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.
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.
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)
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)
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)
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 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
@@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
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
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.
@@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
@@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.
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.
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. ;)
@@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 is more or less standard approach, however, we got very lucky 2 (!) times, otherwise we would have to deal with sixth degree polynomial equation...
@@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 :)
@@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.
Just another solution : Take y = 1/x Then we have - √(x-y) +√(1-y) = x √(x-y) = x - √(1-y) Squaring both sides x - y = x² + (1-y) -2x√(1-y) Cancelling y from both side and rearranging the terms - x² - x + 1 = 2x√(1-y) Dividing both sides by x - x - 1 + y = 2√(1-y) [•.• y = 1/x] Again squaring both sides - x² + 1 + y² -2x -2y + 2 = 4(1-y). [•.•xy =1] Simplifying the terms - x² + 1 + y² -2x + 2y - 2 = 0 => ( -x + 1 + y )² = 0 [•.• 2 = 2xy] => -x + 1 + y = 0 Replacing y by 1/x - -x + 1 + 1/x = 0 => -x² + x +1 = 0 => x² - x - 1 = 0 Rest follows. I think this is straight forward / conventional solution. Taking y = 1/x was only for simplification.
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 !}
Bruuuuh my math knowledge and solving is like hell and I'm in 10 standard. I hate math related things but dude!! When i saw your video in my recommendation it looked interesting and the looks exactly matched the content. You are the first one in my life to make me see maths like a entertaining subject. Keep it up man the videos are awesome!!
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
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.
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.
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.
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?
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!
Nice one. I have not read all the comments. I tackled it a little differently. Started with your equation and then you can invert both sides. 1/x = a-b/(x-1) so 1 - 1/x = a -b add to the original equation and substitute x - 1/x = a again, you get a^2+1 = 2a . so a=1 and you can get values of x as you did
I mentally visualized the graphs of the parts and estimated any possible solution was near two and three up to five. I suppose 1.6 is close enough to two to feel like a good estimate for mental work.
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
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.
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.
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
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?
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
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 🤣😀
You would like this question: Let f:N×N->N be a function such that f(1,1)=2 , f(alpha+1,beta)=f(alpha,beta) + alpha and f(alpha,beta +1)=f(alpha,beta) - beta, for all alpha,beta belongs to N and f(a,b)=2001 then find the number of ordered pairs (a,b) is?
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!!
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.
We can simply rearrange the expression as (x-1/X)^1/2 = x-(1-1/X)^1/2 and square both sides. After simplification we end up with x²-x-1=0 and can find the solution
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.
There are actually 2 solutions. The solution (1-sqrt(5))/2 = -0.618... also solves the original equation if you accept that the second bracket to the power of 1/2 can be the negative value of the "calculator output" you get (1.618). Therefore: 1 - 1.618 = -0.618 --> also solves the equation
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.
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)
Not sure if this is a longer path but I had got the solution by substituting x=sec^2(y) Simplifying the equation could get sin(y)=cos^2(y). Solving we get 2 values for sin(y) : (-1-√5)/2 and (√5-1)/2 Sine would only be (√5-1)/2 Now x=sec^2(y) = 1/cos^2(y) We already have sin(y)=cos^2(y) So x= 1/sin(y) Substituting sine of y value we get x = (√5+1)/2
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
This would have a lot more value if you explained what line of reasoning can lead to that specific choice of equation manipulations. As is, this video proves that your solution is correct, but it doesn't teach much about problem-solving. Would you please consider including this kind of explanation in future videos?
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.
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)
My solution : Take (1-1/x)^1/2 to rhs and square both sides. We're left with x = x^2 +1 - 2(x^2 - x) ^1/2 Take x^2 to lhs and multiply both sides by - 1. Now the equation is : (x^2 - x) = 2(x^2 - x)^1/2 - 1 Take (x^2 - x) ^1/2 =t and solve the quadratic in t. It becomes the same equation as in the video x^2 - x - 1 =0 And it's solution is the Golden Ratio.
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)
Objection. A square root can be a negative number. Example, square rooting of 9 is positive or negative 3. Therefore, the 1 - squareroot(5) divided by 2 answer is legitimate. You just need to take the positive square root for the first square root function but the negative square root for the second.
