I would like to add that this problem actually have a nice geometric solution, because |a-b| has a geometric interpretation of a distance between a and b. |x| can be rewritten as |x-0| and if you take a standard number line the problem will translate to "find all points, such that a sum of their distances to 0 and 2 add up to 3". It's almost obvious that the only points on the number line are -½ and 2½.
If x is expanded to a complex number or vector, then it looks like the equation for an ellips(oid) with foci at 0 and 2 and a major axis of 3. -1/2 and 5/2 are just the vertices, or where the ellipse intersects the ℝeal axis.
This is amazing! I can already tell this is a really nice trick for analytic problems because of the Triangle Inequality(which pops up in things like metric spaces). And this is also a really profound view of what abs() is doing. Thank you for the video! I'm going to play with this some more.
@@chitlitlah You will get 0=0 in the end. Of course you will also get 0≤x≤2 in the process, but it's a nice exercise to try to find where you'll get this restrictions from exactly.
We can also plot the graph of |x| and |x-2| on the x-y plane. They are symmetric on the x=1 line and so we can just find a solution for x+x-2=3, then find it's distance from 1, then use that distance to find the other solution.
Since there are only two absolute value terms, I think it should be easier and less error-free to directly partition the solution. 1. When X >= 2, the original formula becomes 2X-2=3, and X = 2.5 is in line with the assumption and is a positive solution 2. When 0
Just asking if this way of mine was valid: Instead power both side, I just solve the usual linear way but i do this, +(x)+(x-2)=3 +(x)-(x-2)=3 (invalid) -(x)+(x-2)=3 (invalid) -(x)-(x-2)=3 i ignore the invalid equation since +x-x=0, and i get the same ans as above.
I solved this in a different way Finding roots: 2,5 and -0,5 Then write a quadratic: ax²+bx+c -b/a = Sum of roots = 2 c/a = Multiply of roots = -1,25 inserting values to the quadratic x² - 2x - 1,25 = 0 (multiply by four) 4x² - 8x - 5 = 0
I just square it as is. The squaring removes the absolute value notation except the middle term. Then you subject the middle term and square again. I got the same quadratic equation. Actually considering the cases is easier. When x
It results in a quadratic equation of 12x^2-24x+9 when divided by three is 4x^2-8x+3. This is factorable and results in x=3/2 and x=1/2, not complex solutions
|x - 2| = 3 - |x| Right side must be positive, so x is in between -3 and 3. Assume x >= 0 |x-2| = 3 - x Try x = 3, 1 ≠ 0 Try x = 2, 0 ≠ 1 Try x = 2.5, 0.5 = 0.5 ✔️ Set y = -x Assume y > 0 |-y - 2| = 3 - |-y| |y + 2| = 3 - |y| y + 2 = 3 - y 2y = 1 y = 0.5 -> x = -0.5 So solutions are x = 2.5 and x = -0.5
Then it becomes an ellipse. Extend to the quarternions and it becomes a 4D ellipsoid. Technically the original equation just used absolute values, so they don't even need to be complex numbers, but rather arbitrary vectors, as long as you replace the 2 with an equivalent vector for the given vector space, with which it will still plot an N-dimensional ellipsoid assuming standard euclidean space, and even then it will still give an equivalent of an ellipsoid in whatever that space is.
@@kobethebeefinmathworld953 Double checking Wikipedia, it might be more accurate to call it a spheroid, but that's ultimately just a special case of an ellipsoid.
Replace |x| with |x-a|, |x-2| with |x-b| and 3 with c and you can generalize it. For a-b+c =/= 0 and a-b-c =/= 0, the quadratic equation is 0 = 4x² + 4(a+b)x + (a+b)²-c² And the solutions are x1 = -(a+b+c)/2 and x2 = -(a+b-c)/2
BY DEFINITION: abs(x) = sqrt(x^2) _ abs(x) + abs(x-2) = 3 sqrt(x^2) + sqrt((x-2)^2) = 3 sqrt(x^2) + sqrt(x^2 - 4x + 4) = 3 +-x + (+-(x-2)) = 3 case 1: x + x - 2 =3 2x - 2 = 3 x = 5/2 case 2: -x - x + 2 = 3 -2x = 1 x = -1/2 case 3: - x + x - 2 = 3 - 2 = 3 (this case is not possible!) case 4: x - x + 2 = 3 2 = 3 (this case is not possible!) in the end, we get two solutions: x = -1/2 and x =5/2.
probably the next question to ask is find all complex z such that |z| + |z-2| = 3 edit: employing the same method as used in the video, sqrt(x²+y²) + sqrt((x-2)²+y²) = 3 (x-2)²+y² = 9 - 6sqrt(x²+y²) + x²+y² - 4x - 5 = - 6sqrt(x²+y²) 16x² + 40x + 25 = 36(x²+y²) 20x² + 36y² - 40x - 25 = 0 therefore z which satisfy the equation are of the form x+iy that satisfy 20x² + 36y² - 40x - 25 = 0 i did this of the top of my head correct me if im wrong🤗
But √(x²)=|x| works only in real numbers. Try x=i , √(i²)=√(-1)=i But |i|=√(0²+1²)=√(1)=1 . I know that most of questions about absolute value wants only real solutions therefore this is not very important.
