The Quadratic Formula No One Taught You
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- Опубліковано 9 тра 2024
- We derive an alternative version of the quadratic formula, then explore advantages and disadvantages of each version. This includes values for which they are defined, and the effect of rounding on the accuracy of solutions.
00:00 Intro
00:19 Derivation 1
02:29 Derivation 2
04:55 Problems with the original formula
07:40 Problems with the new formula
11:31 Comparison of accuracy
14:17 Why you should use both
Crazy how I never thought of this during my 35 years of solving 2nd degree equations.
Because it doesn't work if x = 0.
Why choose a method that doesn't work for all cases, has roots in the denominator, and takes the same amount of work?
Because of numerical stability? Have you seen the video til the very end?
@@user-jc2lz6jb2e Using the quadratic formula for x = 0 is anyway not sensible at all. In such cases, we have an equation of the form a x² + b x = 0, and by factoring to x (ax + b) = 0, it's obvious that x = 0 is a solution.
@@user-jc2lz6jb2e why would you ever use the quadratic formula to solve for the solutions if one of them is x = 0?
Ditto.
IMHO, I would keep the "minus over plus" in the alternate formula.
Absolutely right. Then you can see that the corresponding root for the two formulae comes about when you add the discriminant in one formula and subtract it in the other.
I was aware of this version. In days of old when computers ran on steam and precision was to 3 or 4 decimal places, the standard formula failed if b was much bigger than c say b=10000 and c is 0.1 then the c is lost in the computing rounding. The alternate formula takes the c into account and contributes to the computation. It would be interesting to see what happens with modern day computations with the two formulas when b is very much bigger than c.
slowly turning into blackpenredpen, I see :)
Just need 2X speed...
This is genuinely so freaking cool. Thank you for sharing, great work as always Dr Barker!
At first, I thought couldn't you multiply the two formulas together, and cancel out the square root entirely, and just end up with x² = something simple?
But then I realised that the root that has the + in one form is the one that has the - in the other form, so they won't actually cancel out if you multiply the same root to itself. But if you multiply the two roots together, you do get a bunch of cancellation... and it ends up collapsing down to x1*x2 = c/a, which is one of Vieta's formulas.
He demonstrated 'systems of equations' between two derivations of the same expression, a nice tactic and one I had NEVER considered in the taken-for-granted Quadratic Formula. And he did it in pieces without saying, directly, "We'll now use a System of Equations to solve for abx." It grew organically and I really dig that.
Rationalising the numerator was really cool for me. I had earlier tried using the quadratic formula to get the roots of a linear polynomial but was not successful because division by zero, but I wasn't able to think of rationalizing the numerator which would allow the quadratic formula to also work for linear polynomial
Great stuff. Enjoying your channel. Always fun to go after problems from every direction ... can optimize for the needs of particular engineering problems. Thanks. Cheers ...
If A=0, then it isn't a quadratic, so no one would use the quadratic formula.
Some problems don't tell you that a=0 , intentionally write the x² term .
So its better IMO .
When a quadratic equation is multiplied by the number making the coefficient a = -1/2 (so we have -0.5x² + bx + c = 0), the familiar quadratic formula simplifies into x = b ± √(b²+2c).
@@keescanalfp5143 Here's two I've learned for factoring or solving:
1. Slide and divide
Solve a related quadratic, x²+bx+ac, and divide the negative roots by a total product of a to get this factor back. If you end up with fractional roots, just clear the denominators.
2. Po Shen Loh's alternative method
Divide out a factor of a.
Looking at the coefficients, recast the quadratic as x²-2mx+p. (m is 1/a times the mean of the roots, and p is the product divided by a)
Now x= m+/-sqrt(m²-p).
yeah .
so then first you divide the whole thing by 2a .
an other view can come up by using an alternative standard form which favours an even value of b .
when solving the form
ax² + 2bx + c = 0 ,
you might get rid of any kind of coefficient 2 and 4 .
you will however keep the denominator,
of course unless a = 1 .
x_¹,² = [-b ± √(b² - ac) ] / a .
@@keescanalfp5143 That's *almost* the same as Po Shen Loh's method- he divides out the factor of a so now a=1 anyway!
