A tricky problem with a "divine" answer!

Can you solve this equation? Thanks to Hrigved for the suggestion!
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[@DallasMay]

It seems like someone started with the golden ratio and then worked backwards to find the most absolute complicated looking equation they could come up with.

[@satyam9267]

First you have to say equation 1 to (((x-1/x)^1/2)+((1-1/x)^1/2))=x...(1) The rationalize left hand side then what you get you have tosay that equation 2 then then equation 1and 2 add them together the assum that (1-1/x)= any veriable then then you will get quadratic solve this quadratic get value of veriable then put value of veriable= (1-1/x)

[@animerecords7953]

Most likely, it just boils down to arbitrary assumptions and tricks

[@Lemda_gtr]

They do backwards scenes & claiming smart 😂😂

[@NestorAbad]

Those equations are usually quite tricky to solve, thanks for sharing! My approach was the following: instead of making substitutions at the beginning, first I manipulated the equation, √(x + 1/x) + √(1 - 1/x) = x √((x²-1)/x) + √((x-1)/x) = x √(x²-1) + √(x-1) = x√x Now square both sides: x²-1 + x-1 + 2√((x²-1)(x-1)) = x³ Expand the product inside the root and arrange terms: 2√(x³-x²-x+1) = x³-x²-x+2 At this point, the substitution is more clear: let's call y=x³-x²-x+1. The equation then transforms to 2√y = y+1 Square both sides and arrange terms: 4y = y²+2y+1 0 = y²-2y+1 0 = (y-1)² y = 1 Back to the substitution: 1 = x³-x²-x+1 0 = x³-x²-x 0 = x(x²-x-1) And since x=0 is not a solution of the original equation, we are left with x²-x-1 = 0. The ending is then like in the video.

[@Username-vn1wx]

Ur comment days 6 days ago

[@khwabm]

your comment is 6 days ago 😱

[@vimleshmaheshwari9018]

Very well solved but how did you commemt 6 days before the video was uploaded without editing your comment

[@theUnmeshraj]

Hey Mr. Time travellers, NASA wants to know your location...

[@FumaxIN]

@@khwabm patreaon

[@MSJ_7]

Saw "divine" in the title and knew straightaway the answer would be the golden ratio 😛

[@caniggiaful]

Same. But I didn't manage to solve it, even though he only used standard steps.

[@Rickety3263]

I knew the answer... but theres no way I could have figured it out. 😅 I guess I still get full credit in California schools 😂

[@bibhuprasadmahananda6986]

Exactly....

[@sohailahabib2225]

What is the golden ratio?

[@EditingHiG1]

Same pinch

[@mixymaxy5403]

My brain left the chat

[@madhavanguha2626]

Your 19 likes and this video 19 dislikes what a coincidence

[@mixymaxy5403]

@@madhavanguha2626 didn't see that coming

[@RedRacconKing]

Was it ever in the chat to begin with?🤔

[@mixymaxy5403]

@@RedRacconKing just kidding I actually love maths

[@aashsyed1277]

@@madhavanguha2626 not now

[@КузебайГерд-ы1е]

"You should be able to solve it" Yeaa... and you should be able to perform the 500 kg deadlift.

[@rajnikyadav]

Lol

[@h00db01i]

just for the love of god don't do _both_

[@slingshot99]

@@h00db01i I'm a gym rat who's studying engineering, so *don't tempt me* . :D

[@h00db01i]

@@slingshot99 math before wrath in that case ;9

[@satyam9267]

First you have to say equation 1 to (((x-1/x)^1/2)+((1-1/x)^1/2))=x...(1) The rationalize left hand side then what you get you have tosay that equation 2 then then equation 1and 2 add them together the assum that (1-1/x)= any veriable then then you will get quadratic solve this quadratic get value of veriable then put value of veriable= (1-1/x)

[@peterkwan1448]

I solved by: Multiplying both sides by sqrt(x), sqrt(x^2 - 1) + sqrt(x - 1) = x sqrt(x) Moving sqrt(x-1) to the right side and taking square on both sides, x^2 - 1 = x^3 + x - 1 + 2 x sqrt(x^2 - x) Subtracting 1 from both sides and dividing both sides by x (since x is not equal to 0), x^2 - x + 1 - 2 sqrt(x^2 - x) = 0 The left side is a perfect square, so we get : x^2 - x - 1 = 0 Solving for x, we get: x = (1 + sqrt(5))/2 or x = (1-sqrt(5))/2 However, x = (1-sqrt(5))/2 does not fit in the original equation since x will be negative but the two square roots are both positive. So x = (1+sqrt(5))/2 is the only solution.

