Solving quadratic equations by completing the square 

It's magic - no human can do that. I'm talking about a seamless marker switch.

The golden ratio. x = 1 + \frac{1}{x} x^2 = x + 1 x^2 - x = 1 Now this is a quadratic equation, so we can complete the square. (x - \frac{1}{2})^2 = \frac{5}{4} x - \frac{1}{2} = \pm\sqrt{\frac{5}{4}} = \pm \frac{\sqrt{5}}{2} x = \frac{1}{2} \pm \frac{\sqrt{5}{2} = \frac{1 \pm \sqrt{5}}{2} = \phi If you don't understand LaTeX notation, \frac{x}{y} just means x/y, \sqrt{x} means the squareroot of x, \pm means plus or minus, and \phi means the Greek letter phi (Φ)

How is this non factorable?

@@user-dh8oi2mk4f Because you can't factor it If we could factor x^2 - x - 1, then we could put it into the form (x + a)(x + b), which equals x^2 + x(a+b) + ab. This means ab must equal -1, while a + b = -1 which is not possible

@@user-dh8oi2mk4f x(x-1) = 1 => x^2 - x - 1 = 0 which is not factorable since there is no way of expressing the -x as two parts with coefficients that sum to -1 and have a product of -1 (i think) also bcz it doesn't have rational roots and nobody wants to factorise irrational roots unless they do

@@videogamemusicguy4208 “not factorable” and “not factorable over integers” are 2 completely different statements, and the first one is just wrong here

Call the three natural numbers a,b,c we'll have: a+b+c=abc Because a, b, c can change values ​​to each other without affecting the problem so let's assume a>=b>=c -> 3a >= a+b+c -> 3a >= abc -> 3 >= bc since b,c are natural numbers so bc=0 or 1 or 2 or 3 if bc=0 then a=b=c=0 if bc=1 then b=c=1 and a+1+1=1.1.a (does not exist a) if bc=2 then b=2 and c=1 so a+2+1=a.2.1 so a=3 if bc=3 then b=3 and c=1 so a+3+1=a.3.1 so a=2 (wrong because a>=b>=c) So (a;b;c)=(0;0;0) or (3;2;1) and other permutations

@@o_o2688 actually I did the same way ...but here we need to put the different values to get the proper solution, but what I think is to find a method to get the values directly as an parametric equation in only one variable in that all the values can be calculated... But i think the type of method i want is not exist...😅😅.. Thanks for your reply...

@panchamsrivastava7386 I know that too, but I want a perfect method to get those numbers as solution, I don't want to do brute force like method

@@b77vedantmore51the word you’re looking for is “rigorous”

Not when you use it in this strange "factoring" way. But if you use actual completing the square thing - lots of times it is. But for that to be faster, I believe, you need to understand why it works. This guy did not explain it neither in this, or previous videos, so it might be hard to achieve just by his videos. But in general, when you have 1*x + cx + d = 0, where c and d are some constants it's faster, then quadratic formula.

In practice, what people did before calculators, was use a slide rule. You could use the change-of-base rule, to write x^y = e^(ln(x)*y). Let y = 1/2, for square rooting. The slide rule has features for calculating ln(x), where x would be the square rooted input. You'd then divide that by 2, to account for the exponent of 1/2. Then, use the other slide rule to calculate the exponential function. There are many other methods, like Heron's method that I think would be the easiest to remember: en.wikipedia.org/wiki/Methods_of_computing_square_roots

There's a method that looks similar to long division, which you can use to calculate the answer precisely to any number of digits. It's in the Wikipedia article that ​@@carultch linked, under "Digit-by-digit calculation", although you have to scroll past some theory to get to the actual description and a worked example.

Division. It should be taught in school tho.

@@tobybartels8426 Did that a few times in the 80's when basic calculators started at 50 dollars or more. It is a slog. Pairing off digits and testing a few because sometimes you guess wrong, and are to large of a number.

