Відео

Great job Very nice explanation

a=1,,b=9

Beautiful Explanation !

x belongs to {0; 4}

a + 1 = 10, b + 1 = 2(At dis point no need to add or subtract the two equations. Rather, solve the linear equations directly to obtain a = 10-1= 9, b = 2-1 =1.

😊 Greetings from Warsaw Poland. There is a basic error at the end with e^ln (2(2-x)). The (2-x) is an exponent so it has to be moved to the front of ln2. Like this e^(2-x)ln2. It is very important. Regards 😊

Two methods with two different answers!

x=0

@ 3:00 / 12:45 x³ - 6x² + 9x - 4 = 0 → before to continue, always try to find an obvious solution (as - 2 , - 1 , 1 , 2) x³ - 6x² + 9x - 4 = 0 ← you can see that (1) is a root, so you can factorize (x - 1) (x - 1).(x² + ax + 4) = 0 → you expand x³ + ax² + 4x - x² - ax - 4 = 0 → you group x³ + x².(a - 1) + x.(4 - a) - 4 = 0 → you compare to: x³ - 6x² + 9x - 4 = 0 (a - 1) = - 6 → a = - 5 (4 - a) = 9 → - a 9 - 4 → a = - 5 → of course, because above (x - 1).(x² + ax + 4) = 0 → it becomes (x - 1).(x² - 5x + 4) = 0 ← you can see that (1) is a root, so you can factorize (x - 1) once again x² - 5x + 4 = 0 (x - 1).(x - 4) = 0

Why isn't (x,y) = (-3,4) a solution? I think that you missed this because you "cancel" squaring and square-rooting operations. But you really need to be careful doing that don't you? It is better to take everything to one side of the equation and then factor. Then you get all the solutions.

try , one solution among many , 162=27*6 , let u=V3^((Vx)/2) , let v=V3^((Vy)/2) , u^2-v^2=27*6 , (u+v)(u-v)=27*6 , let u+v=27 , let u-v=6 , u+v+u-v=27+6 , 2u=33 , u=16,5 , u^2=272,25 , v=u-6 , v=10,5 , v^2=110.25 , V3^Vx=272,25 , Vx=log(272,25)/log V3) , Vx=10.2069 , x=~ 104.181 , V3^Vy=110,25 , Vy=log(110.25)/log V3) , Vy=8.56126 , y=~ 73.2951 , solu , x=~ 104.181 , y=~ 73.2951 , test , V3^V(log(272,25)/log V3)) - V3^V(log(110,25)/log V3)) = 272.25-110.25 , --> 162 , OK ,

let u=3^x , u^3 +/- u^2 - u - 60=0 , (u-4)(u^2+4u+15)=0 , u=4 , 3^x=4 , x=log4/log3 , +1 -4 u^2+4u+15=0 , u=(-4+/-V(16-60))/2 , u= -2+i*V11 , -2-i*V11 , +4 -16 solu , x=log4/log3 , +15 - 60=0 , test , 27^(log4/log3)-3^(log4/log3)=64-4 , --> 60 , OK ,

Let 3^x=a a(a^2 - 1)=60 a(a+1)(a - 1)= 60 (a - 1) a (a+1) = 3×4×5 a=4 log4 to the base 3 =x

Completely wrong

Excellent as usual! 😊

Obviously, wrong decision

You first define the domain of x and y. it should be positive integers. at 2:42 I understand b = √y is integer, but how can you guarantee a = x^(√y/2) is integer? if x is not power of 2 and √y is odd , a is irrational. in that case there are infinitely many combinations of (a+b)x(a-b).

(x + y)² = x² + 2xy + y² → given: x + y = 2 4 = x² + y² + 2xy → given: xy = 1/2 4 = x² + y² + 1 x² + y² = 3 → given: x + y = 2 → y = 2 - x x² + (2 - x)² = 3 x² + 4 - 4x + x² = 3 2x² - 4x = - 1 x² - 2x = - 1/2 x² - 2x + 1 = - (1/2) + 1 (x - 1)² = 1/2 x - 1 = ± 1/√2 x - 1 = ± (√2)/2 x = 1 ± (√2)/2 First case: x = 1 + (√2)/2 y = 2 - x y = 2 - 1 - (√2)/2 y = 1 - (√2)/2 Second case: x = 1 - (√2)/2 y = 2 - x y = 2 - 1 + (√2)/2 y = 1 + (√2)/2

Thank you for explaining. x=66, y=1 is another solution. If 33^(1/16) is admitted, irrational numbers are OK. If so, x=√69, y=4 can be one of the solutions.

