The Quadratic Formula No One Taught You

[@KarlFredrik]

Crazy how I never thought of this during my 35 years of solving 2nd degree equations.

[@f5673-t1h]

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?

[@HoSza1]

Because of numerical stability? Have you seen the video til the very end?

[@bjornfeuerbacher5514]

@@f5673-t1h 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.

[@theimmux3034]

​@@f5673-t1h why would you ever use the quadratic formula to solve for the solutions if one of them is x = 0?

[@NidalDajani]

Ditto.

[@frobozz55]

IMHO, I would keep the "minus over plus" in the alternate formula.

[@RexxSchneider]

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.

[@hiyayahiyaya5645]

in fact , it isn't something new , we have the formula of product of roots , αβ=c/a let α=(-b±√(b²-4ac))/(2a) we can get β=2c/(-b±√(b²-4ac))

[@anonymoususer2756]

Yes, because if you multiply them together you get x^2 = c/a, when it should be x(1)x(2) = c/a.

[@abve2517]

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.

[@wirmaple7336]

As a CS student, you can use a "decimal" (128-bit) or even an arbitrary-precision data type and there will be no rounding errors. However, doubles still have rounding errors (especially with very large numbers)

[@leif1075]

This is basically the same formula..is there any advantage to it?

[@rachit7645]

​@@wirmaple7336They were talking about the old times

[@undercoveragent9889]

@@leif1075 It avoids the 'divide by zero' problem when a=0.

[@edg42]

​@@undercoveragent9889The formulas no longer applies for bx+c=0.

[@shafiandpritha7701]

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.

[@SaidVSMath]

This is genuinely so freaking cool. Thank you for sharing, great work as always Dr Barker!

[@NoNameAtAll2]

slowly turning into blackpenredpen, I see :)

[@mndtr0]

Just need 2X speed...

[@bucc5207]

I was in my thirties when I first encountered this version of the quadratic formula, in a wonderful book called Numerical Recipes. For computational accuracy, you want to choose the radical term with the largest absolute value. So define S = -b + sqrt(b*b-4ac) when b0. Then the two solutions are S/2a and 2c/S.

[@jarrodfrench957]

This comment really helped me understand that you don't necessarily want to use only one of the formulas but using both to help with accuracy! Thanks for posting this!

[@bucc5207]

@@jarrodfrench957 you're welcome!

[@gametimewitharyan6665]

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

[@poetry_pulse2705]

I myself discovered this 4 years ago (while I was in my 10th standard) and showed it to my teachers. But no one gave proper attention. 😞😞😞

[@Shakti258]

Wow brilliant 🥇🏆👏👏

[@poetry_pulse2705]

@@Shakti258 Thank you🙏🙏🙏

[@PluetoeInc.]

interesting indeed , how did you come across this thought ? like what were the previous bogies in the train of thought that led to this .

[@poetry_pulse2705]

@@PluetoeInc. I used concepts of sum and products of roots of a quadratic equation and the original qudratic formula. After some calculations I arrived at that new formula.

[@Μοχντ_Αλί]

I actually came across this when i was in 12th standard i told wy friends and teachers but nobody gave attention to it as well

[@RexxSchneider]

From 5:56 onward, there's no need to consider the two cases where b>=0 and b

[@pritamroy9320]

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

[@RexxSchneider]

@@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.

[@typo691]

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

[@pritamroy9320]

@@RexxSchneider I understand. Thank you for explaining.

[@Bhuvan_MS]

Yes, I was actually confused as to why he took separate cases for values of b.

[@RCHobbyist463]

Having taken a Numerics class, I appreciate how well this video presents the challenges of trying to get accurate numbers out of a computer.

[@mrphlip]

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.

[@quandarkumtanglehairs4743]

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.

[@piedepew]

Are you vieta?​@@quandarkumtanglehairs4743

[@pauselab5569]

Yup exactly what is used for cubics

[@memestrous]

Ohhh that's where the product of the values comes from

[@davidellis1929]

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.

