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We solve for x first, second we solve for y to thrid plog the values on the equations...

You solved it the difficult way😅

Great work. Thanks

Awesome!!!!

What's about factorizing 16^4 out and have 16^4(16-1)

Arigato gozaimasu. You have given me more confidence

y=-i*sqrt(8). To solve you have to take sqrt(y) as c+id, where c and d are real values and i=sqrt(-1)

I started with dividing by 5^x and ended up with 5^2x-5^x-1=0 solving to log base 5 (-1+sqrt(5)/2) any idea why i got root 5 instead?

This is a 30sec problem without pen and paper. Your method is way too complicated. And there is only one solution, not two.

Je ne comprends pas anglais mais j'ai tout compris très bon professeur

The fact that this takes more than two lines eludes me. Twice 5^k = 100, 5^k = 50, k = log5(50). The end...

Too long

at 1:42 where you cancelled out the square root and the raised to the second power, shoudn't you have taken the absolute value of what was under the square root, leaving you with y instead of -y?

that was really smart, can you make more videos with problems that use difference/sum of two cubes?

Solution: f(x) = f(x+3-3) = (x-3)²+3*(x-3)+2 = x²-6x+9+3x-9+2 = x²-3x+2 Checking the result: f(x+3) = (x+3)²-3*(x+3)+2 = x²+6x+9-3x-9+2 = x²+3x+2 everything o.k.

How did you obtain the square root of -1 in your first application of square root?

Great but I think your "Note" at the end has an error in the explanation. Your 2^ log(3) base(2) does not match a^ log(a) base(b)

Is this really necessary ?

I want to be a great mathematician.

wow ok got it

X Is 2

But in exams One had less than a minute to solve this without a calculator 😂

Nice

Guru!! I hail u oo

Simplified?

great explanation! i enjoyed it very much.

that was a really good video i enjoyed it thank you

how do you input the lambert w function into the lambert w function into the calculator?

😂😂😂 i cant make it one of difficult math problem ever

what if you meet it for the first time..would you reach those steps?

You can't do all these in an interview. An easy method is expected from the student. The quicker way is to list the 6th roots of the complex number z=2^6. Which are 2e^{ipik/3} for k=0,1,...,5

fair enough

Why not simply writing : Sqrt(x) = 2 - Sqrt(2) . So (Sqrt(x))^2 = (2-Sqrt(2))^2 and x = 4+2-4Sqrt(2) = 6-4Sqrt(2)

I got lost a bit in the middle but then when you were wrapping up it just clicked. Cool problem

super ! thank you

There is only 1 correct answer that is X2 (X1 is not the correct one but it formed from the "square both side" method)

nice !

Lovely maths… because nude girls are not here twerking people are not interested 😊

Man, you are so stupid

27; 1,1508... .

Simple rootx = 2 - root2 X = (2-root2)2 X = 4+2-4root2 X = 6-4root2

2(3-2V2).

Good video! One thing though, by squaring both sides two times you introduced an extraneous solution, namely 6 + 4*sqrt(2). To show this, you can write 6 + 4*sqrt(2) as 4 + 4*sqrt(2) + 2, which is precisely (2 + sqrt(2))^2. If you plug this solution in the original equation, this happens: sqrt(2) + sqrt((2 + sqrt(2))^2) = 2 sqrt(2) + 2 + sqrt(2) = 2 2*sqrt(2) = 0 sqrt(2) = 0 (absurd) Notice that you can do the same process with 6 - 4*sqrt(2) to show that it is in fact a solution to the original equation: 6 - 4*sqrt(2) = 4 + 4*sqrt(2) + 2 = (2 - sqrt(2))^2 sqrt(2) + sqrt((2 - sqrt(2))^2) = 2 sqrt(2) + 2 - sqrt(2) = 2 2 = 2 I found this out because I tried this problem for myself and this is how I did it: sqrt(2) + sqrt(x) = 2 sqrt(x) = 2 - sqrt(2) By squaring both sides you get: x = 4 - 4*sqrt(2) + 2 x = 6 - 4*sqrt(2) Hope this helps and my explaination is clear, keep up the good work!

why you didn't subtract sqrt2 from both sides and then squared the equation? and also you need to check your answers because one of them in not right. The plus one solution is not right

I like your helmet !

Thanks 🙂

Excellent clear & concise demonstration ! Thanks for sharing ……..😉

cubic equations have 3 roots.

Great video!