The absolute best version of the quadratic formula though is when you divide both sides by a, and change b into 2b and c into c^2. Then what you get is that for the quadratic equation x²+2bx+c² = 0, the solutions are x = -b ± √(b² - c²), which is nice because you have a difference of two squares inside of the square root, and that actually allows you to express the formula in a yet simpler way: let b+c = p and b-c = q. The quadratic formula becomes x = -(p+q)/2 ± √pq, which is kind of cool because it expresses the roots in terms of the arithmetic and geometric means (of the same two numbers) added together
For people in the comments saying what about a,b,c = 0
You should look at what this formulas means. It gives the roots of the 2nd degree polynomial
ax^2 + bx + c = 0
Here, if a,b,c = 0 then what are you even solving for?
0 + 0 + 0 = 0
If a=0 it's not a 2nd degree polinomial;
If c=0 then it is on the form of ax²+bx=0.... x(ax+b)=0... so the solutions are x=0 or x=-b/a;
Given x=2c/(b+- sqrt(b²-4ac)... since c=0 we have x=0/(2b)=0... or x=0/0
I mean... for that last bit the formula still works... you just need to be more careful: evaluate lim_(c -> 0) 2c/(b+ sqrt(b²-4ac)) [assuming b>0.... else use +sqrt(b²-4ac)... since sqrt(b²)=-b if b
An entirely different approach is to verify that the roots of cx^2+bx+a=0 are the reciprocals of the roots of ax^2+bx+c=0. Using either the traditional quadratic formula or the new formula in this video, it's straightforward to demonstrate that the product of each root of the first equation and the opposite-signed root of the second equation is 1.
Reciprocal roots generalize to higher dimensions, as well. I once posed a problem for a high school math competition: One of the roots of ax^4+bx^3+cx^2+dx+e=0 is 2. Name one of the roots of ex^4+dx^3+cx^2+bx+a=0.
@@frobozz55 Also 2.
@@xyannail4678 Nope. Plug 2 into the first equation, and you get 16a+8b+4c+2d+e=0. If you plug 2 into the second equation you get 16e+8d+4c+2b+a=0. These aren't the same thing. The clue was "reciprocal roots".
@@frobozz55 1/2. Please I don't know if you mean conjugate roots, give me a bone here or help or tell me where to try and challenge myself to get the solution.
Am I supposed to Ruffini my way out of this one?
@@xyannail4678 1/2 is correct. 1/2 is the reciprocal of 2. There's no Ruffini here. If you plug 2 into the first equation and 1/2 into the second equation, you can find that they are equivalent.
We looked at the accuracy issue to introduce Vieta's formulas. Quite nice!
First time I have ever seen this. Really interesting. Thank you.
Having taken a Numerics class, I appreciate how well this video presents the challenges of trying to get accurate numbers out of a computer.
I remember coming across this equation myself about two years ago while I was trying to find the proof behind the quadratic formula. Really good to know there are other people out there interested in this stuff.
From 5:56 onward, there's no need to consider the two cases where b>=0 and b
Right. Also in that case the equation is not even quadratic when a = 0. What the hell are we even using a quadratic equation formula for in that scenario.
Then whole point of a quadratic equation is when p(x) = 0 where p(x) is a quadratic polynomial, generally of ge form ax²+bx+c where a,b,c are real numbers and a≠0. So there's no relevance when a = 0. Smh
@@pritamroy9320 Well, perhaps we might want to write a program to solve a family of quadratic equations as part of a mathematical model, where the leading coefficient could vary depending on some as yet unmeasured condition.
It would be good to know that the formula to write the algorithm we used didn't fail to return a solution if a happened to be 0 in some instance. The error would happen using the normal quadratic formula and we would get no solution with that algorithm. However, the alternate formula only throws an error on one branch (which can be caught and discarded), while the other branch would return the correct solution to the resulting linear equation.
Yeah this part really confounded me. The fact that the square root function always returns the posotive root in the reals is why we have that plusminus convention to begin with
@@RexxSchneider I understand. Thank you for explaining.
It's more an informatics than a mathematics topic. If you program it, you'll find indeed that the better results are coming from the mixed formula. Great topic for better understanding of the rounding errors with floating point numbers!
Awesome
Great learning
Thanx bro
Great Job Never really thought to play around with the quadratic equation like, did you find this out on your own or has this been documented before? either way keep up the good work
I love this! It's been over 40 years since last seeing such a careful examination of avoiding subtraction in a numerical calculation. This was one of our numerical methods assignments, but I don't recall anyone at the time noting that rationalization of the numerator could avoid the subtraction.