Amazing 😍.

Ответ не верный По крайней мере х=66, у=1 тоже решение, а его нет как решения.

Dividing by n is not good, since you are eliminating one of the possible solutions. Eg. In solving: x^2 - 2x =0 Dividing throughout by x, we will have: x - 2 = 0 Which gives, x = 2. But the equation should be having two solutions. The other solution was lost in division.

Super superb sir hat's off to you

Sempre bonito assisti aos seus vídeos.

Why second one is incorrect though it comes out of correct equation answering

To all those who are saying to do trial and error first Yes it could have been done that way But it's not about finidng the answer. It's more about making the way of finding the answer easier. If the answer was not 5 maybe something higher , it would be harder. But doing some steps to make the equation look easier is the main motive. So , I feel like we should try enjoying the journey more rather than the destination.

Very very nice

First view : x = 1

Nonsense. If you wanted to do trial and error, you could have started with n. There was no need to juggle things around and to introduce r.

Try for other solutios: # -*- coding: utf-8 -*- """ Created on Tue Sep 17 11:32:54 2024 @author: jwb45 """ f = lambda x=10: print('<> '*x) f() def fact(num: int): fact = 1 for n in range(1,num+1): fact = fact * (n) # print(f'n = {n} and the factorial is = {fact:,}') return fact fnum = int(input(('Enter Number to Verify: '))) cube = fnum**3 print(f'Number is: {fnum}, Factoral is: {fact(fnum)}, Cube is: {cube}') print(f'Left: {fact(fnum)+fnum} and Right: {cube}') print(f'DELTA = {fact(fnum)+fnum - cube}') if (fact(fnum) + fnum == cube): print("True") else: print("False") f()

if a=4,>>>>8^√4---2^√4=8^2--2^2=64---4=60,So,a=4#

For completeness you should check n=0 and n=1 in the initial equation, as you divide by n and n-1 which would be zero in that case.

At least a word about why no other solutions are possible would be nice to have, and of course trying values into original equation as others suggested works faster.

First view : 49 - 9 = 40

let u=2^Va , (u-4)(u^2+4u+15)=0 , u= 4 , /-2+i*V11 , -2-i*V11 / , 2^Va=4 , 2^Va=2^2 , Va=2 , a=4 , test , x=4 , 8^V4 -2^V4 = 8^2-2^2 , 64-4=60 , OK ,

If you are going to put the Value and check the equation for it then why not just put it in the beginning?

This is clumsy! X.(X+1).(X-1) = 60 4.(4+1).(4-1) = 60 4.5.3 =60, then X = 4 And then you solve it with a simple log equation, log2(4).

From the beginning : I started with 3 , then 4 , then 5 which is the answer

First look : 625 - 25 = 600

Many unnecessary steps

64 - 4 = 60 8^2 -2^2 = 60 a = 4

Add eq.1 to eq 2 , you got 2(4^x)=18 4^x=9 (2^2)^x=9 (2^x)^2=9 2^x=3 etc.

Thank you for this video. It is a good method for any positive value of the constant. For this problem, we can see x=4 (a=16) is a solution. We still need to investigate whether there is any other solution. First and east, a should be non-negative and a=0 is not a solution. Next we can see that 2^x+x-20 is strictly increasing for all positive x, which means there is at most one positive solution (and it is x=4, though we can use the W-function in general).

I solved it in my mind x^2 - y^2 = 7 (x+y)(x-y)=7.1 x+y=7 x-y=1 2x=8 x=4 & y=3

3^4 = 81 81 - 4 = 77

5 + 4 = 9 ; 9 x 7 = 63.

X=5,y=4 and z=5 is also a solution. Right? So is x=3,y=9 and z=7….

There are numerous mistakes in this video which, more or less, gets the correct result by luck. For one thing, when you get a result like "0=0", that is satisfied by any value of a or b, so it is incorrect to say that there's no solution there. For another thing, all three equations need to be satisfied, so as soon as one cannot be satisfied, there is no reason to check the others for the same case. As just one more example, there is no reason to-retest in each case the equation that you used to derive the cases -- you already know it will be satisfied.

This is a very awkward way to solve this problem. 😢

4^x=a, 4^y=b; a=9, b=1; x=log(9)/log(4)=1,585; y=log(1)/log(4)=0