Solving x=sqrt(x-1/x)+sqrt(1-1/x) Math Stack Exchange
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- Опубликовано: 7 фев 2025
- I saw this equation with two square roots on Math Stack Exchange and I have a very interesting "triangle" solution to offer. The equation is sqrt(x-1/x)+sqrt(1-1/x)=x and we want to find the real solution for it. Instead of multiplying by the conjugate, we can set two right triangles and use the Pythagorean Theorem to create a quadratic equation. Ultimately, you will see the solution is the golden ratio and you CAN'T get more satisfying than this!
Read other solutions: math.stackexch...
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Solve sqrt(5-x)=5-x^2: ruclips.net/video/BO1T7ebJlO8/видео.htmlsi=aG61fbzPAPJC6Dh3
asnwer=-1 ! asnwer= 1 isit
ans wer=1+-/5 /2 isit
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❤😢.
@@축복-l1l sqrt means "square root". sqrt(4)=2
Sir please make a video on use of and , or in mathematics .means its meaning like in set theory or means union and and meaning intersection.in probability different so please make a video on that
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.
The elegant solutions just feel incredible
I totally agree!
Sir please make a video on use of and , or in mathematics .means its meaning like in set theory or means union and and meaning intersection.in probability different so please make a video on that.
@@blackpenredpen check my comment too please
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*
Is this code? Are yall aliens comunicating😂
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
@@BlackRockrocksnot 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
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,∞)
@@NotGleSkithe same people will join a cult and start movements against science
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.
Had the same thought, but its lovely once you seen it once.
@@ingiford175
What about inverse Pythagorean theorem?
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.
Another way of proving it is equal to 90 degrees:
sin(a+b)=...=x/x = 1 where a+b add up to that angle
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
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) )
i agree
it can be a way, but theres nothing "best" here
That was such an intriguing way to solve it! Thanks a lot again!!
4:58
"I will keep this on the right hand side [...], but i'm gonna put it on the left" ~ blackpenredpen, 2024 😂
This was a delight to watch.
These neat tricks are the essence of solving math problems. Trying to develop this kind of approach now.
Actually, you can square both sides. This leads to a big expression BUT then you can make the substitution y = x^3 - x^2 - x + 1. This quickly yields y = 1 as the only solution. That means 1 = X^3 - x^2 - x + 1, which leads to the same equation solved in the video.
this is the second time youve made a video about using triangles to solve equations and i love it
We can multiply by put a = sqrt(x-1/x) and b = sqrt(1-1/x). So x = a + b.
Multiply the equation by a - b gives x-1/x = a - b, because (a + b)*(a - b) = x - 1.
Substract the second to the first gives 1/x = 2b = 2sqrt(1-1/x) or (1/x+1)^2 = 5
The 2 triangles was a cool way to do that problem. l new that ø or Ø were related to that, but l didnd't know how to show that from scratch. And the way you proved that the triangle was right also blew my mind.
Honestly, that's beautiful, you take a complicated equation and reduce it to something simple.
I love how the videos lately have been about clever tricks to avoid using calculus. Really love it.
Many thanks for teaching us this unique approach, leading to an absolutely brilliant solution. I love this video.
Reverse engineering the area formula to verify perpendicular sides is unexpected yet beautiful. Makes me think this problem was specifically designed for that.
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
I like this one so much i watched it twice. Very nice.
This is such a cool approach. I love it!
I did try the ordinary squaring method first but of course this led to an equation of a much higher degree than we want to see. (Wolfram Alpha spat out the result but that's hardly a satisfying approach.)
Your method to turn this into triangles is supercool!
I stopped it there and tried using the geometric mean theorem. (Actually had to look up the English name. In my native German, this one is called "altitude theorem".)
Didn't know if this would be useful but I thought that multiplying the two roots should get a nicer result that adding them.
Since the altitude is 1/√x, its square is simply 1/x; as the product of the two roots also results in a denominator x this cancels out neatly.
And because on the right hand side, only 1 remains, I could then square the entire equation to get rid of the roots on the left hand side.
This gave me a cubic equation but the constant term canceled out, so the whole thing could be divided by x and which resulted in the Golden Ratio equation x²-x-1=0.
So this would have worked fine, but your approach determining that the large triangle is also a right triangle obviously is way more elegant. (But it only works if it *_is_* indeed a right triangle whereas using the geometric mean theorem would also work for non-right triangles.)
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.
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!
Beautiful stuff. Thank you for sharing.
Great approach! I used:
let u = (x - 1/x) and v = (1 - 1/x)
❶ (√u + √v) = x
(√u + √v)(√u - √v) = x(√u - √v)
(u - v) = x(√u - √v)
(x - 1) = x(√u - √v)
(√u - √v) = (x - 1)/x
(√u - √v) = 1 - 1/x
❷ (√u - √v) = v
add ❶ & ❷together:
❶ (√u + √v) = x
❷ (√u - √v) = v
2√u = x + v
2√u = x + 1 - 1/x
2√u = x - 1/x + 1
2√u = u + 1
u - 2√u + 1 = 0
(√u - 1)² = 0
u = 1
x - 1/x = 1
x² - x - 1 = 0
x = (1 ± √5)/2, but x > 0
x = (1 + √5)/2
Damn nice approach
Good tecnique!