Squaring both sides of the equation twice, and dividing by 𝑥² (𝑥 = 0 is not a solution anyway) gave me a 4th degree polynomial equation. Graphing the polynomial I realized that it has two double-roots, and I could write it as (𝑥 + 𝑎)²(𝑥 + 𝑏)² Expanding this expression and comparing it to the polynomial I could quite easily solve for 𝑎 and 𝑏, giving me 𝑥 = (1±√5) ∕ 2 Then, 𝑥 ≥ 0 ⇒ 𝑥 = (1 + √5) ∕ 2
A straightforward way is to first square both sides, and then assume a^2=x^3-x^2-x+1 to get an expression of a^2-2a +1=0. This means (a-1)^2=0, and a=a^2=1. Then one quickly gets x^3-x^2-x+1=1. Notice that x must be greater than 0, the solution to this expression can only be (1+sqrt(5)/2.
There is an easier way of solving this. After completing some steps, the equation will be alike this- root(x(sqaure)-1)+ root(x-1)=x root x or root(x+1)+1=x root x/root(x-1). After squaring the equation and completing some steps, we will reach this equation- 2 root(x+1)=(x)cube/(x-1) - (x+2) or 2 root(x+1)=((x)cube-(x)square-x+2)/x-1. We will square this equation and will get this equation- 4(x+1)=((x)cube-(x)square-x+2)square/(x-1)square or 4(x+1)(x-1)square=((x)cube-(x)square-x+2)square or 4((x)cube-(x)square-x+1)=((x)cube-(x)square-x+1)+1)square or 4((x)cube-(x)square-x+1)=((x)cube-(x)square-x+1)square +2((x)cube-(x)square-x+1)+1 or 2((x)cube-(x)square-x+1)=((x)cube-(x)square-x+1)square+1 or 0=((x)cube-(x)square-x+1)square- 2((x)cube-(x)square-x+1)+1 or ((x)cube-(x)square-x)square=0 or (x)sqaure-x-1=0 or x= the shown result.
I bruteforced this by eliminating the ^(1/2)'s by squaring until I was left with having to find the roots of 4th degree polynomial, which I realized was the square of (x^2-x-1) after some trial and error.
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
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.
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)
The other possibility can br checked trivially, by seeing that as square root is positive when we are talking about real numbers, so LHS is positive so RHS too.
Yeah I haven't done any formal algebra for probably 40 years. I wrote down the problem and approached it like I would have 45 years ago when I was 15. Took a bit of thought, but in 7 lines and maybe 10 minutes - with NO elaborate substitution tricks - I had the quadratic x**2 - x - 1 = 0. I feel that using your "transform" method you add a level of complexity that no-one really needs, plus you give the impression that it's either an easier or indeed the only way to solve a relatively straightforward algebraic problem.
@@londersito Yikes 😣 I scratched out the piece of paper and it looks like I made a fundamental error on the first line. But my handwriting and "cancelling out" and corrections all make it fairly illegible so I'm going to have to work it all out again. Tomorrow, it's bedtime now!
@@londersito Oops. I started by squaring one side of the equation and not the other, so it's interesting I ended up with the right result. Must have made another stuffup further down. Embarrassing.
Ok i solved this, by squaring the equation and leaving the 2•(...)^1/2•(..)^1/2 alone and squaring up again. In the final form you are left with various degrees of x going up to x^5 which you need horner's theorem to solve but it for some reason to me it feels more intuitive and easier than the solution presented.
maybe i just got lucky this time but i got it right! i solved it differently though: since x^(1/2) is equal to sqrt(x) I just took the whole equation ^2 then you're left with x +(1/x) + 1-(1/x)=x +1/x and -1/x cancel each other out so you have x+1=x^2 and then my steps are the same as the video :)
(1 +/- sqrt(5))/2 I don't see why it's 'tricky' Moderate, at best. Let's multiply both parts on the expression in the left part but with '-' instead of '+' We get two equations, new 1 and initial 2 1.sqrt(x-1/x) - sqrt(1-1/x) = 1 - 1/x 2.sqrt(x-1/x) + sqrt(1-1/x) = x Add them it will be 2sqrt(x -1/x) = x -1/x +1 Which is, if we put y instead of sqrt(x-1/x) y**2 - 2y +1 = 0 (y-1)**2 =0 so x-1/x-1 =0 -> x**2 - x -1 =0 x = (1 +/- sqrt (5))/2
1:39 "We will then divide both sides of this equation by x" If I recall correctly this is not allowed? I remember being taught in school that we must never divide by x when solving an equation. The explanation back then was that we do not know whether x is 0, so dividing by it might take away some solutions. Would somebody please enlighten my why this is possible here?