That is because a modulus of a complex number (which has the same notation as absolute value for real numbers) doesn’t really output a “positive version” of the input, instead it indicates the distance from the origin. You’ll notice that in fact, taking a complex number z and squaring it gives r^2*e^i2theta in polar form, then taking the square root gives |r|e^i2theta/2 And since r is defined as sqrt(a^2+b^2) where a and b are real numbers, it is always positive, so you will always just get your same number back as long as it is non-real. In fact, the reason why sqrt(x^2) gives |x| for negative numbers is because negative numbers have an argument (theta) of pi (or 180 degrees). When squaring x, you are obtaining |r|^2*e^i2pi, but e^i2pi = e^i0, which is 1. Since the root function outputs only one answer (namely, the positive value for real numbers), you will get the positive version.
I kinda hate to be this guy, but... aKtUaLi... You don't need x≥0 in the second equation even when you don't use complex numbers, because √x is defined only for nonnegative arguments when you are dealing with real numbers, so you simply can't plug a negative number in the equation. The problem is in confusing "an equation" with "equivalent transformation" (I don't really know how to say it in English, unfortunately). Here's an example: x+2=5 (an equation) ↔ (equivalent transformation) x+2-2=5-2 There are two equations, but between them is an equivalent transformation of "subtracting 2 out of both sides". Same thing applies to substitutions and... for this as well. If you are given an equation (√x)²=5 you are absolutely allowed to write ↔x=5 (arrow goes both ways), because x is nonnegative from the beginning (if you don't use complex numbers, but that's not the point), however, if you start from equation x=5, you can't write x=5↔(√x)²=5, because x (presumably) might be negative. "But it's five, how can it be negative?" Well, that has to do with the idea of "a domain of a variable", any variable (and I mean ANY) should have one, it should be from some domain, and if your domain is ℝ you don't have this "equivalent transformation"; if it's ℝ₊₀ (or ℂ) (maybe it's specified somewhere before) you do have this equivalence. So there you have it, it's all about domains of variables, you kinda always have to specify them, but we rarely do, because everyone is lazy and it definitely will never confuse anybody... right? Never?
Because the two absolute values here are x and x-2, seems like there are only 3 cases: x>2, 2>x>0 or 0>x. (Assume those include “or equal to” as necessary). Then there is no quadratic to solve, only linear.
Wouldn't you get just 2 equasions? At the same time both x can be positive or negative. If you had x and y you can create 4 equasions because these variables can change independently.
That's not how math works. For instance: |1| + |-1| = 1 + 1 = 2 But |1 + (-1)| = |0| = 0 The sum of the absolute values is not the same as the absolute value of the sum.
I would like to add that this problem actually have a nice geometric solution, because |a-b| has a geometric interpretation of a distance between a and b. |x| can be rewritten as |x-0| and if you take a standard number line the problem will translate to "find all points, such that a sum of their distances to 0 and 2 add up to 3". It's almost obvious that the only points on the number line are -½ and 2½.
That's exactly how i solved this question without paper and pen , but this trick is only limited to |x-a|+|x-b|=c ,format .
If x is expanded to a complex number or vector, then it looks like the equation for an ellips(oid) with foci at 0 and 2 and a major axis of 3. -1/2 and 5/2 are just the vertices, or where the ellipse intersects the ℝeal axis.
@@angeldude101 that is how we learnt to draw an ellipse using a pencil, two pins and a thread.
yay! I also did it that way!
I feel like knowing this is true is way more useful than using it.
This is amazing! I can already tell this is a really nice trick for analytic problems because of the Triangle Inequality(which pops up in things like metric spaces).
And this is also a really profound view of what abs() is doing. Thank you for the video!
I'm going to play with this some more.
Try this same method on |x| + |x-2| = 2. (same left hand side as video but right hand side = 2). Interesting things happen.
All values from 0 to 2 satisfy the equation. I'm not somewhere where I can work out the problem as in the video and see what happens though.
@@chitlitlah You will get 0=0 in the end. Of course you will also get 0≤x≤2 in the process, but it's a nice exercise to try to find where you'll get this restrictions from exactly.
