A very nice solution. I, a 70-year-old pensioner, send you my greetings from Slovakia. When I see such a beautiful solution, all my sadness disappears.
Both Solutions are valid. Because x2 = -0.618 aaprx and 1/X2 = -1.618 apprx. Difference between x2 and 1/x2 = 1 (Exactly). That reminds me what facilitated substitution of the long-term by U at 8:36. -1/x2 appears in both the terms, which is positive. That will make square root positive.
The negative version doesn't actually give you imaginary sqaure roots! The first one is the square root of 1 and the other one gives you the square root of phi sqaured!
I did it without substitution, multiply both side by [(x-1/x)^1/2 - (1-1/x)^1/2] and eventually get (x-1/x)^1/2 - (1-1/x)^1/2 = 1 -1/x , and the two equations to get rid of the second sqrt, then square both sides, simplify and get (x - 1/x -1)^2 = 0.
Thank you for your lecture. If the solution provides us the golden ratio, can we directly derive the first equation from a certain relation in a geometric object related with the golden ratio?? I found the Euclidean way for x^2 - x + 1 = 0 by the similarity in a rectangular. I guess there is the source of the first equation... Please teach that if you know.
Sqrt(5) without calculator! 5= 2.5 * 2. So the square root must be between 2.5 and 2.0. The average of the 2 numbers is 2.25. Now, 5 divided by 2.25 = 2.2222. The average now is 2.2361 and that is very close to the (rounded) value of 2.23607.
Hello sir, is there any difference between writing the square root of some quantity as radical(x) versus as (x)^1/2? Does it affect the domain/range or something? I'm not too clear on this, sorry for the bother. Great video as always :)
Great video! I guess that solution (1-√5)/2 doesn´t work not by x-1/x be negative (because is positive) but is because x is expressed as sum of two square roots
@@abidomar9568 yes, but it's a mistake, it is impossible that x be negative, just look (x-1/x)^(1/2) must be positive, (1-1/x)^(1/2) also must be positive, this implies that x is positive because x = (x-1/x)^(1/2)+(1-1/x)^(1/2) x = (something positive) + (something positive), therefore x = (something positive)
The negative version (x = 1/2(1-sqrt(5)) actually is the solution. Because we want that an expression under a square roots will be positive. So we aren’t want x > 0, we want 1 - 1/x >= 0 and x - 1/x >= 0. For the first one we have: 1 - 1/x >= 0 (x - 1)/x >= 0 x in the interval (-inf; 0) U [1; +inf) For the second one we have: x - 1/x >= 0 (x^2 - 1)/x >= 0 (x-1)(x+1)/x >= 0 and we have that x in the interval [-1; 0) U [1; +inf) and, fortunately (or not xd), the root x = 1/2(1 - sqrt(5)) ≈ -0.618 in the interval.
The negative version can't be a solution, as the result of any root is ALWAYS positive. Since the x on the right hand side of the original equation is equal to the sum of two roots, it HAS to be positive. It literally doesn't matter, what is inside the roots, x is positive.
![Limit of absolute value functions [ lim |x^3 - x|/(x^3 - |x|) as x goes to 0 ]](/img/n.gif)
A very nice solution. I, a 70-year-old pensioner, send you my greetings from Slovakia. When I see such a beautiful solution, all my sadness disappears.
I'm 70 too (but will be 71 later this month). I live in Japan.
Hi prime newtons, letting you know that you uploaded the same video twice!
Thanks. I deleted one. UA-cam probably had a glitch yesterday.
This channel better then my math teacher.
😂
Both Solutions are valid.
Because x2 = -0.618 aaprx and 1/X2 = -1.618 apprx.
Difference between x2 and 1/x2 = 1 (Exactly). That reminds me what facilitated substitution of the long-term by U at 8:36.
-1/x2 appears in both the terms, which is positive. That will make square root positive.
Excellent explanation Sir. Thanks 🙏
The negative version doesn't actually give you imaginary sqaure roots! The first one is the square root of 1 and the other one gives you the square root of phi sqaured!