[@stephan7691]

How did you get from this step: x^2 - x + 1 - 2 sqrt(x^2 - x) = 0 to this: x^2 - x - 1 = 0 Where is the 2qrt() gone?

[@prbprb2]

This is how I did it also, and is the most mechanical way, I think. It requires the least sophistication, which is therefore the best, I think.

[@cedricveinstein6949]

@@stephan7691 As you know (a+b)ˆ2=aˆ2+2*a*b+bˆ2, now using that in reverse with a=sqrt(xˆ2-x), b=(-1) he rewrites the left side as (sqrt(xˆ2-x))ˆ2+2*sqrt(xˆ2-x)*(-1)+(-1)ˆ2 to get (sqrt(xˆ2-x)+(-1))ˆ2 = 0 which is equiv. to xˆ2 - x - 1 = 0

[@peterkwan1448]

@@stephan7691 I think I missed the steps to deduce, but the latter can be deduced by the former one in this way: Since x^2 - x + 1 - 2 sqrt(x^2 - x) = (sqrt(x^2-x))^2 - 2sqrt(x^2-x) + 1 = (sqrt(x^2-x) - 1)^2, x^2 - x + 1 - 2 sqrt(x^2 - x) = 0 => (sqrt(x^2-x) - 1)^2 = 0 By taking square roots of both sides and adding 1 on both sides, we get sqrt(x^2 - x) = 1 Taking squares on both sides, we get: x^2 - x = 1 which is the same as latter equation: x^2 - x - 1 = 0

[@Robbedem]

small typo: x^2 - 1 = x^3 + x - 1 + 2 x sqrt(x^2 - x) should be: x^2 - 1 = x^3 + x - 1 - 2 x sqrt(x^2 - x) (- before the root instead of +) afterwards it's correct again

[@ankitbhattacharjee_iitkgp]

Please keep making these!

[@thanosbabaji1127]

Ok

[@satyam9267]

First you have to say equation 1 to (((x-1/x)^1/2)+((1-1/x)^1/2))=x...(1) The rationalize left hand side then what you get you have tosay that equation 2 then then equation 1and 2 add them together the assum that (1-1/x)= any veriable then then you will get quadratic solve this quadratic get value of veriable then put value of veriable= (1-1/x)

[@P4ExHzLRuuiFMg3X4U3v]

There is also an interesting geometric perspective to the problem. You can construct two right-angled triangles with sides (a, 1/√x, √x) and (b, 1/√x, 1) where a and b are defined as in the video. Now construct a larger triangle by joining these two, so that they share the side with length 1/√x. Since we know that a+b=x, this larger triangle has sides (√x, 1, x). You can prove that this triangle is right-angled (exercise for the reader). Then, by Pythagoras theorem, √(1+x) = x, which gives the correct solution. It is interesting that this triangle (aside from scaling) is the only one with the property that the ratio of the hypothenuse to the largest leg is equal to the ratio of the largest to the smallest leg, and that ratio is √phi.

[@exoplanet11]

Thanks. That's interesting. In other words, is the triangle you are talking about Kepler's Triangle? en.wikipedia.org/wiki/Kepler_triangle

[@P4ExHzLRuuiFMg3X4U3v]

@@exoplanet11 Precisely! I had no idea it was called that, thanks for showing me!

[@IS-py3dk]

@@sharvaripatwardhan426 if its in the ratio 3:4:5 examples 6,8, and 10 have 2x the lengths of 3,4,5 triangle If it keeps on going like lets say 9,12 and 15 you can check if they are in the ratio 3..4..5 and check if they are multiplied with the same digit In 9,12,15 the 3..4..5 are all multiplied by 3 Now reduce 3 from 9..12..15 you will get 3..4..5 so its a right triangle If its not a special triangle Then check by using the GOUGU or The Pythagorean theorem ☺ that is a^2 + b^2 = c^2

[@P4ExHzLRuuiFMg3X4U3v]

​@@sharvaripatwardhan426 Sure. Work with the angles. Call them a,b,c. Notice that they are related by sin(a) = sin(b)/√x = sin(c)/x by the sine rule. From there should be able to show that sin(c) = 1, which tells us that c is a right angle. Of course, if you already know all the side lengths you can plug them in the Pythagorean theorem as the other comment says. But here we do not know them.

