You wouldn’t expect this "quadratic" equation to have 6 solutions!
ฝัง
- เผยแพร่เมื่อ 2 มี.ค. 2024
- Surprisingly, the "quadratic" equation x^2+5abs(x)-6=0 has a total of 6 solutions (2 real and 4 complex solutions) which I did not expect. I came up with this equation purely by accident and I think it is super cool. It will feature complex numbers!
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Solving log-power equation x^ln(4)+x^ln(10)=x^ln(25)
th-cam.com/video/xBpZRWCGw30/w-d-xo.html
Help why is 2ab = 0??
When I was a kid I did something similar to solve a problem, and also to find the general formula for the quadratic (I was like 12 years old and was obsessing with complex numbers) this really brought me memories and a general pleasing sensation: reminiscing thanks for sharing.
If x∈ℍ then how many solutions are there?
oh I just bracketed out (|x| + 6)( |x| -1 ) = 0 and was guessing where are other solutions, but they are complex :d
@@wasordx3245
Because in the equation,
Re(x) + 2abi = 0 + 0i
Here 2ab is the imaginary part and is called Im(x). Since both Re(x) and Im(x) give 0 and 0i (RHS=0), you can equate them to be 0
Love how he explains almost everything from a single problem...Great teacher
your feelings are irrational
25 years ago, final practical math task consisted of rather simple irrational equation. Me, spending preceding preparatory week only by pushing 4 years of history of literature and all book authors into my brain 18 hours per day (as I did ignore it all those years), came with rather unusual solution... logarithms. I got to right solution through many steps, nobody in class would even understand after explanation. My lovely math teacher looked at it for few minutes and then said: "Isn't there simpler solution?" Then I snapped out of my linguistic/literature attuned mind and answered with single middle step required to get to traditional solution. I did pass with "honors" as it used to be called at time.
This random video, YT threw at me today, reminds me of rather unnecessary "mathematical escape" (in this case substitution) from rather direct intuitive "in-mind" solution. (That's as long as one understands principles.)
But would it Math Olympics, it would not ask you to solve this equation. It would substitute number "6" for "a" and say something like:
"a is in R. Define number of solutions based on value of 'a'. And explain why."
Nice to see video which triggers some nostalgia. So I'll finish with: "Why so complex solution?"
@@Fire_Axusits almost like they are complex
My heart stopped when he wrote ±2,±3 without i
Then you need a defibrillator when watching his videos.
For real x
x^2 = |x|^2
So,
b^2-5|b|+6 = 0
Let |b| = u
u^2-5u+6 = 0
u = 2,3
|b| = 2,3
b = ±2,±3
No need to make case of b>0 or b
Yes. I realized that after I finished the video. Lol
Unfortunately, you solved the wrong equation... ;) The equation was x² + 5 |x| - 6 = 0. Not x² - 5 |x| + 6 = 0.
But obviously your approach also works for the correct equation, yielding |b| = - 6 (which is impossibe) and |b| = 1, which gives exactly the real solutions +- 1 from the video. :)
@@bjornfeuerbacher5514
I solved for b taking a=0
As in the video
@@Ajay_Vector Ah, ok, you solved the equation at 5:50, not the original one?
@@bjornfeuerbacher5514
Yeah
Looks like very nice.
0:58 Maybe because the |x| = {-x if x0} doesn't consider complex values.
Absolute values of complex numbers do exist. Since the absolute value is defined as the distance of a number from 0, looking at the complex plane, the absolute value of a complex number is possible.
Sorry, I just remembered that absolute values of complex numbers are real. Haha
Yeah, because the domain(x values) of absolute value function in this case is R
@@sohailansari07289 if you extend the domain to C, the range remains R
Your definition of absolute value is undefined for x=0. It should return x if x >= 0.
Thought it'd be way harder when applying complex numbers but that ab = 0 condition made it trivial, really cool
I love your channel bro, you’re the reason that I can understand calculus and algebra so easily as a freshman, thank you🙏
It's like mathematical poetry. Beautiful.
Too many sol.
