Believe in triangles, not squaring both sides!

[@blackpenredpen]

Solve sqrt(5-x)=5-x^2: kzbin.info/www/bejne/eICUhWqbl6-fhZosi=aG61fbzPAPJC6Dh3

[@와우-m1y]

asnwer=-1 ! asnwer= 1 isit

[@와우-m1y]

ans wer=1+-/5 /2 isit

[@Pramod-y1g]

Bro I need help , I discovered a way to depict tan 90 as a finite value and I want some help in approximating the value using sums I used some square method and some patterns . How can I send u the proof. I wish if u can help me on this. How can I show it to u❤😢.

[@MarieAnne.]

Solution by squaring both sides x = √(x − 1/x) + √(1 − 1/x) First, we multiply both sides by √x x√x = √(x² − 1) + √(x − 1) Next we square both sides: x³ = (x² − 1) + (x − 1) + 2 √((x² − 1)(x − 1)) (x³ − x² − x + 1) + 1 = 2 √(x³ − x² − x + 1) Use a substitution: u = √(x³ − x² − x + 1) u² + 1 = 2u u² − 2u + 1 = (u − 1)² = 0 u = 1 Substitute back √(x³ − x² − x + 1) = 1 x³ − x² − x + 1 = 1 x³ − x² − x = 0 x (x² − x − 1) = 0 x = 0, x = (1 ± √5) / 2 From original equation, we can see that x ≠ 0, and x is not negative *x = (1 + √5) / 2 = φ*

[@eyeswiideshut3289]

Is this code? Are yall aliens comunicating😂

[@NevinBR]

Perhaps slightly easier if you subtract √(1 - 1/x) from both sides at the very start: x - √(1 - 1/x) = √(x - 1/x) Square both sides: x² + 1 - 1/x - 2√(x² - x) = x - 1/x Bring everything over to the left side: x² - x + 1 - 2√(x² - x) = 0 Let u = √(x² - x): u² + 1 - 2u = 0 (u - 1)² = 0 u = 1 Substitute back into the definition of u: 1 = √(x² - x) 1 = x² - x x² - x - 1 = 0

[@NotGleSki]

@@eyeswiideshut3289not a flex that you can’t do basic algebra I mean the thought of a 12 year old being smarter then me is terrifying but ig it’s comforting for someone like you

[@marilynman]

The interesting thing about the function f(x) = sqrt(x - 1 / x) + sqrt(1 - 1 / x) - x is that considering the part sqrt(x - 1 / x) has zeros in x = +-1 but is asymptotic at x = 0 so the domain is [-1,0)U[1,∞). The other part sqrt(1 - 1 / x) has zeros at x = 1 and it's also asymptotic at x = 0 so the domain is (-∞,0)U[1,∞) and the domain of x is (-∞,∞). So the domain of the entire function is [-1,0)U[1,∞)

[@Sandpepe]

​@@NotGleSkithe same people will join a cult and start movements against science

[@ianfowler9340]

This is the kind of stuff that really makes algebra and geometry just so much damn fun. We all know the connections are there, even when they not obvious. But when we actually see/discover one in action you just have to sit back and smile - it has an almost "magical" quality about it. Well done.

[@John-cl8iv2]

The elegant solutions just feel incredible

[@blackpenredpen]

I totally agree!

[@daminkon246]

Checking for the 90 degrees using the area of the triangle, man, i never would have thought about that. Such a nice equation, thank you.

[@ingiford175]

Had the same thought, but its lovely once you seen it once.

[@conanedojawa4538]

​@@ingiford175 What about inverse Pythagorean theorem?

[@BrianGriffin83]

Me neither. I was struggling to prove it, and the answer was right in front of my nose. And even after I saw the video I was still wrapping my mind around it. Maybe the most rigorous way to write it down would be to call a the angle in question and write 1/2*1*√x*sin(a) on the right side of the equation, then solve for a.

[@Mathematiker_KV]

Method without squaring: x = √(x − 1/x) + √(1 − 1/x) Taking reciprocals on both sides and multiplying by the conjugate on the RHS simplifies to: 1/x = [ √(x − 1/x) - √(1 − 1/x) ] / (x-1) So: 1 - 1/x = √(x − 1/x) - √(1 − 1/x) Adding to the original equation: 2√(x − 1/x) = x - 1/x + 1 Setting X = x - 1/x and solving the quadratic yields: X = 1 x - 1/x = 1 Solve to get x = (1 + √5) / 2 since x > 0

[@tanishq_xin]

That was such an intriguing way to solve it! Thanks a lot again!!