[@QuackerDAMG]

@@P4ExHzLRuuiFMg3X4U3v sorry how did you get to sin(c)=1?

[@242math]

I did not even know where to begin. This was complicated. Appreciate and understand your logic. Well done.

[@nickmeale1957]

Presh: "It's quite a divine answer!" Me: ".....Yes! Quite!" *way over my head*

[@pranjalraj7473]

Way too much😂😂

[@marco_gallone]

When I was doing it on my own I ended up trying to factor a sextic function. Well that’s 6 years of engineering down the crapper!

[@timetraveler7]

Some except it came to a quartic function instead of sextic

[@jyothishkumar3098]

you could approximate like an engineer instead

[@zastaphs]

@@timetraveler7 x^4-2x^3-x^2+2x+1 = 0

[@timetraveler7]

@@zastaphs yeah something like that

[@sirlight-ljij]

​@@zastaphs Yeah, I arrived at it by doing the following: t=1/x √(1/t-t)+√(1-t)=1/t √(1/t-t)=1/t-√(1-t) Square both sides 1/t-t=1/t^2+1-t-2√(1-t)/t Multiply by t^2 and eliminate similar terms t=1+t^2-2t√(1-t) 2t√(1-t)=t^2-t+1 Square again 4t^2(1-t)=t^4+t^2+t+2t^2-2t^3-2t Now after you open all of braces and combine everything into a single polynomial you'll get x^4+2x^3-x^2-2x+1=0 Which is equal to (x^2+x-1)^2=0 I actually enjoy this method a lot more. While the original method requires less steps, it is way more obscure, less intuitive; this method of eliminating roots works on a much wider class of equations.

[@aalsii]

Love from India ❤️🇮🇳

[@pengchengwu447]

From China too.

[@enejidjsi5939]

@@pengchengwu447 i thought youtube was banned in china? are you using a vpn?

[@hamidkh5488]

From Iran too.

[@Glinjan]

I'm really proud that I got there by myself 😁

[@satyam9267]

First you have to say equation 1 to (((x-1/x)^1/2)+((1-1/x)^1/2))=x...(1) The rationalize left hand side then what you get you have tosay that equation 2 then then equation 1and 2 add them together the assum that (1-1/x)= any veriable then then you will get quadratic solve this quadratic get value of veriable then put value of veriable= (1-1/x)

[@doublebro_7918]

Love from India bro!

[@seanclough7810]

I knew the answer by the title of the video (at least it's form, kinda) but I'm not savvy on clever substitutions. EDIT: I wonder if phi will become as useful as pi or e in, say, the maths of biology.

[@usptact]

“divine” was a huge hint that golden ratio will be involved. I guessed it :)

[@surajclub5303]

*_क्या तुम्हें पता है ✌️✌️🤣🤣_* 👇👇👇👇👇👇 *_कॉरोना बच्चो का दोस्त है हमेशा पेपर के समय पर ही आता है।🤣🤣😝😝🙈🙈_*

[@satishchaudhary7978]

😂

[@dioptre]

hahha nice one

[@brijeshsrivastava6682]

Love from India 🇮🇳

[@virajmisal651]

He is also indian

[@Username-vn1wx]

Yes 2 months ago....

[@vimleshmaheshwari9018]

How is your comment 2months ago

[@koro-sensei9783]

@@vimleshmaheshwari9018 he has the membership of this channel. Those get the video earlier

[@brijeshsrivastava6682]

@@vimleshmaheshwari9018 because I am HRIGVED who asked presh to upload my question

[@Bry10022]

Very clever solution you have there…

[@3dplanet100]

Wow, that was a really unexpected answer, amazing! I thought it was gonna be a whole number.

[@baze3541]

Really? Unexpected?

[@arctic_haze]

It could not be a whole number. One needs only seconds to see it is between one and two. But I am a physicist and I usually need a rough answer, not a beautiful irrational number.