[@frobozz55]

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.

[@xyannail4678]

​@@frobozz55 Also 2.

[@frobozz55]

@@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".

[@xyannail4678]

@@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?

[@frobozz55]

@@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.

[@philippetrov4881]

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!

[@gametimewitharyan6665]

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?

[@nasdfigol]

0 + 0 + 0 = 0

[@matheusjahnke8643]

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

[@pojuantsalo3475]

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).

[@wyattstevens8574]

@@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).

[@keescanalfp5143]

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 .

[@wyattstevens8574]

@@keescanalfp5143 That's *almost* the same as Po Shen Loh's method- he divides out the factor of a so now a=1 anyway!

[@karolakkolo123]

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

[@pietergeerkens6324]

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) | ]

[@Fire_Axus]

your feelings are irrational

[@violintegral]

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.

[@violintegral]

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

[@pietergeerkens6324]

@@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.

[@pietergeerkens6324]

@@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.

[@khelwood]

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".

[@karolissad.4270]

I have this gripe as well

[@mikeflowerdew7877]

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

[@quandarkumtanglehairs4743]

Because the next decimal place was greater than zero. Still a good observation to consider. +1 for this one.

[@PixalonGC]

yes, this is exactly why teachers tell you to approximate only at the very last step

[@mikeflowerdew7877]

@@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.

[@skilz8098]

Nice job on presenting where each version has its pros and cons due to proportional rounding errors when the precision of floating points numbers is involved. Perhaps extend this topic towards complex analysis where we might try to find the arccos of the angle generated by the dot products of the various roots within a given quadratic even if the roots are complex values... now that would make for an interesting math video!

[@christophernguyen1750]

At 7:07, I’m confused. If b was negative then substituting a -b into the new equation would get 2c/-(-b)+-sqrt(-b)^2 which simplifies to 2c/b+-b which gives two solutions c/b and 2c/0 which is not the same as the 2 solutions if b was positive since it’s not -c/b

[@beatn2473]

We looked at the accuracy issue to introduce Vieta's formulas. Quite nice!

[@Aurochs330]

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?

[@eilonpoem1587]

For both formlas you can get all the solutions, even those where the denominator is zero. You do this by taking the limit a-->0 (for the original formula) or c-->0 (for the new one) and use the taylor expanaion of the square root (or, equivalently, use the L'hospital rule, as in both cases both the denominator and numerator tend to zero).

[@kennethgee2004]

and for good reason. We rationalize the denominator so that the calculation is straight forward. take the simple case of 1/sqrt(2) VS sqrt(2)/2: to work that out using long division the first one is not doable. whereas the second is irrational but if we are willing to cutoff the calculation at some point we can do that calculation.

[@neologicalgamer3437]

Multiply the first formula by the second formula. You get the big b term to cancel our, and you're left with 2c/2a, which ends up being x^2 = c/a, which finally ends with x = ±√(c/a)

[@snowballh1969]

Great observation! However, your final conclusion is unfortunately incorrect, because just like another comment said, the "+-" in the second formula should really be "-+". When both "+-" are the same sign you actually get two different roots using the two formulae. Therefore multiplying the two would give x1 * x2 = c/a, confirming part of the Vieta's formulae that states the product of the two roots in a quadratic is c/a. Interesting idea to multiply them though!

[@Ensign_Cthulhu]

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.

[@major__kong]

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

[@derekschmidt5705]

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.

[@TurdBoi666]

Of course it gives right answer. That's not what the point was tho

[@TurdBoi666]

​@@derekschmidt5705 exactly

[@MasterHigure]

Derivation 3: Vieta's formulas say the product of the two solutions is c/a. Insert the standard formula for one of the solutions (pick either + or - for convenience), and solve for the other.

[@josemieles8764]

A quadratic equation is a second order equation written as ax2 + bx + c = 0 where a, b, and c are coefficients of real numbers and a ≠ 0.

