Відео

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)

Inverted quadratic formula

This is an incredible video. Where do you get your inspiration?

Why would anyone care if the quadratic formula works for an equation that isn't a quadratic. Oh no! Neither form works for a cubic! SMH!

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))

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).

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.

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?

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

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 b<0, and S = -b - sqrt(b*b-4ac) when b>0. Then the two solutions are S/2a and 2c/S.

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.

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. 😞😞😞

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

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??

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.

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)

Interesting question, lol...have to write this algebraically, etc, in equations, but basically you have to make sure that they are within two integers, etc, like if you draw a line at each integers, forming a grid of horizontal lines, you have to make sure they are within two grids, one (x^2 -x or something) above the middle line (so it's floor is on the middle line), the other (1-x^2) below the middle line (so its ceiling is on the middle line), but above the bottom one (so it's ceiling isn't there)...should be easy to write in terms of algebra, but I can't do that in my head, lol, not very good with my head and I'm in a hurry...

interesting, since this gives the correct answer at a=0 unlike the standard form

Thx for another good friday-math.

can this be solved mathematically, without drawing the functions or is graph required

Hmmm, you say "smaller" for "closer to negative infinity"... I try to use larger/smaller for distance or proximity to 0, and higher/lower for the proximity to positive or negative infinity, as I felt that's less confusing. But you're a far better maths educator than me. Am I wrong? Should -1000000 be considered a smaller number than -1? 🤔

1/2 - sqrt(5)/2 = 1 - ϕ 😡😜

omg! Two of my fav irrational numbers in one video!

Fantastic solution! I really appreciate how systematically you approached the problem, making it easy to follow and understand. Your channel is truly a gem for anyone looking to improve their math skills, and your way makes learning enjoyable. Please keep uploading such excellent videos! Your dedication and clear explanations are incredibly valuable. Apart from the maths, I like your English accent, it is amazing. Like from Saudi Arabia 🇸🇦

Very nice explanation! I was able to make the graph, but when it came to plotting the points I got lazy and started looking for an alternative method :P

thanks Mr Barker! Love your visualisations, was confused after the lecture just now and found your videos first

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.

I am not using that. I am gonna just guess. That formula is just too big.

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)

I wish I knew this sooner

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))

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?

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

Could a line exist in the complex plane? Yes. Suppose we choose pairs of complex numbers. We could treat every $(x_{r}+x_{i}\sqrt{-1}, y_{r}+y_{i}\sqrt{-1})$ as points in a pair of Cartesian planes, with a real plane, and an imaginary plane. Simple as coming up with two lines in both planes, then merging the results.

Linear algebra come save us 😭

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

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.

Для большего понимания куда полезней вывести флрмулу для корней, тогда не будет надобности заниматься бесполезными выкладками.

Bro what if x=0 then your method of proving can not be justified

Is there any interesting insight to be gained from thinking in terms of the symmetric matrix of coefficients of the terms? This algorithm you presented certainly seems to work well enough. I am going to bet a beer the eigenvalues have something to do with whether the function has a min or max or saddle.

Great video! You've solved my problem in a very elegant way! Love the w notation!

could this question be solved via the AM-GM inequality as well?

Shreedharacharya Formula❤

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.

I thought it was easier using calculus. Partially differentiate with respect to x, then y, then z, giving three equations in three unknowns. Took about five minutes.

Innovative solution! How can we generalise this for any coefficients and if any yz, x, y, or constant terms are present?

There is another way to solve this, which is perhaps a bit faster, as long as 2n is a positive integer. If it is not, the meaning of the 2n th power is in any case not uniquely defined, since ((1)^{2n} is then already non-uniquely defined. For positive integer 2n, just split the matrix into the sum: {{n,1},(0, -2}}= D + N of a diagonal part D= {{n,0},(0, -2}} and a nilpotent part N:= {{0,1},(0, 0}}. Raising this to the (2n)th power consists of raising D to the (2n)th power, giving the diagonal matrix D^{(2n)} = {{n^{2}), (-2)^{2n}} and adding the sum, from j=0 to 2n -1, of the products D^j N D^{2n-j} = {{0, - 2^(2n-1) (n/2)^j}}, which is a finite geometric series, easily summed. The resulting matrix is the same as yours: {{n^{2n}, (n^{2n} - 4^n)/n+1, {0, 4^n}}. Setting it equal to {{a,b},{0, b- 5b}} gives the special solutions you found.

How would you solve this with calculus ?

Once you said the magic words "complete the square" it was a big enough hint for me. Very cool! Will definitely remember this idea

Ok I'll minimize: You're a trivial problem. You're nothing.