However, haven't you swapped the two cases in the last few minutes? That to be avoided is when the sign being used for sqrt(discriminant) is equal to that of (b); and hence opposite of that for (-b). If programming this, I think I'd simply always select the roots as:
x_1 = [ -b + sign(-b) * | sqrt(discriminant) | ] / 2a
x_2 = 2c / [ -b + sign(-b) * | sqrt(discriminant) | ]
your feelings are irrational
So, to phrase it another way, we want to avoid the cases where +/-b and sqrt(b^2 - 4ac) are both positive? So when b is negative, we ignore the -b - sqrt(b^2 - 4ac) solutions (as Dr. Barker did at the end) and when b is positive, we ignore the -b + sqrt(b^2 - 4ac) solutions.
So, to put it simply, the two equations we choose depends on the sign of b, which Dr. Barker failed to mention at the end of the video
@@violintegral Not really. I find Dr. Barker's description at the end quite sufficient; except I believe he's pointing at the wrong equation of each pair as he gives that summary.
@@violintegral No; other way around. We seek the case where the signs are the same, and a sum (meaning, moving twice in the same direction on the number line) rather than a difference (meaning, moving twice in opposite directions on the number line) is being performed between the two terms in numerator/denominator of the rational expression.
Very clever. Thanks for sharing.
Very well done!
The problems with the new formula seem to arise when (a) the starting equation isn't even a quadratic (a = 0) or when the quadratic has a common term in x that can be factored out and solved trivially (c = 0). In neither case would you reach for the original quadratic formula anyway, so the problem cases are irrelevant.
Also, if you take the original quadratic formula answer for 8 - sqrt(60) and plug that into a calculator, rather than approximating, you do get the right answer.
For the given example, the computer doesn't struggle. There's still enough floating point precision to give a good result. However, that isn't always the case. Sometimes the difference gets close to epsilon, the smallest number the computer can represent. Then results get whacky. But this is all an academic discussion. In the real world, computational libraries are doing these numeric checks to avoid garbage out or the end user does these checks if they're smart. For example, I might check if a ~ 0 before using the quadratic formula. If it is, I'll assume it's a linear equation with a = 0 and use x = -c/b
If you were writing a subroutine that returns values for x, and sometimes a and c are zero and sometimes nonzero, you don't want to have to write a separate subroutine to handle those cases and selectively call the other function.
Of course it gives right answer. That's not what the point was tho
@@derekschmidt5705 exactly
Comparing the accuracy of the two formulas, you calculated the first version to 1 decimal place, then calculated the second, wrote it down to 2 decimal places, and said "look, more accurate".
I have this gripe as well
The extra precision came without increasing the precision of the square root - the remaining digits in the first version were all zero. He also didn't say "look, more accurate", he identified that one solution was actually worse with the new formula, and spent several minutes explaining what was going on. The problem of subtracting two nearly equal numbers is a real issue in any kind of numerical computation, he just used 1dp so it would be easier to explain
Because the next decimal place was greater than zero. Still a good observation to consider. +1 for this one.
yes, this is exactly why teachers tell you to approximate only at the very last step
@@PixalonGC Indeed, though in this case using double precision (15-16 decimal digits) wouldn't have really helped Jane the point. The point is that for _any_ fixed initial precision, one formula gives a better result than the other.
It's Amazing...to see❤️❤️
Muy buen video, se razona todo paso a paso
Very nice work!
I only know about this because of my numerical methods class that featured it since sometimes it gives a more accurate result than the standard formula
Amazing content. 👍
It is smilar formula like Mullers method for finding approximated root of polynomial equation
It makes sense that the expression for the discriminant is unchanged: having an intrinsic geometric meaning ("The scheme cut out by the quadratic polynomial is non-reduced") it should be invariant under the projective transformation x-->1/x .
Indeed a novel perspective into the roots of the ubiquitous quadratic equation.
when dividing by x means we are assuming that x should not be zero right? so if the one of the root of the equation is zero can we still apply this formula?