I think I’m going to solve a polynomial of degree 100,900,000
@@table5584 Go for it :)
The first thing that popped-into my mind was "Nah......gotta be 4 solutions (2 for x^2, and 2 more for |x| )"
Yes, but |x| also applies for complex values, |xi|=x, |-xi|=x. So your logic works.
great video and great solution!
this new video really made my day
so satisfying... literally! I am disappointed that the video has already ended
You should try stuff like z^2 +az* + b = 0 where z* is the complex conjugate of z.
Honestly maths would not be the same for me without these creative problems on this channel
Thank you sir ❤
So, for generalization: An "absolute" term for the "x" in the quadratic equation splits the curve into two separate ones, each having two zero crossings and both symmetrical to the y axis. But since each of them is valid only for either left or right of the y axis, two of the zero crossings are discarded. When going complex, two additional curves with no imaginary part in the resulting value arise along the imaginary x axis, and since along that axis the real part of x - which determines if any zero crossing has to be discarded - remains zero, none of the zero crossings of those two additional curves gets discarded. Thus resulting in 6 solutions as long as the curves along the imaginary x axis have two zero crossings, resp. in 4 solutions if those latter curves have just one zero nudge, or 2 solutions if those latter curves have no zero crossings.
That's a nice generalization of curve discussions for quadratic equations. Thanks.
Wow this is amazing ❤
Please make a video about Bessel functions!
Thank you,sir
Hey what are your thoughts about the 'annals of mathematics studies' series?
Wow, only 12 minutes passed and you already already 500 views 😮 That's the proof of your professionalism ❤
incorrect assertions in the thumbnail is definitely an effective way to engage the math nerds.
Stunning😮
It's easy in polar: x^2 + 5 |x| - 6 = 0, use x = r e^(i theta), r >= 0, -pi < theta theta = pi n => theta in { 0, pi }, and
e^(2 i theta) = -1 => 2 theta = 2 pi n + pi => theta = pi ( n + 1/2 ) => theta in { -pi/2, pi/2 }.
Now r^2 e^(2 i theta) + 5 r - 6 = 0, so
c r^2 + 5 r - 6 = 0, where c = e^(2 i theta) in { -1, 1 }, and so
r^2 + 5 c r - 6 c = 0 (using c^2 = 1), so
2 r = -5c +/- sqrt[ 25 c^2 + 24 c ], so
2 r = -5c +/- sqrt[ 25 + 24 c ].
Thus
c = 1: 2 r = -5 +/- sqrt[ 49 ] = -5 +/- 7 = 2 (since r > 0, can exclude the -12 solution).
c = - 1: 2 r = 5 +/- sqrt[ 1 ] = 5 +/- 1 = { 4, 6 }.
Thus x = r e^(i theta), where
r = 1 and e^(2 i theta) = 1, so theta in { 0, pi },
or
r in { 2, 3 } and e^(2 i theta) = -1, so theta in { -pi/2, pi/2 }.
Solutions: x = (1) e^(i (0)) = 1, or x = (1) e^(i (pi)) = -1,
or x = (2) e^(i (-pi/2)) = -2i, or x = (2) e^(i (pi/2)) = 2i,
or x = (3) e^(i (-pi/2)) = -3i, or x = (3) e^(i (pi/2)) = 3i.
Six Solutions: x in +/- 1, +/- 2i, +/- 3i.
Learned something new yay
¡Felicidades! ¡Qué buen profesor!...
The absolute value sign with obvious potential to add more solutions.
I got the real values of x = +- 1 two different ways, but didn't consider purely imaginary values. I remembered the way to check those solutions is to substitute into the original and then take the magnitude - the coefficient of the imaginary part will be equal and opposite to the real value!
Fun equation!
Write complex root z in polar form, z = r exp(I theta). From the equation, theta = 0, pi/2, pi, 3pi/2 when considering the imaginary part. Then the equation can be reduced to of real variable r for each theta. The answers then follow naturally.
The marker switching in this video is almost as impressive as the math itself.
Now show us the graph, so we can visualise it. You've got a four-dimensional blackboard, haven't you...?
4:30 was the great relevation. It's not 'not too bad', it's genius. I had been trying this for 30 minutes, then I watched this part, and was like, "Ohhhhhhh!"