@@施柏安-b1k Yes, but you can always get the term 1/x in any equation be rearranging and modifying some terms, e.g.: x - 1 = 0 now factor out x: x * (1 - 1/x) = 0 => There is a term 1/x in the equation. In this simple example, we know that x is not 0. But in a more complex example the presence of a term 1/x does not imply that x cannot be 0.
@@Caledoriv From your example, My opinion is that you can only factorize it like that when x not equals to 0. 1*(x-1) = 0 What you do is times x to the first term (1), and divides with x to the second term (x-1). Because you do the step "/x", you need to exclude the condition that "x=0". So, the more accurate way is => x * (1 - 1/x) = 0 for "x not equals to 0" and discuss x=0 respectively. Another example is 2*5=10, If we write it like (2*x)(5/x)=10, x can only be nonzero, or it will be really weird. So, back to the question. If we believe that when the question creator divides with x, he always considers the step I mentioned, then the question won't appear "1/x". So if the question appears "1/x", we will know that x can not be 0.
You've got it wrong. First, you *may* divide by x. Think, if you have an equation in the form x f(x) = 0, it implies either x = 0 or f(x) = 0 or both. Now look carefully, first of them is a part of solution already and latter is like an original equation but with both parts divided by x. So not only you may divide, you *must* divide by x to properly solve such equations. Second part was already explained by other people. In this particular problem we have terms 1/x, which implies x = 0 could not solution. Taking another example, f(x)/x = 1 is equivalent to the system: f(x) - x = 0, x ≠ 0. Even if you get x = 0 when solving f(x) - x = 0, you must discard it.
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.
Ur comment days 6 days ago
your comment is 6 days ago 😱
Very well solved but how did you commemt 6 days before the video was uploaded without editing your comment
Hey Mr. Time travellers, NASA wants to know your location...
@@khwabm patreaon
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.
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)
Most likely, it just boils down to arbitrary assumptions and tricks
They do backwards scenes & claiming smart 😂😂
"You should be able to solve it"
Yeaa... and you should be able to perform the 500 kg deadlift.
Lol
just for the love of god don't do _both_
@@h00db01i I'm a gym rat who's studying engineering, so *don't tempt me* . :D
@@slingshot99 math before wrath in that case ;9
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)
Saw "divine" in the title and knew straightaway the answer would be the golden ratio 😛
Same. But I didn't manage to solve it, even though he only used standard steps.
I knew the answer... but theres no way I could have figured it out. 😅 I guess I still get full credit in California schools 😂
Exactly....
What is the golden ratio?
Same pinch
Please keep making these!
Ok
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)
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.
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?
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.
@@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
@@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
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
I did not even know where to begin. This was complicated. Appreciate and understand your logic. Well done.
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.
Thanks. That's interesting. In other words, is the triangle you are talking about Kepler's Triangle? en.wikipedia.org/wiki/Kepler_triangle
@@exoplanet11 Precisely! I had no idea it was called that, thanks for showing me!
@@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
@@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.
@@P4ExHzLRuuiFMg3X4U3v sorry how did you get to sin(c)=1?
My brain left the chat
Your 19 likes and this video 19 dislikes what a coincidence
@@madhavanguha2626 didn't see that coming
Was it ever in the chat to begin with?🤔
@@RedRacconKing just kidding I actually love maths
@@madhavanguha2626 not now
Presh: "It's quite a divine answer!"
Me: ".....Yes! Quite!"
*way over my head*
Way too much😂😂
Presh bhaijee you are too brilliant..you go straight to the point without unnecessary working. God bless you
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.
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. ;)
@@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
This method was better than the tricky one
@@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...
@@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 :)
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!
Some except it came to a quartic function instead of sextic
you could approximate like an engineer instead
@@timetraveler7 x^4-2x^3-x^2+2x+1 = 0
@@zastaphs yeah something like that
@@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.