[@satyam9267]

First you have to say equation 1 to (((x-1/x)^1/2)+((1-1/x)^1/2))=x...(1) The rationalize left hand side then what you get you have tosay that equation 2 then then equation 1and 2 add them together the assum that (1-1/x)= any veriable then then you will get quadratic solve this quadratic get value of veriable then put value of veriable= (1-1/x)

[@밍밍-b3n]

I squared both sides and reorganized the equation, substituting xㅡ1/x ㅡ1+1/x² = t², x²- x-1+1/x =xt² and then 2t=xt²+1/x, x²t²-2xt +1=0 (xt-1)²=0 xt=1 x²t²=1 x³-x²-x+1=1 x(x²-x-1)=0 x=1+-sqrt5/2 what!!

[@gamingmusicandjokesandabit1240]

The golden ratio really is golden to solving the problem 🙂

[@charlied.4683]

Ik! So cool!

[@bruh-cv7ec]

Spoiler alert must informed beforehand!

[@alphabetagamma4142]

What? No.

[@angrytedtalks]

Ration?

[@gamingmusicandjokesandabit1240]

@@angrytedtalks Well spotted 🙂

[@chrishoggett1375]

There was a time I would have done this very quickly, but after not using algebra for over 20 years my brain required some WD40 to follow

[@theUnmeshraj]

Looks like most of your subscribers are Indians.

[@boberbob1914]

i ended up in a cubic formula bruh

[@surajclub5303]

*Love From India 🇮🇳*

[@pujasingh624]

When this comment will be out, this video will have 700 comments.

[@BiscuitZombies]

Hmm, intriguing. I solved it in the following manner: Rearrange to give: √(x-1/x) = x - √(1-1/x), then square both sides and cancel common terms, x = x^2-2x√(1-1/x) +1 Then do the intriguing substitution, u = √(1-1/x), and we see that x = 1/(1-u^2) by rearranging, sub this in: 1/(1-u^2) = 1/(1-u^2)^2 - 2u/(1-u^2) + 1, multiply through by (1-u^2)^2 and get rid of common terms, put everything to one side, u^4+2u^3-u^2-2u+1 = 0, you can use any method here, but by symmetry we can postulate that this is a perfect square; (u^2+u-1)^2 = 0 ==> u^2+u-1 = 0 ==> u = (-1 +/- √5)/2 Once subbing these to find x, we have that x = (1-√5)/2 and x = (1+√5)/2. We can sub these into the original equation to reject anything extraneous, and we have that x = (1+√5)/2 = phi.

[@iqmathsciencelogicalreason2770]

Excellent tricks! Take care sir.

[@nibaranghosh2202]

Thank you so much sir! Really, math is divine. Math is everywhere. 👍

[@-d3d9ex97]

I wonder if an average person instrested in math is able to figure out these "smart" substitutions in this problem in a short (15 minutes or so) period of time. Is it normal to come up with such solution fast or do they think many hours? I stopped solving after coming up with an equation 0=x^4-2x^3-x^2+2x+1 (with the assumptions that x>1). In fact, wolframalpha gives the golden ratio as a solution but I do not see any way to solve this polynomial

[@nikitakipriyanov7260]

First is to get rid of cubic term (which progresses to Ferrari method). For that you must use this substitution: x=t+½ (so t³ will be cancelled). Try it, you'll be surprised, something else will be cancelled too.

[@IS-py3dk]

This is what makes me feel in peace after a math class early in the morning 😍😍😍 😄😄

[@karangupta1825]

Same

[@satyam9267]

First you have to say equation 1 to (((x-1/x)^1/2)+((1-1/x)^1/2))=x...(1) The rationalize left hand side then what you get you have tosay that equation 2 then then equation 1and 2 add them together the assum that (1-1/x)= any veriable then then you will get quadratic solve this quadratic get value of veriable then put value of veriable= (1-1/x)

[@grabfilm112]

Nice video Fresh!

[@aaronleperspicace1704]

What do you mean Fresh? His name is Prestle Walker.

[@krishna2803]

I thought it was Pringles Alan Walker

[@vishalpandey5326]

Superb content bro. I'm loving it.

[@koro-sensei9783]

Bro???? He is double your age

[@AgneyK]

@@koro-sensei9783 how do you know his age?

[@achillesvgr]

Hmmmmm... Interesting... But why?