Very good. I figured out the +/-2i and +/-3i but didn't find the easy ones!!
very cool, never solved an equation like this before
wow that is very cool
I think factoring the polynomial(function) first and then checking for x=+, x=+, x=+i, x=-i is faster.
Can you please explain why does 6 appear when factoring?
Hi Dr.!
So cool!
Awesome!
你好，有空可以介紹一下Bessel equation 的解與例子嗎？我不太懂甚麼時後用第一類，甚麼時候用第二類
This guy is the only person who could pull an april fools joke on me.
I really want to learn your pen switching technique so that I can flex on my class whenever my teacher calls me on the board. Great video as always though!
there is a video where he explains how he does it. i can't remember the name exactly but you can search it up. good luck c:
He uploaded that trick yesterday
Here you go:
th-cam.com/video/5-nyDHWTJ6c/w-d-xo.html
Forget that, I want the link to the tap eraser white board? Where can I get that!?
@@blackpenredpenCan we do this trick with the left hand?
2:33 I always got confused with absolute values and modulus (ie vectors) as they both use the same symbol. Finally learnt that they are actually the same function but just different names and different applications. Not the same symbol but different application 😂
parentheses meaning either composition or multiplication is hilariously awful if you think about it. or the right superscript -1, which can mean inverse or reciprocal (some ppl even use the WORDS interchangeably). or the most accursed notation in all of quantitative reasoning: when talking about exponential random variables you often see "mu = 1/ mu" which absolutely doesn't mean mu = +-1, but rather that the two instances of mu IN THE SAME EQUATION have different meanings.
Some say math is the art of giving the same name to different things. I often say that math is the art of giving different names to the same thing. Magnitude is magnitude, whether applied to scalars, vectors, matrices, or even p-adics, and whether it's called "absolute value", "norm", "modulus", "weight", "length", or "(square root of) determinant".
Honestly the biggest crime here is that the determinant uses the same symbol, but actually gives the square of the others if applied to the matrix forms of the corresponding objects. (At least when applied to 2x2 matrices.)
@@angeldude101 this is why I find it so confusing 😭
I got scared when you pulled out the blue pen 😂
Actually I can recommended a better way for getting real roots ie |x| is same as x^2 so hence from there we get 6x^2 =6 and x=+-1
Very nice.
funny is that when I solved this myself I made mistakes - wrongly solved cases b=0, so I've got -5, 1 for a>0 and 5, -1 for a
One of those "tap to erase" whiteboards. Yeah, i got one of those
The Michael Jordan refrence is awesome 👏
You should've written "COMPLEX solutions" in the title, otherwise it's easy to see it has exactly 2 real solutions.
When you are asked to give a solution of "f(x)" you gotta give all of them. Whether it is real or complex
Otherwise it will be written that you need to give a real solution..
I can argue the same way ..
This question came in my college Enterance exam!
Awesome
The way I ended up solving it was to recognize that the equation can be written as x^2 = 6-5|x|. Since the right side must be a real number, that means that x^2 must be real. This can only happen when x is real or pure imaginary. Then I split into the four cases x real and positive, x real and negative, x = ai for a positive, and x = ai for a negative.
Anyone know why in the modulus of a complex number , the b part doesn’t have an i ?
When you visualise the complex numbers as a plane with the complex number _a_ + _bi_ represented by the point _(a,_ _b),_ the modulus is just the distance of the point _(a,_ _b),_ from the origin. You can draw a right-angled triangle with length _a_ and height _b,_ with vertices (0, 0), _(a,_ 0), and _(a,_ _b),_ and then you can see that the distance between the origin and the point _(a,_ _b)_ is just the length of the hypotenuse of that triangle. That length is given by Pythagoras's theorem: the square of the length is equal to _a²_ + _b²,_ so the length itself is given by the square root of that.
Integral of x^x dx?
WOW!
for a real number a
*|a| = √a²*
for a complex number z = a + ib (a, b are real)
*|z| = √(a² + b²)*
wow!!
Instead of going for cases, set c = |b|. Then, c^2 = |b|^2 = b^2, so you have -c^2 + 5c - 6 = 0, or -(c-2)(c-3) = 0. So, we have c=2 or c=3. So, b = +/- 2 or +/- 3.