Just another solution :
Take y = 1/x
Then we have -
√(x-y) +√(1-y) = x
√(x-y) = x - √(1-y)
Squaring both sides
x - y = x² + (1-y) -2x√(1-y)
Cancelling y from both side and rearranging the terms -
x² - x + 1 = 2x√(1-y)
Dividing both sides by x -
x - 1 + y = 2√(1-y) [•.• y = 1/x]
Again squaring both sides -
x² + 1 + y² -2x -2y + 2 = 4(1-y). [•.•xy =1]
Simplifying the terms -
x² + 1 + y² -2x + 2y - 2 = 0
=> ( -x + 1 + y )² = 0 [•.• 2 = 2xy]
=> -x + 1 + y = 0
Replacing y by 1/x -
-x + 1 + 1/x = 0
=> -x² + x +1 = 0
=> x² - x - 1 = 0
Rest follows.
I think this is straight forward / conventional solution. Taking y = 1/x was only for simplification.
Love from India bro!
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 !}
Very clever solution you have there…
The golden ratio really is golden to solving the problem 🙂
Ik! So cool!
Spoiler alert must informed beforehand!
What? No.
Ration?
@@angrytedtalks Well spotted 🙂
Bruuuuh my math knowledge and solving is like hell and I'm in 10 standard. I hate math related things but dude!! When i saw your video in my recommendation it looked interesting and the looks exactly matched the content. You are the first one in my life to make me see maths like a entertaining subject. Keep it up man the videos are awesome!!
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
mmmm, I think it s wrong when you write
(1-1/x)^(1/2) = x - (x-1/x)^(1/2) --> (NOT) x² - x + 1 = 2sqrt(x² - x)
(1-1/x)^(1/2) = x - (x-1/x)^(1/2) --> (1 - 1/x ) = x^2 + (x - 1 /x) - 2x (x-1/x)^(1/2) --> 1 = x^2 + x - 2x (x-1/x)^(1/2) -->
x^2 + x - 1 = 2x (x-1/x)^(1/2) --> x^2 + x - 1 = 2 [ x^2 ( x-1/x ) ] ^(1/2) -->
x^2 + x - 1 = 2 [ x^3 - x ) ] ^(1/2) .....
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.
x = x^2 - 2x√( 1-1/x ) +1 => 2x√(1 - 1/x ) = x^2 - x +1 => But x > 0 so : 2√(1-1/x) = x - 1 +1/ x => 2√( 1--1/x ) = x - (1 --1/ x)
Let : u =: √ (1 --1/x ) ; u > 0 ; => 2√ ((1 - 1/x ) = x - (1 - 1/x ) => 2u = 1 / ( 1 - u^2 ) - u ^2
u^4 + 2u^3 - u^2 - 2u + 1 = 0 ; u > 0 so : u^2 + 2u - 1 -- 2/u + 1/ u^2 = 0 / u^2 => u^2 + 2u - 2 + 1 -- 2/u + 1/ u^2 = 0
(u^2 - 2 + 1/ u^2) + 2u - 2/u + 1 = 0 => (u - 1/ u )^2 + 2 ( u - 1/u ) + 1 = 0
Let : u =: ( u - 1/u ) =: Z => (u - 1/ u )^2 + 2 ( u - 1/u ) + 1 = 0 => z ^2 + 2 z + 1 = 0
(Z + 1 )^ 2 = 0 => Z = - 1 => u -- 1/ u = - 1
u - 1/u = - 1 => u^2 + u - 1 = 0 => u = ( --1 + /- √5 )/ 2
if u = ( --1 -- √5 )/ 2 => u ^2 = ( 1 + 2 √5 + 5 ) / 4 = ( 3 + √5 ) / 2
x = 1 /(1 -- u^2 ) = 1 / [ 1 -- (3 + √5 )/ 2 ] = 2 / ( 2 -- 3 - √5 ) = --2 / ( 1 + √5 ) = ... = ( 1 - √5 ) / 2
if u = ( -1 + √5 )/ 2 => u ^2 = ( 3 - √5 ) / 2
x = 1 /(1 -- u^2 ) = 1 / [1 -- (3 -- √5 ) / 2 ] = 1 / ( --1 + √5 ) / 2 = 2 / ( --1 + √5 ) = 2 ( 1 + √5 ) / ( 1 + √5 )( --1 + √5 )
x = 2 ( 1 + √5 ) / ( 1 + √5 )( --1 + √5 ) = 2 ( 1 + √5 ) / 4 = ( 1 + √5 ) / 2
Love from India ❤️🇮🇳
From China too.
@@pengchengwu447 i thought youtube was banned in china? are you using a vpn?