[@reazahmedramim5904]

Brain has left this chat.😂

[@surajclub5303]

*Maths Favourite students*

[@Nishi-wc3zu]

Bro make video on integration please

[@moderneinstein2644]

Me watching presh talwalker's videos,:"I probably could have solved it anyways."😂

[@rsouzaneres]

I knew the trick in 2:02 for the solving of systems of linear equations, but I never asked myself why it works. Now I think that I figured out: a equation is like a weighing scale, so the addition of two equivalent equations is like the addition of the equivalent weights on both plates of the scale that doesn't affect the balance. Am I right?

[@abc-iz9vg]

yeah its especially easier to visualize if you replace variables with numbers

[@mushfikaikfat]

*Love from Bangladesh* 🇧🇩

[@shawnweddle3002]

I had a feeling it was phi immediately

[@aniruddhxie2k215]

Finally a guy who acknowledges that Quadratic Formula was given by BrahmaGupta

[@longlostwraith5106]

Why should anyone care? Are we to acknowledge the inventor of addition and subtraction as well? The quadratic formula is easily derivable.

[@jackmccarthy7644]

@@longlostwraith5106 The pythagorian theorem is also easily derivable, but we still credit him... just saying

[@aniruddhxie2k215]

@Ayush Oh yes

[@PuzzleAdda]

Do checkout Puzzle Adda KZbin Channel for more puzzles and riddles.

[@pickledata7926]

*Congrats to anyone who is early and found this comment!*🏆

[@gm2407]

Comment only been up for three years. Heat death of ths universe is a very long time away. Yay I am early. Thanks for the congrats kind stranger.

[@mathevengers1131]

I knew that the answer is golden when I read word divine.

[@frentz7]

You can also do this problem directly, with no particular "heroics," simply squaring both sides of the equation to eliminate square roots, simplifying to isolate the remaining square root, and then squaring both sides again. And it has a couple neat moments! #1 I multiplied on both sides by sqrt(x), just to be lazy. (This will create the false solution x = 0, so we'll keep that in mind for later.) Now it says sqrt(x^2 - 1) + sqrt(x - 1) = x sqrt(x). You can think of it as sqrt(A) + sqrt(B) = x^(3/2). #2 Now square both sides (caution; may create false solutions). We get A^2 + B^2 + 2 sqrt(AB) = x^3, so x^2 + x - 2 + 2 sqrt((x^2 - 1)(x - 1)) = x^3. #3 Naturally the next step, isolate the " sqrt(AB) " term and square both sides again. But now look! A wonderful thing happens: let's simplify before squaring both sides the last time. We get 2 * sqrt( x^3 - x^2 - x + 1) = x^3 - x^2 - x + 2. Almost the exact same cubic!! Set alpha = x^3 - x^2 - x + 1 (this is super neat, so go ahead and give yourself a greek letter :). .. squaring both sides, from 2 * sqrt(alpha) = alpha + 1, you get 4 alpha = alpha^2 + 2 alpha + 1, which (yep!) simplifies to 0 = alpha^2 - 2 alpha + 1, or 0 = (alpha - 1)^2, so we must have alpha = 1. Plus (bonus!!) check it out : now #4 setting alpha = x^3 - x^2 - x + 1 = 1, we get a SOLVABLE cubic! So x^3 - x^2 - x = 0, or x (x^2 - x - 1). Very cool. Plus #5 we know that x = 0 cannot be a solution, because the original equation contains expressions "1 / x". So that leaves x^2 - x - 1 = 0, or x = (1 +- sqrt(5)) / 2. Finally in the original equation if you plug in, you can see only x = (1 + sqrt(5)) / 2 works.

[@LouisEmery]

I found your method and I stopped at the 2 * sqrt( x^3 - x^2 - x + 1) = x^3 - x^2 - x + 2. and I thought this can't be right. It's not solvable either. If only I did y= x^3 - x^2 - x + 1. Of course it's 4 am now, and I have other things to do. ;)

[@frentz7]

@@LouisEmery hey thanks for the comment! yea same, exactly .. cubic?? that must be a dead end. I think I gave up for a while and then tried again another day

[@doraelog690]

This method was better than the tricky one

[@Hexanitrobenzene]

@@doraelog690 This is more or less standard approach, however, we got very lucky 2 (!) times, otherwise we would have to deal with sixth degree polynomial equation...