[@MikeGz92]

To check if the triangle is rectangular, I think the best way that the triangle on the right is similar to the biggest triangle ( 1/sqrt(x) : 1 = 1 : sqrt(x) )

[@shantanudhiman8194]

This was a delight to watch. These neat tricks are the essence of solving math problems. Trying to develop this kind of approach now.

[@seanoneill2098]

Beautiful stuff. Thank you for sharing.

[@sternli728]

4:58 "I will keep this on the right hand side [...], but i'm gonna put it on the left" ~ blackpenredpen, 2024 😂

[@shannonmcdonald7584]

I like this one so much i watched it twice. Very nice.

[@snejpu2508]

After squaring both sides, -1/x cancels out on both sides and we can write the equation as 2x*sqrt(1-1/x)=x^2-x+1. Then we can square both sides again and get 4x^2-4x=(x^2-x+1)^2. Now we can substitute t=x^2-x and we get 4t=(t+1)^2, 4t=t^2+2t+1, t^2-2t+1=0, (t-1)^2=0, t=1, x^2-x=1, x^2-x-1=0. After solving the quadratic we get the same solution.

[@orangee_blox]

this is the second time youve made a video about using triangles to solve equations and i love it

[@SpaceUA1]

Please do the limit as x -> inf of Γ(x)/subfactorial(n-1) 🙏

[@АртемХапилов-г2ф]

Wanted to check if there are other roots because after squaring two times an equation, we would have 6th degree polynomial. But there are 2 x=0 roots and equation simplified to (x^2-x-1)^2=0. So then the answer is correct, yeah, thanks for the cool solution!

[@indefinite115]

You're an inspiration!

[@bain8renn]

very interesting problem, always very cool when problem-solving involve using geometric intuition, even when its not a geometric problem in the first place!!! pretty cool :3

[@pietergeerkens6324]

I definitely DID want to square both sides; as this gave me first x² = x - 1/x + 1 - 1/x + 2√[ x - 1 - 1/x + 1/x² ] and then, multiplying through by x and rearranging, x³ - x² - x + 2 = 2√[x³ - x² - x + 1]. The latter is easily reorganized as just (√[x³ - x² - x + 1] - 1)² = 0, hence √[x³ - x² - x + 1] - 1 = 0. and x³ - x² - x + 1 = 1 to yield x² - x - 1 = 0 since x = 0 is forbidden in the original presentation. This of course has the well known roots φ and 1/φ, only the first satisfying the original presentation.

[@SidneiMV]

Simply awesome!!

[@inuyasha5521]

It's very creative. awesome!

[@Wakrar]

For checking if the angle is 90°, it would probably me more adequate to write the are formula as a.b.sin(t)/2, and then verifying that sin(t)=1. But that aside, super neat resolution

[@learnscience1250]

Helpful, Thanks you sir.

[@marktikhonov8495]

Question: You are relying on a conjecture that if product of two sides of a triangle a and b is a*b=0.5 * A where A is the area of the rectangle implies that the triangle is a right triangle, but is it really the case? I mean that certainly seems logical, but I have never seen this as a statement.

[@angel-ig]

The area is ½ab*sin(C), and if sin(C) = 1, then C = 90°

[@marktikhonov8495]

@angel-ig good catch

[@homewasus]

Life saver!

[@happy.5]

That's the first thing (GOLDEN RATIO ) come in my mind

[@Viki13]

cool solution!

[@rakib_hoque]

lim (ln(x)-W(x)) ,x→♾️ W(x) is the Lambert W function please do this limit 🙏

[@richardfredlund8846]

can you solve x^3 - 3x = A^3+ 1/A^3 where A = (x + - sqrt(x^2 - 4)) /2 ? answer this is true for all x in the reals although if you plot the r.h.s you will only get the tails on desmos because the intermediate values are complex in the range (-2,2)

[@DMichigan]

That's so cool!

[@g0rgth3b0rg]

I like the triangle solution.

[@kaz7953]

When I saw the thumbnail, I thought this equation was an unexpected identity lol

[@YourFriendlyAlan]

Alright, that was pretty clever!

[@barryzeeberg3672]

Can this be generalized by seeing if there are other combinations of expressions that work out this way, or is this the only one that works? Are there other expressions that make the combined angle into a right angle?