From Iran too.
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.
Magically, you make everything just simple!
Great explanation!
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.
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?
yeah its especially easier to visualize if you replace variables with numbers
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!
Way i did it
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.
Nice one. I have not read all the comments. I tackled it a little differently. Started with your equation and then you can invert both sides. 1/x = a-b/(x-1) so 1 - 1/x = a -b add to the original equation and substitute x - 1/x = a again, you get a^2+1 = 2a . so a=1 and you can get values of x as you did
“divine” was a huge hint that golden ratio will be involved. I guessed it :)
Thank you so much sir! Really, math is divine. Math is everywhere. 👍
The teacher is really good. I will learn from. I will make a video following the teacher to share with everyone.
Major golden ration (sqrt(5)+1)/2 is denoted by capital phi. Small phi is for minor golden ration (sqrt(5)-1)/2.
Excellent tricks! Take care sir.
I'm Soo happy, first time I was able to solve an equation in one of your videos. I used the k method instead, k = 1/x and k^-1 = x
I mentally visualized the graphs of the parts and estimated any possible solution was near two and three up to five. I suppose 1.6 is close enough to two to feel like a good estimate for mental work.
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
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
Not only is the golden ratio the solution, but y=x is tangent at that point to the graph of the LHS of equation.
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.
Gooooood!!
The man is very incredible!
Top. Great !!!
👍👍
Top my Love ♥️🥰
Thanks for your content. When you multiply by (a-b) you should check what is happening when a=b ..........
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.
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
@@dickson3725Oh i see! but it definitely seemed less obvious to me at first.
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?
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
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 🤣😀
You would like this question:
Let f:N×N->N be a function such that f(1,1)=2 , f(alpha+1,beta)=f(alpha,beta) + alpha and f(alpha,beta +1)=f(alpha,beta) - beta, for all alpha,beta belongs to N and f(a,b)=2001 then find the number of ordered pairs (a,b) is?
Superb content bro. I'm loving it.
Bro???? He is double your age
@@koro-sensei9783 how do you know his age?
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!!
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.
We can simply rearrange the expression as (x-1/X)^1/2 = x-(1-1/X)^1/2 and square both sides.
After simplification we end up with x²-x-1=0 and can find the solution
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.
It is simple by multiplying both sides of ² both sides. You will arrive to an equation x³-x²-x=0, dividing all by x. Then there it is.
There are actually 2 solutions. The solution (1-sqrt(5))/2 = -0.618... also solves the original equation if you accept that the second bracket to the power of 1/2 can be the negative value of the "calculator output" you get (1.618). Therefore: 1 - 1.618 = -0.618 --> also solves the equation
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.
You are a genius
Love from India 🇮🇳
He is also indian
Yes 2 months ago....
How is your comment 2months ago
@@vimleshmaheshwari9018 he has the membership of this channel. Those get the video earlier
@@vimleshmaheshwari9018 because I am HRIGVED who asked presh to upload my question
1:21 How'd you find the "difference of squares"
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)
Not sure if this is a longer path but I had got the solution by substituting x=sec^2(y)
Simplifying the equation could get sin(y)=cos^2(y). Solving we get 2 values for sin(y) :
(-1-√5)/2 and (√5-1)/2
Sine would only be (√5-1)/2
Now x=sec^2(y) = 1/cos^2(y)
We already have sin(y)=cos^2(y)
So x= 1/sin(y)
Substituting sine of y value we get x = (√5+1)/2
This was so intriguing couldn't resist watching the solution. Thanks for this amazing problem (Hrigved) and solution (Presh)
jreak
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
Nice one, i like it , we can in the beginning square both sides and then developp the equation
I did it exactly same. What a sum! Thanks to you
This would have a lot more value if you explained what line of reasoning can lead to that specific choice of equation manipulations. As is, this video proves that your solution is correct, but it doesn't teach much about problem-solving. Would you please consider including this kind of explanation in future videos?
A good solution.method of solving is absolute.
your explanation is too smart.
Wow, that was a really unexpected answer, amazing! I thought it was gonna be a whole number.
Really? Unexpected?
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.
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)
Instantly thought of φ. Solved it too and was proved right. Nice problem.
This channel is one of the real ones
@ 3:48 Why should the sum of 2 square roots be a negative number?
My solution :
Take (1-1/x)^1/2 to rhs and square both sides.
