Відео

Можно было начать сразу с комплексных чисел. a^1/2(1+i)...

There is no square roots of negative numbers in the real numbers, you to work a little bit on that, you can't just square both sides, there are some conditions to do that...

이건 a=b가 같아도 되기 때문에 1/a = 1/(13×2) 이 가능하고 답도 무한개임. 바보 같은 문제임. a-b=0이 되는 a,b가 뭐냐고 물어보는 수준.

Nice, complex number

Must set condition for a and b real or whole number

Nice.....

yes,but the solutions of x^2=16 are +sqrt(16) and -sqrt(16 ),that is +4 and -4

I suspect this video was used to generate comments on how it can be done in 3 lines using polar form to get the principle solution. I did it in 2 lines + a circle with triangles to remind myself of sin and cos values. Damn, now I have generated another comment.

There is a simple general expression for this, was commonly asked in the exam CAT which i am preparing. A/x+B/y=1/C => (x-AC)(y-BC)=ABC². One can easily change the reciprocal form into factors.

Sol : 1/a+1/b=1/13 (1-1/2+1/2) 1/13 --> (1/2/+1/2) 1/13 --> 1/26+1/26 a=26, b=26 a+b=52 Ans：a+b=52 for reference only

ㅂㅅ 인가

if you really want to pass the Oxford math admission test, you should improve your solution in several ways: (1) as mentioned by others, you missed the second solution y=-8i. Take into account, that every equation of the form x² = T always has two solutions +/- sqrt (T) (2) IMHO in the verification calculation you cannot square both sides of the equation, because of the same reason. This is not an equivalence transformation. Otherwise you verify -1 = 1 just by squaring both sides of that equation to (-1)² = 1². Instead, You should calculate sqrt (+/-8i) directly, maybe by an approach like (a+ib)² = +/8i. Calculating a and b yields the desired squareroots. (3) you can simplify your calculation by squaring sqrt (y) + sqrt (-y) = 4 right from the beginning. By careful evaluation of the resulting expressions you should obtain the same result +/-8i with much less steps. I hope, this helps to pass the test. Good luck.

4:53 ??, please

Wikipedia: en.wikipedia.org/wiki/Imaginary_unit#Roots

±(1 + i) /√2 is immediate if you see that i = e^iπ/2 hence √i = ± e^iπ/4

Square root of 16 is only 4, not + or -4.

The answer could be a=39, b=39/2, but there are infinitely many solutions

Background song name? Please❤

b . 1.5b = 100 b^2 = 100/1.5 b = 66.66^.5

Can we just use de moivre’s formula?

x=√(y). √(-y)=±xi 4=x(1±i) 4(1-±i)/2=x Differentiating between ± and -± here doesn't really matter x=2(1±i) y=x²=±8i y=8i: √(y)=2+2i, √(-y)=2-2i y=-8i: √(y)=2-2i, √(-y)=2+2i y=±8i

(abc)^2 = 100 x 200 x 300 = 6000000 = 6 x 10000 x 100 = 6 x 100^2 x 10^2 abc = 1000√6 a = abc/bc = 1000√6/200 = 5√6 b = abc√6/ac = 1000√6/300 = (10√6)/3 c = abc√6/ab = 1000√6/100 = 10√6 a + b + c = 5√6 + (10√6)/3 + 10√6 = (55√6)/3

915

15

I can solve the equation x^2 = i, using polar coordinate. But, that dosn't mean sqare root i. How did you chouse a branch of the sqare root ? You must define "sqare root i", at first.

i cannot be equal to plus or minus -1 because i is one number, while plus or minus -1 are two numbers!

ab = 100 bc = 200 ca = 300 200 = 2 x 100 bc = 2 ab c = 2a 300 = 3 x 100 ca = 3 ab c = 3b c = c 2a = 3b b = 2/3 a ca = 300 2a^2 = 300 a^2 = 150 a = sqrt(150) c = 2 sqrt(150) b = 2/3 sqrt(150) a + b + c = = sqrt(150) + 2/3 sqrt(150) + 2 sqrt(150) = = sqrt(150) (1 + 2/3 + 2) = = sqrt(150) (3 + 2 + 6)/3 = = sqrt(150) 11/3 = = 11/3 sqrt(150)

Multiple value of and b

Really painful to watch

You miss another one, solution. √64*(-1) = 8i or -8i So y = 8i or y = -8i

Trivial problem. Assuming principal value of Sqrt, the immediate result is exp(pi.i/2) which can also be written as (1+sqrt(2).i)/2 . The solution path shown gives a bad impression of how advanced math is (to be) applied.

assume 1/a= m and 1/b= n and solve by cramer' rule then resubstitute the value of m and n

The equation Y^2-XY+13X=0, with every X equal or higher than 52, admits solutions Ymin(X)>0 and Ymax(X)>0 (with Ymin<Ymax) where 1/Ymin+1/Ymax=1/13 and Ymin+Ymax=X. The statement does not indicate that Ymin or Ymax must be integers or rationals. In general they are reals.

А ПРИ ЧЁМ ТУТ эЙНШТЕЙН?

This is so painful to watch

correct answers 8i and -8i, because 8i ==> √(8i) + √(-8i) -8i ==> √(-8i) + √(-(-8i)) = √(-8i) + √(8i) = √(8i) + √(-8i) other than that ... your calculation was correct

интересен только конец с множителями

Music's good

good vid.🙂

And like that you were in the 97% that failed... 🤣

Clickbait

(2-1)^6

8i e -8i

Why do you not consider a factorization into (-1)x(-169) or (-13)x(-13) ? Yes, this will lead to solutions with ab<0, but how do you know in advance...

a+b=6

Thank you.

I would write the equation as√y *(1+i) = 4 . Then √y = 4/(1+i) = 4 (1-i)/2 = 2(1-i) -> y= 4(1-i)^2 = 8 i.

Instead of dividing it by (ab) you should devide it by b²

13*(a+b) = a*b In generally, we can let a=13*n, where n is Nature Num 13*(13*n+b) = 13*n*b 13*n+b = n*b b = n*13/(n-1) is integer n = 2 or 14 a+b = 52 or 196

Is this completing the square method? I think the steps are incorrect tho