[@Hexanitrobenzene]

@@LouisEmery I got stuck there, too. Usually, appearance of x^3 term is a sign that one's approach is not clever enough. This time we got lucky, two times even :)

[@bibekanandaswarlearning5224]

The answer is completely wrong because the + value ( (1+√5)/2 ) is not a solution of the equation.... If you draw the graph you will see that the graph doesn't meet at the ' golden ratio ' value..... They have shown the similarity with the golden ratio...but when substituting the value....we can take the negative value tooo...he missed it.... If you took neg value.... The + value cannot solve the given equation.... But the - value ( ( 1-√5)/2 ) can be the root of the given expression... According to GRAPH and the neg value when we vanish the roots..... Plz comment for any queries...

[@mysillyusername]

The substitutions make sense if you take it step by step. First rearrange to have the square-root terms on each side of the equation and square both sides. Then rearrange to have the remaining square-root term on one side and square again: this gets rid of the square-roots and you get: (x^2 - x + 1)^2 = 4(x^2 - x) At this point it makes sense to make the change of variable a = x^2 - x. This gives (a -1)^2 = 0, so a = 1 and finally: x^2 - x - 1 = 0 the famous equation!

[@GodbornNoven]

Way i did it

[@اشکانمحمدی-ز1ث]

I've alway wanted to know how do you know that you should multiply the equation by a particular term to be able to move forward? 1:06 I have seen many difficult algebra problems getting easily solved after multiplying or adding a particular term. Is it all arbitrary or is there a way other than sole creativeness?

[@Erikamil1998]

I'm not a mathematician myself but i think you have to think what is making you solving this problem harder , and after identifying the problem , you think how to get rid of it by using a mathematical tool that suits the situation the best It was my question too , so if anyone else reading this comment knows s.th , please explain your views too

[@schrodingerbracat2927]

Note: x >_ 1 for both square roots to be valid. rewrite original equation as (1-1/x)^(1/2) = x - (x-1/x)^(1/2) Squaring leads to x² - x + 1 = 2sqrt(x² - x) Letting u = x² - x leads to (u-1)² = 0 so u=1 and we have x² - x - 1 = 0 solving, and remembering that x>_1 gives x = [1+sqrt(5)] / 2

[@ractmo]

If that plus would have multiplication it would be the most easy problem 😂

[@cube7353]

When I solved with log, I got a biquadratic equation which I could not solve. Please help by using a biquadratic equation.

[@goodplacetostop2973]

4:15

[@akshatjangra4167]

First penn and now presh

[@goodplacetostop2973]

@@akshatjangra4167 Yeah, I like watching other math channels from time to time lol

[@akshatjangra4167]

@@goodplacetostop2973 well also check out kyle broder,syber maths and maths elite

[@mr_angry_kiddo2560]

@@akshatjangra4167 I follow syber maths ,black pen red pen ,and MYD😍😍

[@karamsalah6256]

Magically, you make everything just simple! Great explanation!

[@lucasaugusto7777]

There is a very interesting way to solve this problem using geometry. since x is positive, we can construct in a triangle ABC with AB = x ^ (1 ÷ 2), AC = 1, and BC = x, so that by plotting the height length (1 ÷ x) ^ 1 ÷ 2, relative to BC, intersects it in D. Thus, the BC segment is divided into BD = (x-1 / x) ^ 1 ÷ 2 and DC = (1-1 / x) ^ 1 ÷ 2. triangle area is given by ((x). (1 / x) ^ 1 ÷ 2) / 2 = ((x) ^ 1 ÷ 2) / 2. But the area can also be calculated by (((x) ^ 1 ÷ 2 ).(1).sin (BAC)) / 2. thus equaling we obtain that sin (ABC) = 1, therefore the triangle is a rectangle in A. therefore, by Pythagoras, x ^ 2 = x + 1.

[@gabrielvaloes200]

Gooooood!!

[@andersonjuliao9451]

The man is very incredible!

[@kamillabarreto509]

Top. Great !!!