We're left with x = x^2 +1 - 2(x^2 - x) ^1/2
Take x^2 to lhs and multiply both sides by - 1.
Now the equation is :
(x^2 - x) = 2(x^2 - x)^1/2 - 1
Take (x^2 - x) ^1/2 =t and solve the quadratic in t.
It becomes the same equation as in the video x^2 - x - 1 =0
And it's solution is the Golden Ratio.
I were thinking to use the hyperbolic Pythagoras's theorem
I'm really proud that I got there by myself 😁
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)
Objection. A square root can be a negative number. Example, square rooting of 9 is positive or negative 3. Therefore, the 1 - squareroot(5) divided by 2 answer is legitimate. You just need to take the positive square root for the first square root function but the negative square root for the second.
2:02 how did you know to add the two equations?
Squaring both sides of the equation twice, and dividing by 𝑥² (𝑥 = 0 is not a solution anyway) gave me a 4th degree polynomial equation.
Graphing the polynomial I realized that it has two double-roots, and I could write it as (𝑥 + 𝑎)²(𝑥 + 𝑏)²
Expanding this expression and comparing it to the polynomial I could quite easily solve for 𝑎 and 𝑏, giving me 𝑥 = (1±√5) ∕ 2
Then, 𝑥 ≥ 0 ⇒ 𝑥 = (1 + √5) ∕ 2
I squared the both sides
Kept it aside
And then multiplied the original question by X
And I equated the 2 equations together
We can also solve it with by assuming terms as sin and cos instead of a and b.And then using sec and tan formula.
This channel is golden as well.
Quick question,where is this applied??
A straightforward way is to first square both sides, and then assume a^2=x^3-x^2-x+1 to get an expression of a^2-2a +1=0. This means (a-1)^2=0, and a=a^2=1. Then one quickly gets x^3-x^2-x+1=1. Notice that x must be greater than 0, the solution to this expression can only be (1+sqrt(5)/2.
The solution is x=[1+sqrt(5)]/2. Sorry for the typo.
There is an easier way of solving this. After completing some steps, the equation will be alike this- root(x(sqaure)-1)+ root(x-1)=x root x or root(x+1)+1=x root x/root(x-1). After squaring the equation and completing some steps, we will reach this equation- 2 root(x+1)=(x)cube/(x-1) - (x+2) or 2 root(x+1)=((x)cube-(x)square-x+2)/x-1. We will square this equation and will get this equation- 4(x+1)=((x)cube-(x)square-x+2)square/(x-1)square or 4(x+1)(x-1)square=((x)cube-(x)square-x+2)square or 4((x)cube-(x)square-x+1)=((x)cube-(x)square-x+1)+1)square or 4((x)cube-(x)square-x+1)=((x)cube-(x)square-x+1)square +2((x)cube-(x)square-x+1)+1 or 2((x)cube-(x)square-x+1)=((x)cube-(x)square-x+1)square+1 or 0=((x)cube-(x)square-x+1)square- 2((x)cube-(x)square-x+1)+1 or ((x)cube-(x)square-x)square=0 or (x)sqaure-x-1=0 or x= the shown result.
wow!!! how an elegant solution!!!!!! beautiful, thx!!!
Nice video Fresh!
What do you mean Fresh? His name is Prestle Walker.
I thought it was Pringles Alan Walker
I bruteforced this by eliminating the ^(1/2)'s by squaring until I was left with having to find the roots of 4th degree polynomial, which I realized was the square of (x^2-x-1) after some trial and error.
Very intelligent! Square root are not easy to manage
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
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.
This is what makes me feel in peace after a math class early in the morning 😍😍😍 😄😄
Same
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)
We can do square on both sides procedure, we get this value
I really like your content , thank you so much for all your efforts
Both solutions can be correct in the complex domain because sqrt(2.618)=-1.618 (one of 2 roots).
The other possibility can br checked trivially, by seeing that as square root is positive when we are talking about real numbers, so LHS is positive so RHS too.
Yeah I haven't done any formal algebra for probably 40 years. I wrote down the problem and approached it like I would have 45 years ago when I was 15. Took a bit of thought, but in 7 lines and maybe 10 minutes - with NO elaborate substitution tricks - I had the quadratic x**2 - x - 1 = 0. I feel that using your "transform" method you add a level of complexity that no-one really needs, plus you give the impression that it's either an easier or indeed the only way to solve a relatively straightforward algebraic problem.