Forman Acton in his book 'Real Computing Made Real' preserves accuracy by solving the quadratic as follows: Divide the standard formula through by a and then redefine the parameters as x^2+2b'x+c'=0. Note b' is not the same as b/a, but has a new definition. Then: x(+ -) = -b' + - sqrt(b'^2-c'). Compute the x for the case where the 2 terms have the same sign. Now the product of x(+) and x(-) = c' or x(+)*x(-)=c'. Use this equation to solve for the other x. Similar idea to the final proposed solution in the video (avoiding a subtraction).
Actually, there's a slightly better (IMO) derivation for this formula, which I absolutely adore, for it relies on an interesting "trick".
The derivation starts by completing the square, but instead of completing the square for ax²+bx you do so for c+bx. So multiply by 4c throughout and after adding a particular zero you get (2c+bx)²=(b²-4ac)x². Taking the square root one gets 2c+bx=±x√(b²-4ac), solving for x yields the formula.
I like it, because it represents a sort of "solving with respect to 1" approach and on a more general note that sometimes it makes sense to swap the role of your parameters (or constants for that matter) and variables.
The trick here is very interesting! Going from 4c² + 4bcx + 4acx² = 0 to (2c + bx)² - b²x² + 4acx² = 0 is not something I would have thought to do, as we have to ignore the 4acx² term when completing the square, but then we get a really clean derivation from there. Thank you for sharing this!
Thank you Sir!
I usually use Po-Shen Lo method to solve quadratic equation. Nevertheless, now I have more formula that I can teach to my students.
intersting video as usual
As a person who programs in physics for practical purposes, this is a constant problem. You want to avoid dividing by a quantity which can be zero or close to zero.... otherwise you need two formulae depending on that quantity. I also use automatic differentiation: this collapses completely if a=0.
The natural way, with this frame of mine, is to use the original formula and multiply top and bottom by -b -+sqrt(Delta)....
It would not occur to me to solve for 1/x.
Very cute alternative point of view.
11:00 Interestingly, given we have an indeterminate form 0/0, taking a limit as c approaches 0 then applying L’Hôpital’s Rule eventually does actually give us -b/a there. So it kind of does give us both solutions, it’s just involved and shaky when it comes to the one we’d be expecting to see.
Of course in practice it’d be much easier to just notice (or if programming, to code in a check for) c=0 and solve directly.
It is written in wikipedia.
Another alternative way of the quadratic formula which is quite elegant is the following:
If instead of starting with ax² + bx + c = 0 we instead start with Ax² + 2Bx + C = 0 (where A = a, B = b/2 and C = c), we can then express the roots with a simpler equation:
x = ( -B ± √(B² - AC) ) / A
I find it interesting because when you do (x + N)² you get x² + 2Nx + N², so the 2 in the middle seems to naturally appear when dealing with quadratics, and to naturally fit in the quadratic formula.
Quadratic equation "a" should not be zero. If a is zero it will become linear
When b^2 = 4ac your formula will give one solution as undefined whereas the original one gives correct
This is a more general, and applicable quadratic equation. Interesting.
Hello, why is it still a quadratic if a = 0? Wouldn’t the highest degree then be just x^1 making the expression a binomial? What am I missing?
Just seeing the thumbnail, this second quadratic formula almost looks like the reciprocal of the first. Almost.
Edit: About the segment when a=0. Can't we just use the formula for a linear equation?
Is there any benefit to not normalizing the equation to begin with by dividing by a?
I was pretty shocked when I found out that this cumbersome formula is taught anywhere at all like this.
When there is no a term then infinity is also a root of the eqn this is highly useful in coordination geometry
The famous alumroF citardauQ!
so now i wonder if there is some sort of regularised formula that works for all the exceptional cases.
Ok, nice. But you could argue that x²=c/a
The error is that the derivation of the new formula consider that ± becomes -+
The absolute best version of the quadratic formula though is when you divide both sides by a, and change b into 2b and c into c^2. Then what you get is that for the quadratic equation x²+2bx+c² = 0, the solutions are x = -b ± √(b² - c²), which is nice because you have a difference of two squares inside of the square root, and that actually allows you to express the formula in a yet simpler way: let b+c = p and b-c = q. The quadratic formula becomes x = -(p+q)/2 ± √pq, which is kind of cool because it expresses the roots in terms of the arithmetic and geometric means (of the same two numbers) added together
I have been taught in my junior high school:
x_1 + x_2 = -b/a ...(1)
x_1*x_2 = ac ...(2)
By the 2nd formula, you get the formula of 1/x.