[@gilmaraguimaraes8550]

👍👍

[@gilmaraguimaraes8550]

Top my Love ♥️🥰

[@Harry-kc1qk]

go down u will find something interesting....... that is a commnet that is posted 6 days ago video was uploaded xD 😅

[@xlvii6994]

It still baffles me how you can use x-1/x at first with no exponent and yet use the whole thing later plus the exponent!!! Is that a trick to solve this type a problem and what is it called ? Thanks

[@TruthOfZ0]

So their fixed point f(x)=x is a point (x,x) where x=φ the golden ratio !!! ok

[@amdc2]

instead of implication you may (at each step) work ny equivalence : that means a≥0 and b≥0 and when you square an equation you add a sign condition ... then in the end (for sure you can still check if it you works fine to be sure but) you don't need to check, you have already eliminated the negative solution because of the conditions at each step... {TY for the video !}

[@williamadams137]

I also solved it, but the method in the video is way cleaner. From the original equation, we obtain (x - 1/x)^0.5 = x - (1 - 1/x)^0.5 . Squaring both sides and then simplifying, we get 2x·sqrt(1 - 1/x) = x^2 - x + 1 . Squaring both sides again we get 4x^2 · (1 - 1/x) = (x^2 - x + 1)^2 . After some simplification and moving everything to one side of the equation, we get x^4 - 2x^3 - x^2 + 2x + 1 = 0 ………(*) Knowing that x = 0 is not a solution to that equation, and inspired by Dr.Peyam’s video a few months ago on factoring “nice”polynomials, we can divide both sides of (*) by x^2 , obtaining the following: x^2 - 2x - 1 + 2/x + 1/x^2 = 0 (x^2 + 1/x^2 ) + (-2x + 2/x) - 1 = 0 (x - 1/x)^2 + 2 - 2(x - 1/x) - 1 = 0 (x - 1/x)^2 -2(x - 1/x) + 1 = 0 (x - 1/x + 1)^2 = 0 x - 1/x + 1 = 0 x^2 + x - 1 = 0 x = φ or x = -φ (reject) So only x = φ is the only real root to the original equation.

[@dickson3725]

You don't need to square both side again Line 2 can be simplified to (sqrt(x^2-x)-1)^2=0 x^2-x-1=0 The aswer is golden ratio but the smaller one doesnt work because x is negative and 1/x is negative

[@williamadams137]

@@dickson3725Oh i see! but it definitely seemed less obvious to me at first.

[@Elhaj778]

I didn't understand why a2-b2=x-1

[@SONUKUMAR-vr2jg]

No Gogu theorem!

[@ryujinzzz6050]

me, substituting x with 0 and 1 first: *task successfully failed*

[@androlsaibot]

I squared everything until I got x^4 - 2x^3 - x^2 + 2x + 1 = 0. Didn't see that this is (x²-x-1)², so I plotted the graph and saw it touches zero at 1.6-something. The rest was easy. Fun fact: this polynomial goes through (-1,1), (0,1), (1,1) and (2,1)

[@kasnarfburns210]

This just shows me how long ago I had studied math!!

[@Catman_321]

I looked at this problem and immediately knew it was the golden ratio because 1/phi=phi-1 so phi-1/phi is 1 and 1^0.5 is still 1 and 1-1/phi is phi^2 so this^0.5 is 1/phi 1+1/phi is phi

[@7ogical]

Is it true golden ratio is everywhere?

[@bot24032]

3:44 interesting fact: this solution showed up because if you take the first quantity (x-1/x)^½ with the minus sign, that would actually be true

[@reindorflance9378]

Quick question,where is this applied??

[@xtraPathshala]

your explanation is too smart.

[@JLvatron]

Nice solution, except why is it "divine" ?

[@WahranRai]

I were thinking to use the hyperbolic Pythagoras's theorem

[@JustAPersonWhoComments]

The answer is x^5 - 2x^4 - x^3 - 2x^2 + 4 + 5x = 0 The resulting equation is a quintic equation, which does not have a general algebraic solution for finding its roots. Therefore we need to solve it numerically using approximation methods such as numerical methods or graphing tools.

[@1Patient]

one of my all time favorite "Divine" moments in the art of mathematics is multiplication by Zero. I enjoy watching Presh's manipulation of the equation... my eyes get tired quickly watching the numbers flying across the equals sign 🤣😀

[@kvignesh7041]

We can do square on both sides procedure, we get this value

[@warpdrive9229]

Many don't know that the Golden Ratio in Fibonacci series was actually invented in India. Fibonacci discovered this Indian invention and like a gentleman he was, he has humbly accepted this and had written in his book that he did not invent the series and that it was an Indian invention.