Can you give us your method?
@@londersito Heh heh I can try
But mathematical notation using text only is tricky 😝
@@christopherbedford9897 or you can take a photo and send it to any website and put the linl here,
@@londersito Yikes 😣
I scratched out the piece of paper and it looks like I made a fundamental error on the first line. But my handwriting and "cancelling out" and corrections all make it fairly illegible so I'm going to have to work it all out again. Tomorrow, it's bedtime now!
@@londersito Oops. I started by squaring one side of the equation and not the other, so it's interesting I ended up with the right result. Must have made another stuffup further down. Embarrassing.
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
Ok i solved this, by squaring the equation and leaving the 2•(...)^1/2•(..)^1/2 alone and squaring up again. In the final form you are left with various degrees of x going up to x^5 which you need horner's theorem to solve but it for some reason to me it feels more intuitive and easier than the solution presented.
maybe i just got lucky this time but i got it right! i solved it differently though:
since x^(1/2) is equal to sqrt(x) I just took the whole equation ^2
then you're left with x +(1/x) + 1-(1/x)=x
+1/x and -1/x cancel each other out so you have
x+1=x^2 and then my steps are the same as the video :)
X= (1+square root 5)/2
(1 +/- sqrt(5))/2
I don't see why it's 'tricky'
Moderate, at best.
Let's multiply both parts on the expression in the left part but with '-' instead of '+'
We get two equations, new 1 and initial 2
1.sqrt(x-1/x) - sqrt(1-1/x) = 1 - 1/x
2.sqrt(x-1/x) + sqrt(1-1/x) = x
Add them it will be
2sqrt(x -1/x) = x -1/x +1
Which is, if we put y instead of sqrt(x-1/x)
y**2 - 2y +1 = 0
(y-1)**2 =0 so
x-1/x-1 =0 -> x**2 - x -1 =0
x = (1 +/- sqrt (5))/2
You're operating right at the limit of what my brain can comprehend in Algebra, and way above what I can figure out in my own.
This just shows me how long ago I had studied math!!
1:39 "We will then divide both sides of this equation by x"
If I recall correctly this is not allowed? I remember being taught in school that we must never divide by x when solving an equation. The explanation back then was that we do not know whether x is 0, so dividing by it might take away some solutions.
Would somebody please enlighten my why this is possible here?
x can not be 0, since the question has the term 1/x.
@@施柏安-b1k Yes, but you can always get the term 1/x in any equation be rearranging and modifying some terms, e.g.:
x - 1 = 0
now factor out x:
x * (1 - 1/x) = 0
=> There is a term 1/x in the equation. In this simple example, we know that x is not 0. But in a more complex example the presence of a term 1/x does not imply that x cannot be 0.
@@Caledoriv
From your example,
My opinion is that you can only factorize it like that when x not equals to 0.
1*(x-1) = 0
What you do is times x to the first term (1), and divides with x to the second term (x-1).
Because you do the step "/x", you need to exclude the condition that "x=0".
So, the more accurate way is
=> x * (1 - 1/x) = 0 for "x not equals to 0"
and discuss x=0 respectively.
Another example is 2*5=10,
If we write it like (2*x)(5/x)=10,
x can only be nonzero, or it will be really weird.
So, back to the question. If we believe that when the question creator divides with x, he always considers the step I mentioned, then the question won't appear "1/x".
So if the question appears "1/x", we will know that x can not be 0.
You've got it wrong. First, you *may* divide by x. Think, if you have an equation in the form x f(x) = 0, it implies either x = 0 or f(x) = 0 or both. Now look carefully, first of them is a part of solution already and latter is like an original equation but with both parts divided by x. So not only you may divide, you *must* divide by x to properly solve such equations.
Second part was already explained by other people. In this particular problem we have terms 1/x, which implies x = 0 could not solution. Taking another example, f(x)/x = 1 is equivalent to the system: f(x) - x = 0, x ≠ 0. Even if you get x = 0 when solving f(x) - x = 0, you must discard it.
∆ABC, AB= 1, AC=√x, BC= x (>0)
H: on BC and AH⊥BC, AH=1/√x
P: on AB and CP⊥AB
BC•AH= AB•CP=> CP=√x= AC
=> A, P: concurrent
=> 1+x= x²
=> x= (1+√5)/2