. the axis of symmetry, is just -b/(2a); the distance U from the midpoint to one of the solutions is then the square root of (M^2 - c), a.k.a. the square root of the quantity of the discriminant D divided by 4a^2.
This is Po Shen lo method
There is also another formula that not many people know which is useful when b is a large even number:
D₁ = (b/2)² - ac
I couldn't even find this formula on the internet when I tried. Maybe it's called something else instead of D₁ in other countries.
personally i love factoring and will try to use it in any case unless necessary for the quadratic formula
I use the -p/2+-root of (p/2) squared-q. p and q are b and c. These points can also be used in the (x-x1)*(x-x2) formula ( sry for the writing dunno How to Write root of something on iPhone)
Honestly I feel like quatratic equation are much easier for me……the overall equation of this whole thing is -b/2a+-((-b/2a)^2-c)^0.5, which seems to be much longer
Isnt there an easier way?
In this case, the formula you mentioned can be sovled using the
Fact that a+b+c=0 in which case one of the roots is 1 and the other is c/a
I know this only works for this equation but equations are usually given in a form to make it easy for you to find the roots and this formula is really easy to specify the roots.
one thing to note is that , a should be equal to 1 to use this method
im guessing it only works if the coefficient of x^2 is 1
I discovered this myself when I didn’t understand the way it was being taught. I’m really intelligent and work for like 5 hours to come up with this
From my and a few friends impressions this is indeed taught in many places, we have evidence of it being tought in germany, austria, india, bangladesh, china, vietnam, new zeeland, hungary, ukraine, russia, kazachstan and probably a lot more countries (almost always in middle/high school)
Aren't those also mostly education systems that emphasize calculation at the expense of applications? This method works, but it's unnecessary to mask parts of such a short calculation into separate calculations. When I was in school before multiline calculators with pretty displays were a thing, there was some point to it, but these days not so much.
@@SmallSpoonBrigade now speaking only for Germany and Austria: it is taught that you should solve ax²+bx+c=0 by dividing by a first: x²+px+q=0 with p=b/a and q=c/a. Then use the formula x(1/2)=-p/2 ± √((p/2)²-q). However the formula using a,b and c is also taught, but barely used in practice, because the so called pq-Formula is a lot easier.
This really reminds me of how you find the eigenvectors of a 2x2 matrix where M is the mean of the diagonals and the C term is the product of the diagonals. 3b1b did a great video on it in their literary algebra series
*linear algebra, phone keyboards suck lol
I think U^2=1/16
U=+1/4 , -1/4
Roots are
Midpoint (1)+1/4
And
Midpoint (1)+(-1/4)
This is the right way
Plz Comment
When calculating U², you are essentially competing the term in the square root of quadratic formula. It is easier to remember, but not very helpful computational wise.
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My high school math teacher asked me if I rode the short bus to school when I presented that to my class.
yeah i mean, this is kind of a derivation of the dharacharya formula, still pretty impressive, gives more visualisation of "roots" of equation
U can also prove it
like this
ax^2+bx+c=0
divide both sides by a
=> x^2+(b/a)x+c/a=0
=> (x+b/(2a))^2+c/a-b^2/(4a^2)=0
=> x+b/(2a)=+-sqrt((b^2)/(4a^2)-c/a)
=> x=-b/(2a)+-sqrt((b^2)/(4a^2)-c/a))
actually u can easily derive the quadratic formula from this
I was singing this in my mind, and then this video pops up the first thing when I open yt first time in the day, fucking hilarious
so it's M=-b/2a and U^2=M^2-c/a?
Yeah, this is my go to method to solve quads. It's pretty easy to quickly do in your head in most situations. They should definitely teach it in schools.
no I don't think so
it's not possible to get roots when we have quadratic equation with complex roots
I'd say that the original formula is much easier to do in the head, and it works in everything
Amazing formula I’ve never thought about this and I haven’t been teaches this trick
Thanks a lot man ❤️🌹🙏
When I didn't know the quardatic formula I used something like this (about 5 month ago)
The quadratic formula is a easy one I personally remember the formula by heart and it is basically easier to just use it rather than using any other thing
" I personally remember the formula by heart " - which presumably means the clumsy Vieto formula and not what is presented in the video.
this is basically the same as completing the squares.
I was taught another Quadratic Formula (aside from the 1st one) and its (b÷2) ± square root of(b÷2)²-ac though please note thats not what the equation really looks as keyboard has a lot of limitations
we have been taught -p/2 + / - sqrt(p²/4-q) which practicalli is the solution here but without the geometry stuff ^^
And from -p/2 you simply have to square it to get the p²/4, so it's fast.
Funny, I've been using this as a kind of private shortcut forever. I didn't know it was some official technique.
A very good method. Fairly obvious if you use to sketch the graphs though.
This looks like a form of completing the square really. Interesting, but I think it's kind of slow because of the multistep process.