[@AruberutoCh]

It is not second order, it is second degree. The order of the differential equation is the highest order of derivative in the function.

[@schmetterlingsjaeger]

11:00 In the c=0 and 0/0 case we still can get the solution -b/a by applying de'Hospital rule on the new formula for c->0.

[@vladislavanikin3398]

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.

[@DrBarker]

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!

[@Bhuvan_MS]

This derivation is probably more rigorous because it avoids dividing by x² and assuming that x≠0.

[@tunneloflight]

Dividing by a number with 1 significant digit gives a result with 1 significant digit. i.e. ~ 10 to ~20. 13.3 and 15.7 expressed to one significant digit are identical ~10 to 20.

[@keescanalfp5143]

could be that the doctor doesn't know very well the meaning of the concept of significant digits . so many people just count the number of digits after the decimal dot , in our continental europe the comma .

[@algorithminc.8850]

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 ...

[@david_porthouse]

Many real world quadratic equations are only slightly quadratic and we are only interested in one of the roots, the one closest to -c/b. This is when this approach comes into its own. There’s an example concerning the blockage in a wind tunnel in my Thesis.

[@marcgriselhubert3915]

We have x1 = (-b -sqrt(delta))/2.a and x2 = (-b +sqrt(delta))/2.a (for a0, c0 to have x1 and x2 0, and naturally delta = b^2 -4.a.c >= 0). As x1.x2 = c/a, we have x1 = c/(a.x2) = c/((-b +sqrt(delta))/2) = 2.c/(-b+sqrt(delta), and same for x2, x2 = 2.c/(-b - -sqrt(delta)) These formulas are scarcely taught probably because the ordinary formulas are sufficient for students and may be because it is not a good idea to put "a priori" a radical at the denominator.

[@johnmorey2633]

First time I have ever seen this. Really interesting. Thank you.

[@matteoallegro5491]

The concept explained at around 9:00 could be very useful when solving a parametric equation with a (or c or both) depending on the parameter. I'll keep this in mind!!!!

[@senthilsoundara7836]

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

[@robertbachman9521]

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).

[@Ansam__]

The final part is very interesting ! I was thinking u were going to use the taylor series for the sqrt to have a better approximation, but you backsided the problem with the other formula

[@3snoW_]

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.

[@anonymoususer2756]

There’s a symmetry between the two formulae for when a = 0 and c = 0. When a = 0: Old formula gives 1/0 (undefined) and 0/0 (indeterminate) New formula gives 1/0 (undefined) and -c/b (solution) When c = 0: New formula gives 0 (solution) and 0/0 (indeterminate) Old formula gives 0 (solution) and -b/a (solution)

[@andrewm6424]

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.

[@Sanguinium_Light]

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.

[@keescanalfp5143]

​@@Sanguinium_Light, 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 . .

[@Sanguinium_Light]

@@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.

[@krozjr5009]

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.

[@alex_ramjiawan]

Can i just ask, why did he keep rounding 7.75 to 7.7?

[@Punklorde_Mentality]

I think the 2nd case at 6:35 is wrong because the square root of anu number should always be positive. Yes, x^2=4 has 2 solutions 2 and -2 but if we take square root on both sides, we have to add a +- sign that is +-x=2 which gives us the desired solution. The plus minus is there to handle the case that b is negative in the first place so it's unneccessary(wrong) to take the 2nd case. But apart from that, great video, learnt new concepts about something I'm familiar with😁😁

[@jceepf]

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.

[@freddiepage6162]

This also shows if you know one root x+, then the other is x- = c/(a x+)

[@karolakkolo123]

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

[@ToothLin]

A negative number squared is positive. ((-b)^2)^(1/2) is b not -b. ((-b)^2)^(1/2) =/= (-b)^(1) ((-b)^2)^(1/2) = abs((-b)^(1))

[@ToothLin]

From 7:02

[@slippinchillin]

Is the -b - sqrt(b^2 - 4ac) argument general? Or does it apply only to this particular case and thus for each cases, we have to determine which one is close to zero to apply the appropriate formula?