We can also plot the graph of |x| and |x-2| on the x-y plane. They are symmetric on the x=1 line and so we can just find a solution for x+x-2=3, then find it's distance from 1, then use that distance to find the other solution.
因為只有兩個絕對值項式, 我覺得直接分區間來解應該更容易也比較不會出錯
1. 當 X >= 2 時, 原式成為 2X-2=3, 得 X = 2.5 符合假設, 是一個正解
2. 當 0
Since there are only two absolute value terms, I think it should be easier and less error-free to directly partition the solution.
1. When X >= 2, the original formula becomes 2X-2=3, and X = 2.5 is in line with the assumption and is a positive solution
2. When 0
this is just for fun i believe and for the purpose of saying this is possible
正體中文就給讚~😍😍😍
Just asking if this way of mine was valid:
Instead power both side, I just solve the usual linear way but i do this,
+(x)+(x-2)=3
+(x)-(x-2)=3 (invalid)
-(x)+(x-2)=3 (invalid)
-(x)-(x-2)=3
i ignore the invalid equation since +x-x=0, and i get the same ans as above.
That's the usual way to do it. He was doing it in a more creative, alternative way.
Exactly, this was my solution as well and is much faster than the presentation.
Yup that right
I solved this in a different way
Finding roots: 2,5 and -0,5
Then write a quadratic: ax²+bx+c
-b/a = Sum of roots = 2
c/a = Multiply of roots = -1,25
inserting values to the quadratic
x² - 2x - 1,25 = 0 (multiply by four)
4x² - 8x - 5 = 0
I just square it as is. The squaring removes the absolute value notation except the middle term. Then you subject the middle term and square again. I got the same quadratic equation.
Actually considering the cases is easier.
When x
It results in a quadratic equation of 12x^2-24x+9 when divided by three is 4x^2-8x+3. This is factorable and results in x=3/2 and x=1/2, not complex solutions
When you said, "let's assume that we can have imaginary numbers", you can also only allow real numbers because a square is never negative anyways.
Of course you can take your car on the highway, but sometimes it’s great to discover the scenery by a longer route by bike!
|x - 2| = 3 - |x|
Right side must be positive, so x is in between -3 and 3.
Assume x >= 0
|x-2| = 3 - x
Try x = 3, 1 ≠ 0
Try x = 2, 0 ≠ 1
Try x = 2.5, 0.5 = 0.5 ✔️
Set y = -x
Assume y > 0
|-y - 2| = 3 - |-y|
|y + 2| = 3 - |y|
y + 2 = 3 - y
2y = 1
y = 0.5 -> x = -0.5
So solutions are x = 2.5 and x = -0.5
Lol that was completely ridiculous. Well done!
Next question: change x to z and extend the domain from the real line to the complex plane.
Then it becomes an ellipse. Extend to the quarternions and it becomes a 4D ellipsoid. Technically the original equation just used absolute values, so they don't even need to be complex numbers, but rather arbitrary vectors, as long as you replace the 2 with an equivalent vector for the given vector space, with which it will still plot an N-dimensional ellipsoid assuming standard euclidean space, and even then it will still give an equivalent of an ellipsoid in whatever that space is.
@@angeldude101 yes, that is correct
@@kobethebeefinmathworld953 Double checking Wikipedia, it might be more accurate to call it a spheroid, but that's ultimately just a special case of an ellipsoid.
Replace |x| with |x-a|, |x-2| with |x-b| and 3 with c and you can generalize it.
For a-b+c =/= 0 and a-b-c =/= 0, the quadratic equation is
0 = 4x² + 4(a+b)x + (a+b)²-c²
And the solutions are
x1 = -(a+b+c)/2 and
x2 = -(a+b-c)/2
BY DEFINITION:
abs(x) = sqrt(x^2)
_
abs(x) + abs(x-2) = 3
sqrt(x^2) + sqrt((x-2)^2) = 3
sqrt(x^2) + sqrt(x^2 - 4x + 4) = 3
+-x + (+-(x-2)) = 3
case 1:
x + x - 2 =3
2x - 2 = 3
x = 5/2
case 2:
-x - x + 2 = 3
-2x = 1
x = -1/2
case 3:
- x + x - 2 = 3
- 2 = 3 (this case is not possible!)
case 4:
x - x + 2 = 3
2 = 3 (this case is not possible!)
in the end, we get two solutions: x = -1/2 and x =5/2.
Another way to do this is to multiply both sides by plus-minus 1
That's quite a supply of Expo markers! :)
Hello, this man is called Steve!!!
