Solving a golden equation

In this video , I solved an equation which turned out to have a golden solution

[@ernestdecsi5913]

A very nice solution. I, a 70-year-old pensioner, send you my greetings from Slovakia. When I see such a beautiful solution, all my sadness disappears.

[@dougaugustine4075]

I'm 70 too (but will be 71 later this month). I live in Japan.

[@CharlesShorts]

Hi prime newtons, letting you know that you uploaded the same video twice!

[@PrimeNewtons]

Thanks. I deleted one. KZbin probably had a glitch yesterday.

[@qetuoa13579gjlxvn]

This channel better then my math teacher.

[@harshplayz31882]

😂

[@isaacbunsen5833]

The negative version doesn't actually give you imaginary sqaure roots! The first one is the square root of 1 and the other one gives you the square root of phi sqaured!

[@sunil.shegaonkar1]

Both Solutions are valid. Because x2 = -0.618 aaprx and 1/X2 = -1.618 apprx. Difference between x2 and 1/x2 = 1 (Exactly). That reminds me what facilitated substitution of the long-term by U at 8:36. -1/x2 appears in both the terms, which is positive. That will make square root positive.

[@surendrakverma555]

Excellent explanation Sir. Thanks 🙏

[@HollywoodF1]

Good problem- good suggestions. Thank you!

[@xyz9250]

I did it without substitution, multiply both side by [(x-1/x)^1/2 - (1-1/x)^1/2] and eventually get (x-1/x)^1/2 - (1-1/x)^1/2 = 1 -1/x , and the two equations to get rid of the second sqrt, then square both sides, simplify and get (x - 1/x -1)^2 = 0.

[@dirklutz2818]

Both parts of the equation do exist in the range [-1,0[ and [1, inf[. So x could be negative! But (1-sqrt5)/2 is too negative.

[@danielbergman1984]

Nice! 👏

[@appybane8481]

x2=(1-sqrt5)/2 don't work because (x-1/x)^1/2+(1-1/x)^1/2 is always positive (if it's real)

[@dirklutz2818]

Sqrt(5) without calculator! 5= 2.5 * 2. So the square root must be between 2.5 and 2.0. The average of the 2 numbers is 2.25. Now, 5 divided by 2.25 = 2.2222. The average now is 2.2361 and that is very close to the (rounded) value of 2.23607.

[@user-sw7rw3px6n]

sir can you please do the cauchy sequence prove?

[@lukaskamin755]

interesting that the LHS is also defined for -1

[@rimantasri4578]

Life is beautiful!

[@SuperTommox]

This was beautiful

[@9adam4]

Should have factored out the x-1 term near the beginning.

[@Dr.keitan]

Thank you for your lecture. If the solution provides us the golden ratio, can we directly derive the first equation from a certain relation in a geometric object related with the golden ratio?? I found the Euclidean way for x^2 - x + 1 = 0 by the similarity in a rectangular. I guess there is the source of the first equation... Please teach that if you know.

[@vestieee5098]

Hello sir, is there any difference between writing the square root of some quantity as radical(x) versus as (x)^1/2? Does it affect the domain/range or something? I'm not too clear on this, sorry for the bother. Great video as always :)

[@PrimeNewtons]

Yes. Unless clearly stated, you can only get non-negative outputs from square-root (x). But x^½can give anything including imaginary outputs.

[@vestieee5098]

@@PrimeNewtons Ah, makes sense. Thank you so much for the explanation!

[@qwertyuiop2161]

what is the source of this problem?

[@5Stars49]

Golden ratio 🎉

[@Charles-ef5vs]

where can I find problems in this level of difficulty?

[@ramunasstulga8264]

In practice books

[@ibrahimmassy2753]

Great video! I guess that solution (1-√5)/2 doesn´t work not by x-1/x be negative (because is positive) but is because x is expressed as sum of two square roots

[@PrimeNewtons]

You are ✅️

[@vasanthalakshmi6734]

Need help sir why can't the opposite of golden ratio is a root.?

[@rafaelsimoes2974]

x cannot be negative, because x is the sum of two root squares, positive + positive = positive

[@abidomar9568]

Please look at the question "x belongs to real number "

[@rafaelsimoes2974]

@@abidomar9568 yes, but it's a mistake, it is impossible that x be negative, just look (x-1/x)^(1/2) must be positive, (1-1/x)^(1/2) also must be positive, this implies that x is positive because x = (x-1/x)^(1/2)+(1-1/x)^(1/2) x = (something positive) + (something positive), therefore x = (something positive)

[@lilypad429]

​@@abidomar9568should be "real POSITIVE number instead of just real

[@neclis7777]

IR*....

[@hba12]

it will be easier if you write x^2 -1 as (x+1)(x-1)

[@comdo777]

asnwer=1x

[@anonakkor9503]

let’s gooo haha

[@formarce]

I have a beaultifull solution for this equation using geometry

[@flamewings3224]

The negative version (x = 1/2(1-sqrt(5)) actually is the solution. Because we want that an expression under a square roots will be positive. So we aren’t want x > 0, we want 1 - 1/x >= 0 and x - 1/x >= 0. For the first one we have: 1 - 1/x >= 0 (x - 1)/x >= 0 x in the interval (-inf; 0) U [1; +inf) For the second one we have: x - 1/x >= 0 (x^2 - 1)/x >= 0 (x-1)(x+1)/x >= 0 and we have that x in the interval [-1; 0) U [1; +inf) and, fortunately (or not xd), the root x = 1/2(1 - sqrt(5)) ≈ -0.618 in the interval.

[@m.h.6470]

The negative version can't be a solution, as the result of any root is ALWAYS positive. Since the x on the right hand side of the original equation is equal to the sum of two roots, it HAS to be positive. It literally doesn't matter, what is inside the roots, x is positive.

[@Taric25]

Reduce[Sqrt[(-1 + x)/x] + Sqrt[-x^(-1) + x] == x, x, Reals] gives the positive answer.

[@nothingbutmathproofs7150]

I got the same results but did not take into account that the sum of two sqrts can't be negative

[@kiro9291]

you might have uploaded the same video twice

[@ScrewY0UguyS]

Wait a second. x can be negative. Within following bounds: -1