The Better Quadratic Formula You Won't Be Taught
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- เผยแพร่เมื่อ 25 ก.ค. 2022
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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
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
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?
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
It's the same formula but simplified to a complicated equation 😅
@@JeeAdvancerause this when b is even
Midnight equation
This is what we learn in Sweden as well!
Ach ja, und natürlich auch das schöne "pq-Formel Lied" vom DorFuchs
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
Dorfuchs?
die jute alte pq formel
Simpliflied quad equation is just easier to remember
@@toyb-chan7849 Absolut zuverläßig. :)
@@jofx4051 from first hand experience, that is not true. The pq formula is messed up more often by forgetting to divide b and c by a.....
Most states in Germany prefer the "larger" formula for a reason, it also is easier to use when solving physics problems that usually come with crooked numbers.
Introducing the a factor into this equation is surprinsingly simple.
Thank you for the guidance ; i hadnt realised what the b factor represented.
This is essentially equivalent to dividing the entire equation by a, and replacing the linear term bx with 2Mx.
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.
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.
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)))
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).
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'm going to start using this. Thanks Brian!
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?
This is life changing!
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!
@Reio4 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.)
@@danieltemelkovski9828 you don't have to know how the black box works to use the black box
Completing the square has other uses outside of just finding zeroes, such as when representing rational functions as partial fractions in a form suitable for integrating. But this does look shorter, you just have to simplify to a more specific form first.
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
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
I had never heard of this before. Thanks Brian!
You bet!
its actually called Viete's theorem or Viete's equations , they relate solutions for anypolynomial and are really useful, I wrote a paper and had like 20 usages of them in Olympic exercises
Awesome video! Thank you! This very interesting from a computational standpoint. Seems much faster to implement compared to the usual formula. That being said, I wonder if this is still applicable for quadratic formula with complex roots. It seems that the graphical intuition won't be applicable then.
It does still work, the only difference is you get U^2 to be a negative term instead of positive. Then when taking the square root on both sides, i pops up
is this really faster for computers? because this is literally the quadratic formula
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.
Beautiful!
Amazing formula I’ve never thought about this and I haven’t been teaches this trick
Thanks a lot man ❤️🌹🙏
So glad I'm not the only one using xtop and distance delta plus/minus from xtop. With this formula you will know xtop, the discriminator and delta, each important to understand a parabola.
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?)
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 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.
"M plus or minus square root of M^2-p!"
-Tim Blais (he made a short tune for it, used in some 3Blue1Brown videos)
This formula reminds me of a variant of the quadratic formula to use when b is even, if you set M=b/2 then the overall formula can simplify to (-M+-sq(MxM-ac))/a, removing the clunkiness of the x4 in the square root and the x2 in the denominator
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!
Isn't that difference of roots √Δ / a
your feelings were irrational
the difference of roots for a quadratic is : √∆/ |a|
I'm actually proud of myself rn because I already came up with this exact thought before
That's really interesting! We had to factor for quite a while as we hadn't learnt the quadratic formula yet last year, so it got me thinking about an easier way to do it rather than just in my head. I attempted to use simultaneous equations but eventually it lead me right back to square 1.
That's valuable, thank you ☺
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.
personally i love factoring and will try to use it in any case unless necessary for the quadratic formula
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
In India we do this at the very beginning of the quadratic equation chapter
I was never taught the vertex formula my entire life, I had a test a couple months ago and used completing the square for finding the vertex. I know, such a waste of time, but now I actually know the vertex formula and proved it using calculus. Thanks.
A very great method to solve quadratics
Love You!!!
I will hold on the completing the square method
admiring the simplicity of this, i am left wondering why this is not how everyone is taught this formula.
thank you for opening my eyes to this wonder.
god bless you, stay healthy.
Simplicity ? This method is probably the hardest way to do it, why struggle with this graph-based algorithm when you can just remember one goddamn formula, that's even easy to find again if you forget since the proof is quite simple
I wish i had seen this video 1 day before I had my test
This method much way faster than the original. Thank you so much!
After following this channel I started falling in love with Maths
normally completing the square (x+b/2)^2 + B is the go to for quadratics with even 'b' value and then factorising (x+A)(x+B) for prime 'b' values
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.
Dude this save me from the factoring quadratics exam thanks!
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 😀
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.
I actually had derived this on my own to save time 😭😭😭
What an amazing video!!