[@samarthasamartha4778]

It is smilar formula like Mullers method for finding approximated root of polynomial equation

[@alphalunamare]

So if b^2 = 4ac+1 or 4ac then and only then is the same answer is given by both formulations?

[@starrysky5190]

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.

[@avmathtech6162]

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?

[@drowzeerutherford6037]

I discovered this version on my own.

[@johnspeno8163]

Great treatment of the approximations. Barker yer the man!!

[@shauas4224]

From computational point of view, I guess you could use this to find inverse of roots without having to do division so its kinda cool. Never thought about this

[@nuranichandra2177]

Indeed a novel perspective into the roots of the ubiquitous quadratic equation.

[@chayansarma4443]

If b=c=0 how can we get solutions using this form Also if c=0 there is a solution x= -b/a which we will not get from this form.

[@Zersetzor]

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.

[@MonzurRahman-eb2vv]

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

[@trueriver1950]

It comes into it's own when a is close to zero but it's still not negligible. For example a ballistic trajectory in a very low gravity environment, or with a very fast projectile

[@MrSummitville]

​@@trueriver1950He started by explicitly saying, "... when a *is* equal to 0 ...'". So, right there, that is the end of the discussion ... The formula is *not* a quadratic equation.

[@zacharyyebuah4673]

why didn’t u change sqrt (b^2) to abs (b) and let b always be positive 10:26

[@rv706]

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 .

[@trytea]

I'm a little confused. Since when is squaring a negative number still remains negative? Sqrt((-b)^2)) should result in b. Sure the end result is the same but it's just the principle.

[@Orion_Fritz]

Yeah, it was rather confusing, especially since sqrt(x²) is the literal definition of the absolute value function |x|.

[@marcoottina654]

Is it, therefore, x = c/a ?

[@TamissonReis]

Ok, nice. But you could argue that x²=c/a The error is that the derivation of the new formula consider that ± becomes -+

[@Bhuvan_MS]

It is more appropriate to mention that (x1)(x2)=c/a and not x²=c/a This is one of Vieta's formulae for sum and product of roots.

[@AndrejPanjkov]

so now i wonder if there is some sort of regularised formula that works for all the exceptional cases.

[@_SikanderWrites]

Bro can we compare both formulas and substitute -b±√b²-4ac as some variable and get x=±√c/a , but it is not correct idk i am just 12

[@rob876]

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

[@DrBarker]

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.

[@ethancheung1676]

oh nice. didn’t realise we can apply either sum of roots or product of roots here

[@rob876]

@@ethancheung1676 Thanks Ethan. Slide rules and log tables were in use when I was at school.

[@ethancheung1676]

@@rob876 i also learned about log tables. didnt use a slide ruler but neither a calculator

[@DutchMathematician]

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.

[@kmyc89]

(12:05) 'just' say _2sqrt(15)_ or next time prepare a different equation

[@soyoltoi]

It's kind of interesting how the signs flip, giving you the second solution where the normal quadratic formula gave you the first. I wonder if this relates to some kind of symmetry, maybe the fact 1/x is an involution?

[@sheikhAbdelrahman]

When a = 0, it is no more quadratic equation! Then it makes no sense to use quadratic formula!

[@ApresSavant]

I agree. But I suppose this is a useful thought exercise for a programmer where one would sweep values through zero and you would not need special logic to detect that.

[@SummerFrost23]

It is written in wikipedia.

[@ccmplayer87]

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.

[@randomrandomizer]

If a is zero it’s not a quadratic any more and the solution is staring you in the face, so you don’t need a formuka??

[@nathanbarnes3969]

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

[@SuperRousku]

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.

[@SuperRousku]

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.

[@cxpKSip]

This is a more general, and applicable quadratic equation. Interesting.

[@Lordmewtwo151]

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?