Is it, therefore, x = c/a ?
or:
x1.x2 = c/a
x1,x2 = [-b±√(b^2-4ac)]/(2a)
x2,x1 = c/{[-b±√(b^2-4ac)]/(2a)}
= 2c/[-b±√(b^2-4ac)]
for accuracy:
x1 = [-b + √(b^2 - 4ac)]/(2a), b < 0 or [-b - √(b^2 - 4ac)]/(2a), b >= 0
x2 = c/a/x1 or x2 = -b /a - x1
Very neat! I like the simplicity of just using c/x_1 or -b - x_1 to avoid having to do the whole calculation over again.
oh nice. didn’t realise we can apply either sum of roots or product of roots here
@@ethancheung1676 Thanks Ethan. Slide rules and log tables were in use when I was at school.
@@rob876 i also learned about log tables. didnt use a slide ruler but neither a calculator
Some minor corrections...
You did not make any assumptions on the value of a. But let's assume that a ≠ 0 (otherwise we do not actually have a quadratic equation).
Then from elementary high school mathematics we know that x1+x2 = -b/a and that x1·x2 = c/a.
Hence, in your reply, in the second and last line, the values of b and c must be replaced by b/a and c/a respectively.
If you plot both graphs you'll see that both formula are well integrated together
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X/y is asking which number times y=X. So 0/0 is asking which number times 0=0.
Just Found this formula around 2 years ago. When I first got it I thought I was a genius and going to become a mathematician or something 😂 then I realised it's already been discovered long ago but simply nobody teachers it
This is the only version of the quadrant formula I use in software engineering. It’s superior to the other version because “a” can be zero, which is the case when computing the trajectory of a projectile in zero gravity.
It seems a bit moot to be exploring the case when a = 0, given that, by definition, a ≠ 0 for ax² + bx + c to be a quadratic
Need c/=0 for this version.
My comment is that when a=0, we don't have a quadratic. We have a linear equation instead.
Sure when a is a constant, fixed value. But what about when we have a family of quadratic equations where the value of a can change? It would be nice to be able to program an algorithm to give us a usable solution even in the case when a becomes 0, and the alternative formula allows us to do that (we just have to catch the error when one of the roots is undefined).
@@RexxSchneider l don't know that the value of "a" changes in a quadratic because it is always a constant. If it is 0 (zero), then then it is not a quadratic.
@@EE-Spectrum Look, I quite clearly said a _family of quadratics._ If you need some examples to understand it, try these:
3x^2 + 5x - 2 = 0
2x^2 + 5x - 2 = 0
x^2 + 5x - 2 = 0
See how the value of the constant a is different ("changes") between the set of equations?
So what might the next equation be in that family? This might be a mathematical model of some process where the leading coefficient can take on different values. Maybe someone might want to write a program to solve the equations in this family? They would have to use a quadratic formula.
The point I'm making is that if you use the usual formula, it will generate an error when a takes on the value zero and you'll get no solutions. Whereas if you use the alternate formula, only one branch produces a error and the other gives the correct solution to the resulting linear equation. See the difference now?
Isn't it the case that sqrt(b^2) is always equal to |b|, since b^2 is always positive, and the square root of a positive number is always the principle root, which is also positive?
Yes, but if b is negative, |b|=-b because it switches sign (-b is positive btw)
@@julianbruns7459 Good point. I guess I had a brain hiccup.
@@julianbruns7459 But surely √(b^2) is conventionally taken to be positive, therefore √(b^2) is *always* equal to |b|, no matter what sign b has. You'll find that @ib8rt is right.
The problem is due to the casual approach of the so called western modern mathematicians or mathematics towards using decimal places perfectly incorrectly.Whereas in Ancient indian mathematics or rather newly unearthed form of the archaic style of mathematics called Vedic maths this will not arise.You guys can check for Yourself.Interestingly the method to find tge solution to quadratic equation was first formulated by an ancient indian mathematician .If i deserve to remember correctly..it is Brahmagupta who did this first.It is even known very well that tge concept of Zero,Decimal places were first used by Indian mathematicians.