[@Eichro]

Isn't it dumb luck that the (x - sqrt(x) - 1) triangle turned out to be right? You replace that 1 with a 2 and the whole trick falls apart.

[@Kishblockpro]

yes

[@ethohalfslab]

Yep. It works simply because it's the golden ratio.

[@JDO-u5f]

No. That’s not a coincidence. That is, because it was set up that way. Simple as that.

[@Masteg.]

is it possibe to solve sqrt(x^2-1)+sqrt(x-1)

[@Davics02]

Hello bprp, how can I evaluate this integral (2xsin(x))/(3+cos(2x)).dx from 0 to pi. ?

[@fabianortiz4200]

hey! I've been stuck on this equation I made up when I was bored, 10^x = x^2 I tried taking the natural logarithm of both sides but I can't seem to solve it can you help?

[@proguyz78]

Golden

[@JayFang-i6p]

"x is so much cuter compared to that right" 🤣

[@herbie_the_hillbillie_goat]

I always get so confused when he breaks out the dreaded BLUE pen. 😁

[@MIIIM-7]

1.618

[@medaliberrada8285]

can someone explain how we know the topvangle is 90 degrees

[@ناصريناصر-س4ب]

We calculate the area of triangle ABC in two different ways. The first is [ABC]=(AB*AC*sinA)/2=(1*√x*sinA)/2=(√x*sinA)/2. The second is [ABC]=(BC*AH)/2=(x*1/√x)/2=√x/2 where AH is the height. Comparing the two methods, we find that sinA=1, so A=90°.

[@captnmaico6776]

so cool

[@DuyThanh-ff7dl]

Bruh 11 yrs ago how did you find this

[@blackpenredpen]

Suggested posts after suggested posts

[@scottleung9587]

Cool!

[@penguincute3564]

By similar triangles, that angle is 90° by default lol.

[@beansprugget2505]

Can you elaborate?

[@trueriver1950]

​​​ yeah, but how do you know the triangles are similar? (You are right that they are, but tell me how you know they are) The method shown was to compute the area two ways, which seems to me quicker than any geometric argument that says that all three triangles are similar.

[@daniel_77.]

"Asked 11 years ago" lol you're late

[@eyeswiideshut3289]

*reads half of title* ILLUMINATUS COMFIRMED

[@FireStormOOO_]

That's a clever solve - can't just do "algebra autopilot".

[@cdkw2]

omg how do people think of these!?!?!

[@jwangosho]

ChatGPT is struggling with it

[@SurendraReddy-ey5wv]

Nice

[@oneshot7456]

I had an AP maths exam today and one question asked to find the other factor of a quintic equation for 7 marks. I would have had no idea how to do it had I not watched this guy. Got the idea to brute force it and solve for each coefficient of x individually from one of your videos. You’re a life saver bro!

[@user-db4lk7yg3o]

what ap class is making you factor a quintic?

[@Traw-ve7qf]

how to focus in Math

[@emontrailers]

❤

[@papakwamekuttin2978]

It's always the golden ratio😪

[@cuongvd]

blackpenredpenbluepen

[@Pramod-y1g]

Bro I need help , I discovered a way to depict tan 90 as a finite value and I want some help in approximating the value using sums I used some square method and some patterns . How can I send u the proof. I wish if u can help me on this. How can I show it to u❤😢

[@bjornfeuerbacher5514]

"I discovered a way to depict tan 90 as a finite value" That makes no sense. Using the _definition_ of tan, it's easy to see that no such finite value exists. I. e. your calculation obviously is wrong somewhere.

[@Theraot]

My guess is that you found an infinite sume that in the limit should converge to tan 90, except it does not converge.

[@6310-c5h]

охренеть. лайк!

[@SigfriedNothung]

😄

[@gabest4]

The golden ratio? What are the chances? I think this equation was rigged.

[@nyclegendz4665]

Second comment daddy ❤ Always love for u 💗

[@forcelifeforce]

Write sentences.

[@pue0]

i didnt asked

[@endersteph]

*ask

[@pue0]

@@endersteph i didnt asked

[@rattyoman]

hellow, im somewhat curious about something and haven't found anything about it; what is the relation between a sine wave and a circular wave (as in, a wave made from opposing semicircles)? I know sine and cosine can form a circle together, and that they're closely related to π, so I'm curious how relevant that is to a circular wave.

[@yogeshkumar69693]

think that the domain of this equation must be x ∈ R-{0}