Similarly, when solving for a, we can set d = |a|. Since d^2 = |a|^2 = a^2, we have d^2 + 5d - 6 = 0, or (d-1)(d+6) = 0. So, d = 1 or d=-6. But d = |a| so must be non-negative, and we can scrap d=-6. So, d=1 and a = +/-1.
This still gives you the result that x = 1, -1, 2i, 3i, -2i or -3i.
Can you tell me the limit for lim x->inf W(x!)/(W(x)^1/W(x))? Im just curious i put it into wolfhamalpha it didn’t show me the answer
Is that related to this video?
No? Im just asking, can I?
@@Softcap I think the comment section should be for discussing the video. If comment sections are filled with random chatter about unrelated matters, it makes it harder for people who want to discuss the video to find relevant discussions.
@@Softcap I think the limit is infinity, since W(x)^(1/W(x)) goes to 1 (i think)
Awesome! I'm wondering if it's a coincidence that your solutions are all the positive and negative factors of the last term, 6 (+- 1,2,3, and the 6 you rejected).
Does taking the absolute value, and using complex numbers for solutions provide all the positive and negative factors of c as solutions?
I think the coefficient on the linear term is also important
5:27 I am not sure if I understand the next step of -b^2 + 5root(b^2) -6 = 0 completely. When b is a real number, then root(b^2) should always return a positive number (principal value of square root). I would have just cancelled the square and the square root. So I don’t understand why it should be treated like an absolute value and divide cases upon that
because your way disregards negative values of b
It is the fact sqrt(b²)≥0 makes you not able to "cancel" sqrt and square. If you put b=-1, you will have sqrt(b²) = 1, but if they're "cancelled" naively, the result will be just b = -1, a different number, meaning cancellation was not allowed since it gave different result. Instead, we write sqrt(b²) = |b| = {b if b≥0, -b if b
Instead of making 2 cases for, wouldn't it be more efficiente to resole the polynôme in absolute value of b and say b=+or - its absolute value?
Absolute value doesn't tell you the value of x^2 when x is complex.
@@Skyler827b is the imaginary part
Today in college I ended up trying to solve the equation
ln(x)-1/x=0
I eventually ended up trying to use the w function that I came across on this channel:
xe^x=e
xe^x=(1)e^1
W()=W()
x=1
This is clearly not a solution, and I am now left wondering how it should be solved, any help would be greatly appreciated.
Is x allowed to be complex?
Look at the boxes and boxes of markers behind you!
Is there a general formula for equations of this form?
if you only want the real solutions you can use the quadratic formula two times.
ax² + b|x| + c
x
6 solutions! 6 championships! And if he hung the e picture upside down, another 6!
Three sixes, the cursed number.
And accidental factorial got in your comment too.
6! = 720, don't forget
OK so I always try before I watch. I came up with only x = 1 and x = -1.
I used x²+5x-6 = 0 for x greater than or equal to 0. I used x²-5x-6 for all x less than 0.
Looks like there were complex solutions. I never would have thought...
By assuming | x|= t then solving for t we get t=1, -6 that is |x|=1, -6
X= +,-,1 any four imaginary solution
Edit: is there any mistake and if yes please point it out
Bro what if we just let mod x = y and we know that x² = modx² so eq become y²+5y-6 then we get( y-6)(y+1)= 0 then mod = 6 and mod x = -1 after that if we open mod it becomes +-6and -+1 so there are four solutions?
I was like huh lets see whst this is... Then a few minutes into thr vid im like wait my brain is to small for this😅
How do we know that a and b can't be both different than 0?
How come if b
Same question
Can you do sin(sin(x)) = sin^2(x). I completely forgot the complex form of sine 😢
Very nice problem!
Here is my solution, which is a bit different.
x²+5|x|-6=0
⇔x²=6-5|x|
So x² is real
If x²≥0, then x is real, so x²=|x|²
|x|²+5|x|-6=0
(|x|+6)(|x|-1)=0
As |x|≥0, |x|=1
So x=±1
If x²
I was really just lucky to create this equation with 6 solution by accident. Thanks for the solution on why such an equation cannot have 8 solutions!
I was disappointed when I found out the 4 other solutions are complex numbers
I don't know why my previous comment is not shown... A better way to approach this problem is to write complex root z in polar form, z = r e(i theta), r > 0. Deduce that theta is multiple of pi/2, then the equation can be transformed into quadratic of positive r, and be solved easily.