I wouldn’t have solve that this way but it’s a clever way to do it
probably the next question to ask is find all complex z such that |z| + |z-2| = 3
edit:
employing the same method as used in the video,
sqrt(x²+y²) + sqrt((x-2)²+y²) = 3
(x-2)²+y² = 9 - 6sqrt(x²+y²) + x²+y²
- 4x - 5 = - 6sqrt(x²+y²)
16x² + 40x + 25 = 36(x²+y²)
20x² + 36y² - 40x - 25 = 0
therefore z which satisfy the equation are of the form x+iy that satisfy 20x² + 36y² - 40x - 25 = 0
i did this of the top of my head correct me if im wrong🤗
Thanks
very clever approach
But √(x²)=|x| works only in real numbers. Try x=i , √(i²)=√(-1)=i
But |i|=√(0²+1²)=√(1)=1 .
I know that most of questions about absolute value wants only real solutions therefore this is not very important.
That is because a modulus of a complex number (which has the same notation as absolute value for real numbers) doesn’t really output a “positive version” of the input, instead it indicates the distance from the origin.
You’ll notice that in fact, taking a complex number z and squaring it gives r^2*e^i2theta in polar form, then taking the square root gives |r|e^i2theta/2 And since r is defined as sqrt(a^2+b^2) where a and b are real numbers, it is always positive, so you will always just get your same number back as long as it is non-real.
In fact, the reason why sqrt(x^2) gives |x| for negative numbers is because negative numbers have an argument (theta) of pi (or 180 degrees). When squaring x, you are obtaining |r|^2*e^i2pi, but e^i2pi = e^i0, which is 1. Since the root function outputs only one answer (namely, the positive value for real numbers), you will get the positive version.
You are correct, in general |x| = √(xx*). It's just that for ℝeal numbers, x* = x and therefore xx* = x².
I kinda hate to be this guy, but... aKtUaLi...
You don't need x≥0 in the second equation even when you don't use complex numbers, because √x is defined only for nonnegative arguments when you are dealing with real numbers, so you simply can't plug a negative number in the equation. The problem is in confusing "an equation" with "equivalent transformation" (I don't really know how to say it in English, unfortunately). Here's an example:
x+2=5 (an equation)
↔ (equivalent transformation)
x+2-2=5-2
There are two equations, but between them is an equivalent transformation of "subtracting 2 out of both sides". Same thing applies to substitutions and... for this as well. If you are given an equation (√x)²=5 you are absolutely allowed to write ↔x=5 (arrow goes both ways), because x is nonnegative from the beginning (if you don't use complex numbers, but that's not the point), however, if you start from equation x=5, you can't write x=5↔(√x)²=5, because x (presumably) might be negative. "But it's five, how can it be negative?" Well, that has to do with the idea of "a domain of a variable", any variable (and I mean ANY) should have one, it should be from some domain, and if your domain is ℝ you don't have this "equivalent transformation"; if it's ℝ₊₀ (or ℂ) (maybe it's specified somewhere before) you do have this equivalence.
So there you have it, it's all about domains of variables, you kinda always have to specify them, but we rarely do, because everyone is lazy and it definitely will never confuse anybody... right? Never?
Because the two absolute values here are x and x-2, seems like there are only 3 cases: x>2, 2>x>0 or 0>x. (Assume those include “or equal to” as necessary). Then there is no quadratic to solve, only linear.
What is the method you used for the last equation? What is it called? I've never seen such way to solve it before
Legend, my idol❤️🫀🫀🫀🫀
Thank you
Perfect
Wouldn't you get just 2 equasions? At the same time both x can be positive or negative. If you had x and y you can create 4 equasions because these variables can change independently.
Thanks sir, great 😃😃😃
I've never seen this before. I love it!
link for tic tac toe factor , please
Niiccceee... 👏👏👏👏👏👏👌👌👌👌👌👌👍👍👍👍👍👍
👍
|x|+|x-2|=|x+x-2|=|2x-2|=3
Will give you the same answer. So why to complicate it
2x-2=3 or 2x-2=-3
2x=5 ......2x=-1
X=5/2 or x=-1/2
Same thing much less complicated
That's not how math works. For instance:
|1| + |-1| = 1 + 1 = 2
But
|1 + (-1)| = |0| = 0
The sum of the absolute values is not the same as the absolute value of the sum.
Did i really never realize that you can write |x|=(x^2)^(1/2)?
I knew the other way around it but I never think about it.
😁😁😁😁
👋👋👋👋hi
The first
very rudimentary
This probably is the worst method
Тупое решение простеишее уравнение раскрыли модули с умом и решили линеиные уравнения
why tf am i subscribed? i havent even seen this dude before. unsubbed.