Thank you so much! I'm just a dad who hasn't done any math in 20 years and spent an hour last night helping his daughter with really basic trigonometry. But I can do that stuff in my head, so I thought I would indulge in some more challenging problems. I don't care how young or old someone is, this kind of content is _always_ a good use of one's time.
I've been watching your videos for the past two months. Before watching the solution, I always try to solve the questions on my own. This is the first question I've successfully solved so far! I feel really good :)
Great video! -k is the sum of the roots of the quadratic equation and x is the conjugate of 1/x. There is no conjugate of pair of integers that sums to 0, 1 or -1. The rest will satisfy the condition.
When I tried to do it, the first thing I tried was to use a function to figure out what the set of solutions would look like. I represented the function f(x) = x + 1/x, and plotted the graph. With that, it is easy to see how there is a minimum in x=1, where f(1)=2. Also, due to the inverse symetry of the function, you can analyze the positive side of the graph and afterwards extrapolate that to the negative side in order to figure out the other half of the solutions. To see what the solutions of the equations look like, all you have to do is see where the graph cuts the horizontal lines y=2, y=3, y=4, etc... The points where those horizontal lines cut the graph are your solutions, and you can clearly see that, besides the minimum point in (1,2), all other solutions come in pairs that complement each other, with one being greater than 1 and the other being smaller than 1. Due to the inverse symetry of the function, you can also expect the same type of distribution of the solutions in the negative side of the graph. Knowing that the solutions exist and how they are distributed, you can solve the equation in the same manner you did on this video. I guess I took a bit of an unnecessary step in the process to ensure the existence of the solutions and the distribution of them, but still, it gives another perspective of the results that can be interesting. :) Nice video, with very intuitive explanations. Keep up the good work. Cheers!
Before watching video. ±1 obviously. Then the graph behavior will be constantly approach y = x (because 1/x becomes increasingly negligeable as x goes to infinity). Since all integer values of y are on the y=x line, and the graph is approaching but never touches the line (at least in the x>0 direction) there will be no solutions outside of the area near zero where (x + 1/x) - x ≥ 1. Just kind of a sketch for how to start. .... edit just saw that the teaser says there are infintely many more.... and I thought there wouldn't be any more... uh oh. O wait. Close to zero the graph goes to plus infinity so it will cross all integer y values past a certain point on the way up! (I'm still just looking at the positive side) The solutions will probably be algebraic irrationals. Edit.... I like the solution in the video very much
x + 1/x = k , where k is an integer x² - kx + 1 = 0 ... apply quadratic formula ... x = [k ± √(k² - 4)]/2 Since we want x to be real, (k² - 4) must be greater than or equal to 0 ==> |k| ≥ 2 All real solutions: x = [k ± √(k² - 4)]/2 where k is an integer and |k| ≥ 2
x doesn't need to be an integer only x+1/x = 'Ans' this Answer(Ans) needs to an integer and for that condition he solved the equation and got the values. He never claimed x needs to be an integer.
Thank you so much! I'm just a dad who hasn't done any math in 20 years and spent an hour last night helping his daughter with really basic trigonometry. But I can do that stuff in my head, so I thought I would indulge in some more challenging problems. I don't care how young or old someone is, this kind of content is _always_ a good use of one's time.
you're right !
I'll be 70 yo. in a few months and can't stop watching these videos!
🤣🤣
I've been watching your videos for the past two months. Before watching the solution, I always try to solve the questions on my own. This is the first question I've successfully solved so far! I feel really good :)
Great video! -k is the sum of the roots of the quadratic equation and x is the conjugate of 1/x. There is no conjugate of pair of integers that sums to 0, 1 or -1. The rest will satisfy the condition.
We may as well find the complex x as well.
If x is complex, you only need to remove the restriction for k
When I tried to do it, the first thing I tried was to use a function to figure out what the set of solutions would look like. I represented the function f(x) = x + 1/x, and plotted the graph. With that, it is easy to see how there is a minimum in x=1, where f(1)=2. Also, due to the inverse symetry of the function, you can analyze the positive side of the graph and afterwards extrapolate that to the negative side in order to figure out the other half of the solutions.
To see what the solutions of the equations look like, all you have to do is see where the graph cuts the horizontal lines y=2, y=3, y=4, etc... The points where those horizontal lines cut the graph are your solutions, and you can clearly see that, besides the minimum point in (1,2), all other solutions come in pairs that complement each other, with one being greater than 1 and the other being smaller than 1. Due to the inverse symetry of the function, you can also expect the same type of distribution of the solutions in the negative side of the graph.
Knowing that the solutions exist and how they are distributed, you can solve the equation in the same manner you did on this video.
I guess I took a bit of an unnecessary step in the process to ensure the existence of the solutions and the distribution of them, but still, it gives another perspective of the results that can be interesting. :)
Nice video, with very intuitive explanations. Keep up the good work.
Cheers!
Great video. Amazing work as always!
Before watching video. ±1 obviously. Then the graph behavior will be constantly approach y = x (because 1/x becomes increasingly negligeable as x goes to infinity). Since all integer values of y are on the y=x line, and the graph is approaching but never touches the line (at least in the x>0 direction) there will be no solutions outside of the area near zero where (x + 1/x) - x ≥ 1. Just kind of a sketch for how to start. .... edit just saw that the teaser says there are infintely many more.... and I thought there wouldn't be any more... uh oh. O wait. Close to zero the graph goes to plus infinity so it will cross all integer y values past a certain point on the way up! (I'm still just looking at the positive side) The solutions will probably be algebraic irrationals. Edit.... I like the solution in the video very much
Great Video and very nice explanation.
I never would have guessed there are infinitely many solutions!
His smile means, killed the equation by them 😂
Is it just me more he looks like dr foreman from house md 😂😂
It's not a video, just feels like, it's goosebumps in me..😅
x + 1/x = k , where k is an integer
x² - kx + 1 = 0
... apply quadratic formula ...
x = [k ± √(k² - 4)]/2
Since we want x to be real, (k² - 4) must be greater than or equal to 0 ==> |k| ≥ 2
All real solutions:
x = [k ± √(k² - 4)]/2
where k is an integer and |k| ≥ 2
Tank you. Now I understand.
That was really cool.
my idea was to set sin((x+1/x)pi)=0 and solve for it, but that's definitely not an easy way. No idea why that was my first thought tho
I ❤ this!
Réponse inattendue !!!
Très amusant !
Good
This guy r un...🗿🗿
Infinity 1/infinity =0 but + infinity= infinity
Good😂
If x=7 then x + 1/× is not an integer. (7+1/7). What is my fault?
Here your x does not follow the property x= (k±√k²-4)/2 7:05
k is the integer where you fill in 7, which gives a value of x. Fill that x into x + 1/x and it will be an integer.
x doesn't need to be an integer only x+1/x = 'Ans' this Answer(Ans) needs to an integer and for that condition he solved the equation and got the values. He never claimed x needs to be an integer.
He solves for k=7 to get x. Then, using that x, he reverses the process to get back to k=7.
Too much words and calculus for succes a easy problem.