[@TGRRohit]

When there is no a term then infinity is also a root of the eqn this is highly useful in coordination geometry

[@thidasvinnath8017]

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

[@narayansareekuthir5330]

I found it when i was in 6 th class i have proof also i shared it with my teacher i proved it also but my teacher said me to rationalize

[@Grassmpl]

Need c/=0 for this version.

[@AdityaKantKushwaha]

Shreedharacharya Formula❤

[@alberttatlock1541]

Why is a=0 a special case quadratic equation? Surely if a=0 its not a quadratic, its a linear equation.

[@innovationsanonymous8841]

nice. since you showed me yours, I'll show you mine: (-b +- sqrt(b**2 + (y-c)4a))/2a combining the two should work (haven't checked, but x=x): 2c/(-b +- sqrt(b**2 + (y-c)4a))

[@afsarrashid2967]

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[@chibimentor]

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[@ib9rt]

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?

[@julianbruns7459]

Yes, but if b is negative, |b|=-b because it switches sign (-b is positive btw)

[@ib9rt]

@@julianbruns7459 Good point. I guess I had a brain hiccup.

[@RexxSchneider]

@@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.

[@byronwatkins2565]

If a=0, then we have a much simpler linear equation and not a quadratic.

[@alexmeanin8049]

To have radical in the bottom is BOLD solution, no doubts. Reason, that noone hear about this, is obvious. 😅😅😅😅😅 I can bring a NUMBER of such stuff, no problemo.

[@gamingnorm9369]

If you plot both graphs you'll see that both formula are well integrated together

[@viveksaxena8430]

Awesome Great learning Thanx bro

[@usernameisamyth]

another good work

[@MichaelMaths_]

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

[@trueriver1950]

Intuitively obvious, if you reflect for a moment on the equation ax² + bxy + cy² = 0 which is to be solved for the ratio of x/y. Clearly x/y is found by the traditional formula. By symmetry, y/x must be found by the inverse after swapping a and c. That's the formula given here. Setting y=1 we get the usual quadratic formula for x. Setting x=1 we get the "new" formula. As given, this is just a plausibility argument, but it can easily be made rigorous.

[@chrisikaris5891]

Very nice work!

[@anujrajput7649]

Amazing love from❤❤❤❤ indian

[@sdspivey]

If A=0, then it isn't a quadratic, so no one would use the quadratic formula.

[@mb59621]

Some problems don't tell you that a=0 , intentionally write the x² term . So its better IMO .

[@xdragon2k]

Since it isn't a quadratic formula but instead just a linear one, it only have ONE valid solution.

[@apan3324]

@@mb59621 Bro if they intentionally write the x² term then the coefficient would automatically be one? Then you wouldn't need to use the 2nd one anyway. When a is set to 0, the formula changes into basically just solving basic algebra, where you do inverse operations and try to have x by itself. I mean if you don't even remember how to do basic algebra but you remember the odd 2nd version of the quadratic equation that no one really talks about, then ig you can say that you were glad you remembered it.

[@apan3324]

The 2nd quadratic formula can't even solve it when C is set to 0, WHICH IS STILL A QUADRATIC EQUATION. It literally fails to do it's job of solving quadratic equations, whereas the more recognizable formula can do it.

[@mb59621]

@@apan3324 well the more competitive mathematics you solve , people use sophisticated terminology such as just polynomial equation etc . Them using the word quadranic yields a weakness in the question. To answer your opinion , they can always have like an ax²+bx+c .. where they don't tell you anything about a . Many problems will have an entire function as a coefficient of a term and you have to check whether there is a maxima/minima of some sorts for that coefficient lets say vaguely . You will encounter plenty of such problems in calculus with dead ends but then it is useful if you have a special weapon like such in inventory.

[@meccamiles7816]

Very clever. Thanks for sharing.

[@N00byEdge]

why don't you look the alternative formula where you have to divide the entire polynomial by a first? then you don't have to worry about the pesky a = 0 either. In my opinion, it's also easier to remember. x^2 + bx + c = 0 x = -b/2 +- sqrt((b/2)^2 - c)