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.

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.

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))

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

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)

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

​@@wirmaple7336They were talking about the old times

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

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

Just need 2X speed...

@@mndtr0 bprp is on crack and meth at the same time

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

Wow brilliant 🥇🏆👏👏

@@Shakti258 Thank you🙏🙏🙏

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

@@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.

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.

Are you vieta?​@@quandarkumtanglehairs4743

Yup exactly what is used for cubics

@@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

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!

@@jarrodfrench957 you're welcome!

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.

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

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!

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.

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.

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

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

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.

​@@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 . .

@@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.

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.

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!

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

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

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

@@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.

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.

@@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.

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 .

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.

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?

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

I have a marvellous proof but my mirror does not have wide enough margins😊

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.

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.