The vedic maths people are some of the most obnoxious
You have to assume that x = 0 is not a solution to the equation as this invalidates the derivation of the alternative formula, where it is necessary to either divide by x^2, or use the term 1/x, both undefined when x = 0
When x = 0 is a solution, c is also zero. In that case you can anyways save a lot of computation by checking for the special cases of any of the of the coefficients being zero:
- a = 0: x = -c/b
- b = 0: x = ±√(-c/a)
- c = 0: x = 0 or x = -b/a
Checking for c = 0 as a special case is especially advantageous, as it avoids performing an entirely unnecessary square root operation.
You can clear this case by checking for c = 0. That has the additional advantage that you can get the other root simply as x = -b/a without the unnecessary square root operation.
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I am genuinely confused by you saying that Sqrt(b^2) is positive for positive b and negative for negative b. Is this some form of convention to inject analytical continuation consistency into the sqrt function? This is the first time ever I see this. To the best of my knowledge, b^2 is a positive number all the times , regardless if b is positive or negative. In a way, squaring b removes all kinds of "memory" of the sign of b. The sqrt function acts on b^2, which is always positive regardless of the sign of b, and has no way of "bypassing" the power function and somehow figure out what the original sign of b was. Think of it in an algorithmic way, as if you were to implement this in a computer: unless you endow the power function with a memory kernel that retains the information of the original sign of b, b^2 will always be positive and the very same value regardless of the original sign of b. If that's not the case, I am honestly questioning everything I know about math...
Edit: in fact, the sqrt(x^2) function was one of the quick and dirty way I would implement the Abs function in some of my early programming. It is also the way things like standard deviation or chi square are calculated to ensure having a positive sum regardless of the original sign of the error
You don't have to worry about the a=0 or c=0 conditions because the formula is never used in those cases.
But we may not know initially when plugging in the values for a and c that they are zero or not, they could be parameters too
by hand sure, but by a computer it can be
@@will-jh7yi Well, that's the goal here, then, isn't it? To know by doing it yourself! That's what's so FUN about it!
First, solve the problem. Then write the code.
- John Johnson
"... because the formula is never used in those cases."
Give it time. lol
It's another tool in your toolkit.
Nice
I much prefer Vieta’s formulas for deriving the quadratic because it’s much more interesting and easy to understand IMO
If a = 0, would there not be an x^2? Which would mean you no longer have a quadratic. Therefore, “the new quadratic formula” is not actually a quadratic formula at all. Without an X^2 variable, you do not have a quadratic equation.
What he wanted to show is that if you ever want to use the usual quadratic formula when a = 0, you can't, because there is a 2a in the denominator. The "new" formula can still provide you the single solution the equation has, because a = 0 won't be a problem.
@@lucasfelipe8147,
think we've just to catch what is actually the question in this .
how can we notice that coefficient
a is equal to zero ?
try to understand that it's hardly possible to say, „well a=0 ” , when no term with x² is to be seen . which should be the case to be allowed to ascertain the zero value of a .
so meeting qx + r = 0 ,
how would we come upon proposing
well, we notice that p = 0 here , so let's take yet the quadratic formula .
.
@@keescanalfp5143 It's not that hard to notice. If it's a first degree polynomial, we can suppose a = 0 from ax² + bx + c = 0 because it will give us bx + c = 0, which is indeed a first degree polynomial. The fact is that we can say a lot of things are 0. For example: what if we wanted to use the cubic formula? So let a and b equal to 0 in a polynomial of the form ax³ + bx² + cx + d = 0.
I have discovered the same formula during class 9 .I also thought that why nobody teach it .
А зачем? При таком подходе у нас с≠0, т.е. формула не универсальна, к тому же, если дискриминант отрицательный, то появляется комплексность в знаменатели, которую довольно тяжело убрать.
Еслибы я делил 2 x на бесконечность повторений в прогрессии, то куда бы делся этот умник????
I just use poh sen lau one, been many years now, seems fast enough, though i also knew this from wiki😅
Any number times 0=0. So 0/0 has no wrong answer.
My physics sir had already taught this formula to us
The way he makes an x. c and backwards c. I've never seen it that way.
(12:05) 'just' say _2sqrt(15)_ or next time prepare a different equation
In Germany we use the pq equation
sqrt(b^2) = |b| so -b is 2c/(b(+-)b)
The second formula is not valid for c=0 where as first formula is famous for all quadratic eqns
9 standard questions
Signifigant digits? Ignore them at your own risk.