Now, I will make a little mental exercise here: (just for curiosity)
n=(x-(1/x))½
n²+b²=a² or
n²= a²-b²
n²=a²-b²
x-(1/x)=a²-b²
(x½)² - (1/x½)² = a²-b²
consider that way
=> a=x½ and b=1/x½
m=(1-(1/x))½
m²+c²=d²
m²=d²-c²
1-(1/x)=d²-c²
the same for m,c,d:
Then, c=1/x½ and d=1
Note: b=c; (we'll make h= b=c)
And we will considered only the positive values for a,b,c,d.
x=n+m
x²=n²+2mn+m²
=> x² = (a²-b²) + 2(a²-b²)½(d²-c²)½ + (d²-c²)
=> x² = (a²+d²) - (2h²)+2[a²d²-a²h²-d²h²+h⁴]½
make H²=a²+d²
=> x² = H²-2h²+2[a²d²-H²h²+h⁴]½
=> x²-H²=-2h²+2[(ad)²-(Hh)²+h⁴]½
make A=Hh/2 and S=ad/2
=> x²-H²=-2h²+2[(2S)²-(2A)²+h⁴]½
supose S=A
//////////////////////////////////////////////////
If, S=A {then, Hh=ad => H²h²=a²d²=> (a²+d²)h²=a²d² => (1/d²)+(1/a²) =(1/h²) => 1+(1/x)=x => (x+1)/x=x => x²= x+1 => x = (x+1)½
but, x=n+m
=> n²+2mn+m²=x+1
=> x-(1/x)+2mn+1-(1/x)=x+1
=> 2mn=2/x
=> mn = 1/x
=> m²n²= 1/x²
=> [1-(1/x)][x-(1/x)] = 1/x²
=> x-(1/x)-1+(1/x²) = 1/x²
=> x-(1/x)-1=0
=> x-1= 1/x; x ≠ 0
=> x² -x -1 = 0
=> x= ф or x= [1-(5½)]/2
That means the veracity of our supose imply that x must be equal these values to the next equation be true.
///////////////////////////////////////////////////
next => x²-H²=-2h²+2[h⁴]½
=> x²-H²=-2h²+2h²
=> x²-H²=0
So, if A=S, then:
=> x²=H²
but, H²=a²+d²
=> H²=(x½)²+(1)²
=> H²=1+x
=> x²=H²
=> x²=(1+x)
=> x²-x-1=0
=> x=ф or x=[1-(5½)]/2
Considering x > 0, then:
=> x = ф
The solutions are the same, therefore the method used to solve it is true. 😊
You're an inspiration!
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
You don't need to check a lot to verify the top angle of the big triangle, simply notice that the sum of the rest of the 2 angles of the big triangle is 90 degrees, so the top angle definitely will be 90°.
When I saw the thumbnail, I thought this equation was an unexpected identity lol
Simply awesome!!
I like the triangle solution.
Please do the limit as x -> inf of Γ(x)/subfactorial(n-1) 🙏
cool solution!
Amazing method 😅😅😅😊😊😊
It's very creative. awesome!
"x is so much cuter compared to that right" 🤣
Helpful, Thanks you sir.
Life saver!
This is certainly very neat. However, at 2:00 you are implicitly assuming that x>1, because otherwise you could not form that triangle. Now, there aren't any solutions with x
That's so cool!
Multiply both sides of sqrt(x-1/x) + sqrt(1-1/x) = x by sqrt(x-1/x) - sqrt(1-1/x). Use difference of squares to simplify and then divide by x on both sides to get sqrt(x-1/x) + sqrt(1-1/x) = 1-1/x. Add this to the original equation to get 2sqrt(x-1/x) = x-1/x+1. Set u=x-1/x to get 2sqrt(u) = u + 1. This is the same as (sqrt(u)-1)^2=0. So x-1/x=u=1. Therefore, we have x^2-x-1=0. This has roots (1 +sqrt(5)) /2 and (1-sqrt(5))/2. Reject negative x since the original equation tells us that x is nonnegative.
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.
The area is ½ab*sin(C), and if sin(C) = 1, then C = 90°
@angel-ig good catch
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!
what ap class is making you factor a quintic?
Bro the area of the triangle part was kinda guessing
Can you give a more generalised method?
Anyway big fan
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.
I used sin theorem to find that the angle is 90 i called the angle between 1and sqrt(1-1/x) a that made the one thats on the other side 90-a and did the same with the other side so i got 90-b then said that sin(180-a-b)/x is gonna equal sina/sqrt(x) then just found sina which was 1/sqrt(x) so when you substitute it'll be 1/sqrt(x) times 1/sqrt(x)= sin(180-a-b)/x so 1/x= sin(180-a-b)/x so sin(180-a-b)=1 meaning sin(a+b) is 1 and since both of these angles are smaller than 90 their addition must just be 90 😊
Alright, that was pretty clever!
