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.
Thanks for posting great math videos! I think there may be a little error at the end of this one, though. The reason why the second solution is extraneous is not because one of the radicals is not defined (the domain of the equation is [-1, 0) U [1, inf) and the second "candidate" is about -0.62). The problem is that the left side is a sum of two non-negative numbers (square roots) so it cannot be equal to a negative number.
I always really enjoy Prof. Prime Newtons' videos. They are excellent, and he's a great teacher. In this problem, however, he is wrong. Some other repliers got it partially right. Prof. PN correctly noted that there is ambiguity in the literature between the difference between x^(1/2) and sqrt(x). In my world, if x is real and >0, then sqrt(x)>0, the positive root. As such (as one commentator correctly noted), the domain where t1=x-1/x>0 and where t2=1-1/x >0 is the union of x in x>1 and x in (-1,0). These give, correctly, x=(1+/- sqrt(5)/2 as solutions. But there is one more solution, namely x=0sup-. I.e. the limit as x->0 from below. I think that Prof. PN missed the former solution because given the +/- ambiguity in ()^(1/2) , there are actually four checks to see if sqrt(t1)+sqrt(t2)=x, namely with signs (+,+), (+,-), (-,+), and (-,-). I do not know if he will admit 0sup(-) as a solution. Good problem!
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!
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.
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.
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 , add the two equations to get rid of the second sqrt, 2(x-1/x)^1/2 = x - 1/x +1 which can be written as [ (x-1/x)^1/2 - 1]^2 =0 , hence x-1/x = 1
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 :)
What is the domain of x? (x - 1/x)^1/2 => x - 1/x ≥ 0 => -1 ≤ x < 0 or x ≥ 1 ---(1) (1 - 1/x)^1/2 => 1 - 1/x ≥ 0 => x < 0 or x ≥ 1 ---(2) (x - 1/x)^1/2 + (1 - 1/x)^1/2 = x => x ≥ 0 ---(3) so domain of x is x ≥ 1
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.
![Find (x+y+z) [Harvard-MIT] Guts contest](http://i.ytimg.com/vi/Pe0s5N_jkLA/mqdefault.jpg)
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.
Excellent explanation Sir. Thanks 🙏
Thanks for posting great math videos!
I think there may be a little error at the end of this one, though. The reason why the second solution is extraneous is not because one of the radicals is not defined (the domain of the equation is [-1, 0) U [1, inf) and the second "candidate" is about -0.62). The problem is that the left side is a sum of two non-negative numbers (square roots) so it cannot be equal to a negative number.
I always really enjoy Prof. Prime Newtons' videos. They are excellent, and he's a great teacher. In this problem, however, he is wrong. Some other repliers got it partially right. Prof. PN correctly noted that there is ambiguity in the literature between the difference between x^(1/2) and sqrt(x). In my world, if x is real and >0, then sqrt(x)>0, the positive root. As such (as one commentator correctly noted), the domain where t1=x-1/x>0 and where t2=1-1/x >0 is the union of x in x>1 and x in (-1,0). These give, correctly, x=(1+/- sqrt(5)/2 as solutions. But there is one more solution, namely x=0sup-. I.e. the limit as x->0 from below. I think that Prof. PN missed the former solution because given the +/- ambiguity in ()^(1/2) , there are actually four checks to see if sqrt(t1)+sqrt(t2)=x, namely with signs (+,+), (+,-), (-,+), and (-,-). I do not know if he will admit 0sup(-) as a solution. Good problem!
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!
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.
domain of x is x ≥ 1
So, both square roots are positive and the sum of those should be a negative number? I am calling both sides on that.
Good problem- good suggestions. Thank you!
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.
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 , add the two equations to get rid of the second sqrt, 2(x-1/x)^1/2 = x - 1/x +1 which can be written as [ (x-1/x)^1/2 - 1]^2 =0 , hence x-1/x = 1
Thanks a lot ... teacher
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.
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.
interesting that the LHS is also defined for -1
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!
What is the domain of x?
(x - 1/x)^1/2 => x - 1/x ≥ 0 => -1 ≤ x < 0 or x ≥ 1 ---(1)
(1 - 1/x)^1/2 => 1 - 1/x ≥ 0 => x < 0 or x ≥ 1 ---(2)
(x - 1/x)^1/2 + (1 - 1/x)^1/2 = x => x ≥ 0 ---(3)
so domain of x is x ≥ 1
Anybody else missing the 'Happy birthday 2 u' jokes? :)
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.
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 ✅️
Nice! 👏
where can I find problems in this level of difficulty?
In practice books
what is the source of this problem?
sir can you please do the cauchy sequence prove?
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
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
domain of x is x ≥ 1
Life is beautiful!
it will be easier if you write x^2 -1 as (x+1)(x-1)
This was beautiful
IR*....
I have a beaultifull solution for this equation using geometry
asnwer=1x
you might have uploaded the same video twice
let’s gooo haha
Wait a second. x can be negative.
Within following bounds:
-1