🙄
But 0/0 is an indeterminate form, so couldn't you use something like L'Hôpital's rule to see if we can recover the -b/a solution?
l'Hopital is done with derivatives
Yeah you could try taking a limit instead of plugging in directly
@@violintegral take the limit with respect to what? l'Hopital is about functions.
Assume that b ≥ 0, so sqrt(b^2) = b. Then, one could find the limit of 2c/(-b+sqrt(b^2-4ac)) as c->0. The derivative of -b+sqrt(b^2-4ac) with respect to c is -2a/sqrt(b^2-4ac) (using the chain rule and the fact that the derivative of sqrt(x) is 1/(2sqrt(x))). So, using L'Hopital's rule, one gets the limit of 2/(-2a/sqrt(b^2-4ac)) as c->0. Simplifying, one gets the limit of 2sqrt(b^2-4ac)/(-2a) (multiplying by the reciprocal) = -sqrt(b^2-4ac)/a (dividing the numerator and denominator by the common factor of -2) as c->0. That limit (evaluating at c = 0) then simplifies to -sqrt(b^2)/a = -b/a, as expected. So, L'Hopital's rule does in fact lead to the correct solution.
2c/0 works fine.
Must be something wrong with your definition.
Just doing surds
We are not taught to do it that way because its not necessary. Sheer swaggering.
√(x²) = |x|, for every real x value.
not to demean the observation, but it kind of feels like we learn nothing from it, it could be used while explaining quadratic equations to a class for the first time, because well all the observations or results are kinda obvious
When a = 0, it is no more quadratic equation! Then it makes no sense to use quadratic formula!
@05:52-07:40
When considering the cases "b>=0" and "b
No, its still -b.
@@julianbruns7459
Why?
🍌🤔
Once you realise that √(b^2) is |b| (and not simply "b") by convention, you'll see that you don't need two cases, and the algebra is actually identical , no matter what sign b has.
@@RexxSchneider
@7:37 I am referencing the "-b" that appears as the first variable in the denominator of "(2c)/(-b ± (b²)^½)".
In the two cases of "b>=0" and "b
@@HeinrichDixon The alternative formula has "-b" as the first term in the denominator -- it's a "term", not a "variable". That is algebraically derived to be "-b" regardless of the actual value of b. There is no reason whatsoever why the first term of the denominator should be anything other than "-b".
You don't need two cases because that term is "-b" whether b is positive, negative or zero. Nevertheless, if you insist on considering two cases, when the value of "b" is positive, the value of "-b" is negative, and when the value of "b" is negative, the value of "-b" is of course positive. That doesn't mean we should be altering the sign of the "-b" term.
That's how the term "-b" works: it means the negative of the value of "b" . The "-" is an operation that negates the value of the following expression or variable. It is *not* an "indicator" of the sign of the following variable.
If a=0, then we have a much simpler linear equation and not a quadratic.
Fun fact: This quadratic formula isn't applicable when c = 0.
Edit: nvm, it would only made one of the root DNE while having another solution equals to x = -b/(2a), which is correct.
Second Edit: Oh wait, it shouldn't be only one root when c = 0, instead there is x = 0 as a solution for the quadratic equation where c = 0.
Third Edit: I commented before watching the entire video lol, Dr Barker has already pointed the problem out in the formula.
Noob overrated comment no one cares
Wow this is absolutely useless... AMAZING! Actually this is beautiful
But this implies that x isn’t 0, which isn’t always the case.
What if a b and c all equal 0?
0/0=0
you wouldn't be able to use the quadratic equation if a was equal to zero anyway (the quadratic formula only applies when a ≠ 0)
Really interesting video. I guess in the electronic age the approximation issues are less important. I am of the age where I used logarithm tables! The alternate method would have been relevant then!
@@lox7182 And this alternate framing does when c *and* -b+/-sqrt(b²-4ac) are both 0
Then you're trying to solve for the value of x that makes 0 + 0 + 0 = 0. Good luck with that.
Or you can use a calculator.
I thought "-b" should be read as "opposite of b", and not "negative b", since "b" could be negative...
Saying negative b still works, even if b is negative. If b = -3, then "negative b" is -b = -(-3) = 3. This is just semantics anyway
@@lumina_ Negative negative 3 is not the proper way to go about saying the number three. It isn't proper to say negative b. Semantics or not.