Let (x-1/x)^1/2=a and (1-1/x)^1/.2=b. Then, a+b=x and a^2-b^2=x-1. So, a-b=1-1/x. Thus, 2a=1+x-1/x=1+a^2. So, a=1, i.e., x-1/x=1 and so x=1/2[√5+1], as x is positive.
What about doing this with the trigonometry
That's the first thing (GOLDEN RATIO ) come in my mind
Wow, that was crazy
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?
So if c,d,e>0 then SQRT(ec/d)x = SQRT(cx-d/x) + SQRT(e-d/x) should solve the same way giving x = (d/e + SQRT(d^2/e^2 + 4c/d))/2, right?
I allways think every problem with phi as the solution is a good problem
Me, an engineering student: ITERATIVE METHOD, GO!
lim (ln(x)-W(x)) ,x→♾️
W(x) is the Lambert W function please do this limit 🙏
By similar triangles, that angle is 90° by default lol.
Can you elaborate?
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.
@@trueriver1950don’t overthink it. Your low IQ is showing
Hello bprp, how can I evaluate this integral
(2xsin(x))/(3+cos(2x)).dx from 0 to pi. ?
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?
"Asked 11 years ago" lol you're late
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)
I always get so confused when he breaks out the dreaded BLUE pen. 😁
Bruh 11 yrs ago how did you find this
Suggested posts after suggested posts
Golden
is it possibe to solve sqrt(x^2-1)+sqrt(x-1)
can someone explain how we know the topvangle is 90 degrees
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°.
*reads half of title*
ILLUMINATUS COMFIRMED
Squaring both sides is my automatic reflex, lol, but the title of the video seems to discourage that...but, squaring both sides, would create only one radical, then put it alone on one side and square both sides again, lol, and then keeping track of what had been done the problem should be easy to solve...as for without doing that, lol, hmm...
Cool!
so cool
1.618
let sqrt(x-1/x)=a, sqrt(1-1/x)=b ( a,b>0 or a,b are not real number (ig) )
then a²-b²=x-1/x-(1-1/x)=x-1
and a+b=x
which is urghh same
omg how do people think of these!?!?!
That's a clever solve - can't just do "algebra autopilot".
ChatGPT is struggling with it
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.
yes
Yep. It works simply because it's the golden ratio.
No. That’s not a coincidence. That is, because it was set up that way. Simple as that.
Bad logic
Then you would use the sine area formula and cosine theorem instead of the pythagorean theorem, it would require trigonometry knowledge though, and that angle being 90 degrees makes it easier for viewers
Nice
how to focus in Math
I don't think big triangle is a right triangle, the product of red and blue parts must be equal to 1/x.
I didn't understand why x cannot be negative.
In the original question, x is equal to the sum of two square roots which can never be negative so x can neither.
❤
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❤😢
"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.
My guess is that you found an infinite sume that in the limit should converge to tan 90, except it does not converge.
@bjornfeuerbacher5514 well if you could think of it some outer way u will get it, lol I can't even u show u the proof here
@Theraot well ur guess I kind of wrong
I did use a tan sum formula for an n number of angles and some of the patterns I could see in those numbers, I can't explain it here
@@Pramod-y1g " if you could think of it some outer way u will get it"
No. By _definition_, the value of tan 90° is infinite. If you get a finite value, your calculation is simply wrong. It's that simple. There is no "outer [other???] way" in this. A value cannot be infinite and finite at the same time, that's self-contradictory.
Plug it into Newton's method or bisection method and be done with it xd
It's always the golden ratio😪
blackpenredpenbluepen
sqrt(x - 1/x) + sqrt(1 - 1/x) = x
so x - 2/x + 2sqrt[(x-1/x)(1-1/x)] + 1 = x^2
so 2sqrt[x - 1 - 1/x + 1/x^2] = x^2 - x - 1 + 2/x
so 4x - 4 - 4/x + 4/x^2 = x^4 - x^3 - x^2 + 2x - x^3 + x^2 + x - 2 - x^2 + x + 1 - 2/x + 2x - 2 - 2/x + 4/x^2
so 4x - 4 - 4/x + 4/x^2 = x^4 - 2x^3 - x^2 + 6x - 3 - 4/x + 4/x^2
so x^4 - 2x^3 - x^2 + 2x + 1 = 0
so (x^2 - x - 1)^2 = 0
so x = (1+sqrt5)/2
😄
I will find god
охренеть.
лайк!
Second comment daddy ❤
Always love for u 💗
Write sentences.
The golden ratio? What are the chances? I think this equation was rigged.
i didnt asked
*ask
@@endersteph i didnt asked
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.
think that the domain of this equation must be
x ∈ R-{0}