I love how I have been taught this from the beginning (I live in Sweden)
I have a SECRET method first
No. of A × No. of C then
Find the 2 no.s that add or subtract to make No. Of B
Then you will get total 4 no.s
Then just do it idk i will explain through example
2x^2 - x + 6x = 0
A=2, B=1, C=6
Now A×C=2×6=12
Find the 2 no.s that multiply to be equal to 12
And add or subract to make B which is 1
It will 4 and 3 (if u use brain)
Then you will get, 2x^2 -4x + 3x - 6=0
2x (x - 2) +3 (x - 2)=0
Then, (2x+3) (x-2)=0
2x+3=0 , x-2=0
x = - 3/2 , +2
Hope it helps some kids
Acho lindo como só no Brasil a primeira forma é chamada de fórmula de Bhaskara
you can also use a simpler formula for quadratics with even B rather than the general one, cause it is little bit faster. k=b/2, x12=(-k+-sqrt(k^2-ac))/a
This method much way faster than the original. Thank you so much!
The content you deliver is awesome!
Can you please let me know that from which textbook source did you study all the methodologies of this particular lecture?
I think this is same as split in the middle term taught in India
Dude this save me from the factoring quadratics exam thanks!
Typing x=1 and x=3 is not logically correct. x cannot be both 1 and 3 at the same time. x1 = 1 and x2 = 3 is correct though. or "x=1 or x=3" is correct aswell. At 0:46
I taught my students this method too. But they're used to memorization which is a pity.
x²-4x+3=0
Midpoint=-(-4)/2=2
U²=midpoint ²-c²
U²=2²-3=4-3=1
X=2-1=1
And,
X=2+1=3
In the real world, many equatIons are only slightly quadratic and we are only interested in the solution which is slightly different. It has the complicated bit in the denominator and avoids destroying accuracy by subtraction.
the quadratic formular seams like a much easier to memorize algorithm.
quadratic formula is also called Bhaskara in some spanish countrys
My math teacher taught it to the tune of row row row your boat
The thing is, once you learn linear algebra, matrices are just easier. Especially as the you get to more than just 2 dimensions
It was given in 3blue1brown .
Well if you really want to minimize the effort of learning the quadratic formula, then I guess you can do the following:
1. Make the value of a = 1 by dividing the equation by a on both sides.
2. Differentiate the quadratic and equate it to zero and to get a value of X (Call it B, where it stands for Bolt's Constant)
3. Calculate J = (√∆)/2
(∆ = discriminant)
(J stands for Jack's constant)
4. Now the roots of the equation is given by:
X1,2= B ± J
( im sure you can remember this easily!😂)
Note: is Jack's constant (J) has a real value, the root(s) with be real aswell given that a,b,c were real.
If J is an imaginary number, roots will also be imaginary.
You can call this The "BJ" Theorem.
I use the Quadratic Formula to factor
Oh,man,how did you find this splendiferous technique😳😳👌👌🙏🙏,God bless you forever!You deserve piles of likes
By watching Po Shen Loh's Video .he didn't discover it. 😂
Many thanks but I didn't invent it 😅
@@BriTheMathGuy Pardon,my pal,Brian,but can I ask what discipline you're trying to major in at university?
@@amn4921 I studied math when I was in school :)
@@BriTheMathGuy Yeah,that's what I'm talking about;I purely appreciate and look up to the people like you who are honest and not trying to simply plagiarize other ideas of sages;keep your scientific path up! May you be lively and successful in your entire life!👌👌🙏🙏
Try solving x^2+ 5x+6 using this method.
This only works for a =1 but for other replace B with b/a and c for c/a and formula works again
That's just the pq formula, isn't it. Never understood why people use the ab formula that has much higher numbers than the pq one.
Calculators have the quadratic formula built in so I wouldn't say this is better.
This is exatly the pq-Formula wich you can find in every formulary for mathematics. This is nothing else than the normalized a,b,c- Formula for a=1.
I wasn't taught it at all and I had to figure out how to do it myself lmao
We only know this method in germany
Cool!
We use this method at my school
But it is exactly the pq formula or the abc when I use divide by a.
You can never convince me *ax² + bx + c = 0 x = [-b ± √(b² - 4ac)]/2a* is difficult to compute. For being 100% accurate every single time (with a≠0 obviously) at solving an equation otherwise challenging, it is so simple and actually some very simple arithmetic, it cannot be improved.
Yeah I mean if a = 0 it's even easier, x = -c/b linear done.
No like seriously, who in the world ever struggled with the quadratic formula? Even my borderline-illiterate classmates who keep forgetting what (a-b)² is can remember and apply this...
So basically just way more complicated than the normal formula, which you probably remember anyway after having to use it like 500 times... Half of it is discriminant, which is still pretty useful to know and calculate first so you know if you even have to bother with the rest of the equation.
can someone explain to me how this would work on a different formula like 16x^2 - 12x - 100. Maybe i’m overseeing smth, but with this formula I end up with U = +/- sqrt(6.3906) and that’s incorrect…
isn't this just compleating the square ?
Wait that was the cuadratic formula? I thought it was for advanced physics or something in spanish it’s called the general formula
Mans just did the quadratic formula in pieces and said it was easier💀 it does help explain where the equation comes from tho
3/4+5/4=2
But we want -2
Can you explain that
we were taught the difference of squares method at my school. Do other schools not go over it?
... this is solving a quadratic equation by completing the square, and using the axis of symmetry fact. This method is well explained in IBS math textbook. Thank you for sharing.
If this is completing the square method then my school must’ve taught me something else
Is this will valied if there was imaginary number roots in the equation?
yo we can skip the process by putting c/a in equation which makes it M+rtM2-c/a
I’ll be honest, this was super confusing. I understand in theory but I think I would need to do it in practice and I’m 26 and haven’t touched the quadratic formula in years