Good problem- good suggestions. Thank you!
I did it without substitution, multiply both side by [(x-1/x)^1/2 - (1-1/x)^1/2] and eventually get (x-1/x)^1/2 - (1-1/x)^1/2 = 1 -1/x , and the two equations to get rid of the second sqrt, then square both sides, simplify and get (x - 1/x -1)^2 = 0.
Both parts of the equation do exist in the range [-1,0[ and [1, inf[. So x could be negative! But (1-sqrt5)/2 is too negative.
Nice! 👏
Thank you for your lecture.
If the solution provides us the golden ratio, can we directly derive the first equation from a certain relation in a geometric object related with the golden ratio??
I found the Euclidean way for x^2 - x + 1 = 0 by the similarity in a rectangular.
I guess there is the source of the first equation... Please teach that if you know.
interesting that the LHS is also defined for -1
sir can you please do the cauchy sequence prove?
Life is beautiful!
Sqrt(5) without calculator! 5= 2.5 * 2. So the square root must be between 2.5 and 2.0.
The average of the 2 numbers is 2.25. Now, 5 divided by 2.25 = 2.2222. The average now is 2.2361 and that is very close to the (rounded) value of 2.23607.
x2=(1-sqrt5)/2 don't work because (x-1/x)^1/2+(1-1/x)^1/2 is always positive (if it's real)
Should have factored out the x-1 term near the beginning.
Hello sir, is there any difference between writing the square root of some quantity as radical(x) versus as (x)^1/2? Does it affect the domain/range or something? I'm not too clear on this, sorry for the bother. Great video as always :)
Yes. Unless clearly stated, you can only get non-negative outputs from square-root (x). But x^½can give anything including imaginary outputs.
@@PrimeNewtons Ah, makes sense. Thank you so much for the explanation!
This was beautiful
what is the source of this problem?
where can I find problems in this level of difficulty?
In practice books
Great video! I guess that solution (1-√5)/2 doesn´t work not by x-1/x be negative (because is positive) but is because x is expressed as sum of two square roots
You are ✅️
Golden ratio 🎉
Need help sir why can't the opposite of golden ratio is a root.?
x cannot be negative, because x is the sum of two root squares, positive + positive = positive
Please look at the question "x belongs to real number "
@@abidomar9568 yes, but it's a mistake, it is impossible that x be negative, just look (x-1/x)^(1/2) must be positive, (1-1/x)^(1/2) also must be positive, this implies that x is positive because x = (x-1/x)^(1/2)+(1-1/x)^(1/2)
x = (something positive) + (something positive), therefore x = (something positive)
@@abidomar9568should be "real POSITIVE number instead of just real
IR*....
it will be easier if you write x^2 -1 as (x+1)(x-1)
let’s gooo haha
asnwer=1x
I have a beaultifull solution for this equation using geometry
The negative version (x = 1/2(1-sqrt(5)) actually is the solution. Because we want that an expression under a square roots will be positive. So we aren’t want x > 0, we want 1 - 1/x >= 0 and x - 1/x >= 0.
For the first one we have:
1 - 1/x >= 0
(x - 1)/x >= 0
x in the interval (-inf; 0) U [1; +inf)
For the second one we have:
x - 1/x >= 0
(x^2 - 1)/x >= 0
(x-1)(x+1)/x >= 0
and we have that x in the interval [-1; 0) U [1; +inf)
and, fortunately (or not xd), the root x = 1/2(1 - sqrt(5)) ≈ -0.618 in the interval.
The negative version can't be a solution, as the result of any root is ALWAYS positive.
Since the x on the right hand side of the original equation is equal to the sum of two roots, it HAS to be positive. It literally doesn't matter, what is inside the roots, x is positive.
Reduce[Sqrt[(-1 + x)/x] + Sqrt[-x^(-1) + x] == x, x, Reals] gives the positive answer.
I got the same results but did not take into account that the sum of two sqrts can't be negative
you might have uploaded the same video twice
Wait a second. x can be negative.
Within following bounds:
-1