[@arnauab25_03]

I like maths a lot, but I gave up. *NOW IMAGINE THE PEOPLE WHO HATE MATHS*

[@vishalmalviya89]

Same I like maths but couldn't do it , but I haven't gave up.

[@evehead713]

The only problem is not knowing basic calculus so i really don't understand a single thing you did.

[@HUEHUEUHEPony]

He didn't use any calculus

[@evehead713]

@@HUEHUEUHEPony what it is called? algebra or something

[@maxwellwright5292]

Very intelligent! Square root are not easy to manage

[@lesliecruzado2793]

I guess I was to much of a noob to think I could solve this one!

[@idrisdanismaz6155]

çok uğraştırıyon be reis bi tık kısa yap da millet anlasın

[@jimv9210]

Maybe someone has already pointed this out, but the equation also has the real solution -0.618033... (i.e., 1- phi). There's nothing in the presentation of the problem that excludes using the negative root of one of the terms, which in this case is the second term. I believe the other two combinations of +/- on the terms yields complex solutions, though I haven't worked them out.

[@aditya00078]

Me:- X will be nearly equal to 1.61

[@agustinvisini989]

You said "give it a try" like if it was that simple 😅. I'm pretty sure that I wouldn't be able to solve it for days if I tried to without watching the answer previously.

[@RihabSOUSSAN]

Yesh... I even tried to solve it because of how easy he made it look... also have you seen Attack on titan 's end... [spoiler without context:🐦]

[@agustinvisini989]

@@RihabSOUSSAN yes I've seen it. I thought it was going to be more impressive because AOT had the potential to have an unforgettable end, but I liked it by the way. I would've liked to have Jean, Connie, Gabi, Falco, Annie, Pieck, Reiner and Armin dead. Isayama made a lot of unnecesary deaths during the anime and in the final everyone live so I didn't like it very much

[@pyrite2060]

@@agustinvisini989 same, although tbf this was not the original ending as he changed it back in 2017. Apparently the original was much darker and more in line with the atmosphere in s3

[@孙林可]

I can get the first step, but I have a bad habit of not writing down the substitution so I guess I can't finish the whole solution.

[@arneshpal7702]

hi

[@parikshithkv2542]

How can i ask my question to presh Can someone tell??

[@bhushandhawane4755]

Email him

[@alihedayati5704]

The answer is wrong , it is out of our domain

[@ThatJosiahGuy]

Another interesting fact about the Golden Ratio, for those who don't know, is that it relates to the Fibonacci Sequence (1, 1, 2, 3, 5, 8, 13, 21...): φ = Lim (k → ∞): [k+1]/k

[@uvraviz113]

it ramanunjann formula not brahama goop

[@Aranwaar]

I found in 1 second because of the title. I just tryed x = 1 and thought: Maybe it's the golden ratio

[@PublicEnemynu1]

I think that , as a-b is equal the zero, we can not multiply both sides with 0. When we try the x value you find, it does not give the correct answer.

[@whendreamismathematics4709]

Not Brahmagupta Sridhar Acharya

[@ghyllezymethtagulalap9867]

Here I was thinking the answer was 1

[@blabla-hb4fe]

Ahh easy bro.. I solved it

[@peace7439]

Hey I solved it without video 👍

[@Amoeby]

Pretty simple equation. Solved it the same way, but didn't even do the substitution.

[@justnowi8967]

How tall is presh *Tall* walkar

[@nikhildev3869]

You could easily solve the question by taking log on both sides and getting the solution

[@MuffinsAPlenty]

I don't see how this would make it any easier. The logarithm of a sum doesn't have a "simpler" form.

[@jimcameron6803]

I multiplied out the quartic and got x^4 - 2x^3 - x^2 + 2x + 1 = 0. Which I totally failed to notice was a perfect square and instead factored as (x+1)(x)(x-1)(x-2) + 1 = 0. But the symmetry of that expression clued me in to the idea that x=1/2 was a significant point, so I substituted y = 1/2 - x and got y^4 - (5/2)y^2 + 25/16 = 0. Even I could spot that that's a perfect square: (y^2 - 5/4)^2 = 0 and taking the negative square root gives y = -sqrt(5)/2 and x = (1+sqrt(5))/2 as required.

[@adalbertoantelami9434]

You are a genius

[@CRSGAMEZONER]

Love from Kerela Love from India