Depends on whether the square part is inside or outside the square root. If it's √-̅2², then it's (i√̅2)² = -1*2 = -2. If it's √̅-̅2̅² , then it's √̅-̅4 = -2i. (And if it's √̅(̅-̅2̅)̅², you get √̅4 = 2.)

That's the quadratic formula; in the USA, we learn it as (−b±√(b²−4ac))/(2a) (but with horizontal bars so no brackets are needed). Completing the square is worth knowing, however, because it can be used in other situations, such as finding the standard form of an equation for a conic section.

Another reason for knowing completing the square, is integration and Laplace transforms in calculus, when working with irreducible quadratic denominators. It's easy to integrate the function 1/(x^2 + 1), because the given expression directly matches the derivative (opposite operation of integration) of arctangent. However, what if you were given 1/(x^2 + 6*x + 10)? It turns out, we can transform this expression to look like 1/(x^2 + 1), by using the method of completing the square. x^2 + 6*x + 10 = (x + 3)^2 + 1 This means we can write it as: 1/((x + 3)^2 + 1) Now it is just a horizontally shifted version, of an integral we know evaluates to arctan(x). Thus, the solution is arctan(x+3) + C.

The US does not teach the method based on Vieta's coefficients. if you have x^2 + px + q = 0, and to make it easier to read I usually create an h = p/2 (which is the negative average of the two roots, while -p is the total of the two roots) the formula becomes -h plusMinus sqrt (h^2 -q) Edit the - and the plus minus sign causes odd printing

Rewrite them using Euler's formula: e^(i*x) = cos(x) + i*sin(x) Construct one with the negative input, and subtract one from the other, to determine what sine is, as an equivalent exponential: e^(-i*x) = cos(x) - i*sin(x) e^(i*x) - e^(-i*x) = cos(x) + i*sin(x) - [cos(x) - i*sin(x)] Simplify, and solve for sin(x): e^(i*x) - e^(-i*x) = 2*i*sin(x) sin(x) = -i*[e^(i*x) - e^(-i*x)]/2 Do the same, for cos(y). But add, instead of subtract to solve for cos(y): e^(i*y) = cos(y) + i*sin(y) e^(i*y) + e^(-i*y) = 2*cos(y) cos(y) = [e^(i*y) + e^(-i*y)]/2 Let y = x + pi/2: cos(x + pi/2) = [e^(i*(x + pi/2)) + e^(-i*(x + pi/2))]/2 Use properties of exponents, to separate terms: e^(i*(x + pi/2)) = e^(i*pi/2) * e^(i*x) e^(-i*(x + pi/2)) = e^(-i*pi/2) * e^(-i*x) Evaluate e^(i*pi/2) and e^(-i*pi/2): e^(i*pi/2) = i e^(-i*pi/2) = -i Thus: e^(i*(x + pi/2)) = i*e^(i*x) e^(-i*(x + pi/2)) = -i*e^(-i*x) Add together, and pull the i out in front: cos(x + pi/2) = i/2*[e^(i*x) - e^(-i*x)] Recall out earlier expression for sin(x), which is this the negative of the above expression: sin(x) = -i/2*[e^(i*x) - e^(-i*x)] Conclusion: cos(x + pi/2) = -sin(x), QED

@@carultch thanks ✨

use cos(a+b) identity

No. But, if you have an equation in both x and y which represents a circle in the cartesian plane, then you can use completing the square to bring this equation in a standard form (x − p)² + (y − q) = r² which lets you find both the centre and the radius of the circle, because this equation represents a circle with centre (p, q) and radius r.

Important step for it to not be complicated and chaotic like in the video is to understand why does it works, and he misses that point in both videos. When you know that (a +- b)^2 = a^2 +- 2ab + b^2, and therefore 10x is the 2ab (where a is x), you can easily do it like you wrote. He does not explain this, so he needs this factoring thing to determine what the square actually will be.

His notation is less complicated than yours. But, yes, it would be nice if he'd explain where the "magic number" comes from.