This is super cool. I think this can be linked to Vieta's formula right??
Isn't this just like another way to think about/write completing the square? Because that IS something that they teach at least in my state.
I LOVE this, and I am very sad that I have never seen this in my high school
I'm in middle school and on top of the quadratic formula we were taught another method: make an x shape and put the product of a times c into the top. Put b into the bottom. Figure out what 2 numbers add to b but multiply to that other product(if you can't then there's no solutions).Then, put the original a into the top left of a 2x2 box and the c into the bottom right. Fill in the other 2 boxes with those factors from earlier but they're both multiplied by x. Find the greatest common factor in each row so one row was one factor and the the other row was the other.
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
Is this the same or similar to the one that 3b1b showed on his livestreams from about a year ago? This method is super cool, glad to see it again as a refresher :D
I didn't see the stream but it very well could have been!
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
In Turkey, we are learning these in high school starting from 11. grade.
You are a freaking genius
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
This is literally the quadratic formula but simplifying the 2a from the denominator within the square root
yeah i mean, this is kind of a derivation of the dharacharya formula, still pretty impressive, gives more visualisation of "roots" of equation
I wish I had learned about this starting Algebra II
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.
A very good method. Fairly obvious if you use to sketch the graphs though.
In fact, we can make it general
x^2+bx/a+c/a=0
midpoint = -b/2
x1=-b/2a-U
x2=-b/2a+U
(b/2a+U)(b/2a-U)=c/a
b^2/4a^2-U^2=c/a
b^2/4a^2-c/a=U^2
sqrt(b^2a/4a^2-c/a)=U
x1=-b/2a+sqrt(b^2/4a^2-c/a)
x1=-b/2a+sqrt(b^2-4ac)/sqrt(4a^2)
x1=-b/2a+sqrt(b^2-4ac)/2a hang on a minute.... This seems.... familiar.
i love your vids keep it up
Thanks for the support!
nice!
Btw please do more videos like these since I’m going to 10th grade and I need fast methods to solve problems since I’m my country after year 12 we have a big exam in order to go to university
I’d really appreciate it ❤️🌹
That's really cool!
Glad you think so!
I remember learning this in school, but just looked a little harder(we didn't use the formula just factoring ones), so I never tried it.
It's so cool and obvious when you see it. haha
I love it!
I love how I have been taught this from the beginning (I live in Sweden)
2:53 "Quickly solved for u"
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).
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
I always knew this as pq-Formel and in Germany it is taught as the main method to solve quadratic equations
The discriminant comes an alternate formula of the quadratic function: f(x) = a(x + b)^2 + c. The three scalars here determine the precise location of the extremum point and the slope of the curve. a determines the slope, b determines the inverse of the abscissa offset of the extremum and c determines the ordinate offset. The discriminant comes from taking the f(x) = ax^2 + bx + c function and transforming it into the above. I'm not going to do the full derivation in plain text, but it comes out to f(x) = a(x + b/(2a))^2 + (-b^2 + 4ac)/4a.
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)
We learned both and called it abc formula and pq formula
To those who are interested in the derivation of formulas for polynomial equations, I have a playlist on my channel
with the derivation of all the formulas from order 2 up to 4. Please check it out!
This is quite interesting!
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
Does this only work with equations we *can* factors out?
When we learned to graph parabolas this seems so obvious yet so clever. Thanks bae
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 only works when a = 1 (the quadratic equation has its coefficient of x^2 as 1)
been using the factoring method ever since i started learning about quadratic it is good but when things get complex the quadratic formula is better (in india it is normal to use both we were taught this in school )
we were taught the difference of squares method at my school. Do other schools not go over it?
(x+1)^3:”I think I forgot something.”
x^3+1:”If you forgot, it probably wasn’t that important.”
(x+1)^3:”Yeah, you’re right.”
3x^2+3x: “-_-“
√-1 2^3 Σ π
You gained a subscriber from India ❤❤❤
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 is the formula I learned in school in Sweden, the formula with a, b and c always seemed complicated to me
I think my entire brain fell out of my head about 45 seconds into this video
Please tell is there a formula for finding roots of quartic equation
in germany we learn the p-q Formel which is a formula where you need the x squared part to be 1x squared to apply it find the p which is the number with x and the q which is the number without any x then plug it into -p halves plus minus the square root of phalves squared minus q it is very easy for me to remember due to a song made by the youtuber dorfuchs
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...