Hey Blackpenredpen, can you find the limit x->0((x!)^(((1)/(x))))?
pls
Is it possible to make 8 solutions? I don't think so because it seems the 4 cases are just sign flipped versions of the original equation. And the solutions will only both be valid if they are both positive or both negative. I don't think it's possible to maintain this restriction across all four cases.
No, you can't get 8 solutions. Please see my comment for a proof.
Has unlimited solutions. Let me exp.
First if u want the solution you should know |x|²=x². Then you will have (|x|+6)(|x|-1) you know x=1 and -1 for real. İf u want the solution for complex x ypu should think |x|=1 , |a+bi|=1, a²+b²=1. this is a circle and has unlimited solutions.
wow that's so bizzar. I'd think that there r 4 solutions, 2 for each of the +/- version.
x = 1 or -1 ?
7:11 that's literally the exact same equation you started with, but now you approach it from the angle of considering "a>0 or a
There's one slight difference, you can only take the condition y>0 and y
You can do this at the start too, but you'll get only real answers, i.e., x=±1
OK so the reason this is the only way that works is because supposing |x| = root(x^2) is not true for complex numbers. Therefore you are solving the equation for x in R. Same for x^2 = |x|^2 as well as if you consider cases. (|x|=x or -x is not true for complex numbers)
This is almost identical to the video's solution, but just to see it with a different spin:
First note that x^2 + 5 |x| - 6 = 0 means that x = s is a solution implies x = -s is also a solution.
Big observation is that x^2 + 5 |x| - 6 = 0 implies x^2 = - 5 |x| + 6, which is real.
But x^2 real implies x is either real or pure imaginary (x^2 real means doubling its argument puts you on the x-axis).
In the real case, x = a, and since know x = -a also works, solve it assuming a >= 0, so have: a^2 + 5a - 6 = 0.
Thus a^2 + 5a - 6 = (a - 1)(a + 6) = 0, so a in {1, -6} and a >= 0, so a = 1.
Thus the real solutions are x = +/-1.
In the pure imaginary case, x = a i, a real,
so the equation x^2 + 5 |x| - 6 = 0 becomes -a^2 + 5 |a| - 6 = 0, or a^2 - 5 |a| + 6 = 0.
Again, since know x = -ai also works, solve it assuming a >= 0:
a^2 - 5 a + 6 = 0 => (a - 2)(a - 3) = 0 => a in { 2, 3 }.
Thus get the 4 imaginary solutions +/- 2i, +/- 3i.
Whiteboard erasing technique is awesome!
Thanks!!!
He is best mathematical scientist in today’s Era.❤
What if we put the abs part somewhere else, like maybe |x^2 + 5x| -6, or even nest them like |x^2 + 5 |x| |
@teelo12000 That's really interesting wtf
The first one will turn into x²+5x±6=0, ==> x=+1,-2,-3,-6
And the second will turn into |x|²+5|x|=0, ==> |x| (|x|+5)=0, ==> x=0
Try solving using graph
What about (floor(x^2)) + 5x - 6 = 0
😂
Alternative solution - it's easy to see from the equation that x^2 is a real number - which can only happen if x is either real or imaginary (only).
If x is real then x^2 = |x|^2 and then solving |x|^2 - 5|x| + 6 yields |x|=1 or |x|=-6. As |x| must be positive we get x=+-1.
If x is imaginary then x^2 = -|x|^2 which yields the equation |x|^2-5|x|+6=0. This has the solutions |x|=2 or |x|=3, and therefore x=+-2i or x=+-3i.
0:12 - 0:17...bprp, cut it out, you're silly...!! 😂😂🤣🤣 So you're an MJ era Bulls guy, are ya?? 😁😁
Anyways, back on topic I thought about the possible solutions for the equation and right off the bat I knew complex solutions had to be involved; I was considering a few cases where x
wow that was nice could you do this integration x^2 sin(x^2) in 3 methods or more and if you can do it through Feynman Technique?
I don't understand why -2 and -3 work like that. Why does 5|b| have different value when b is negative? Did you mean to write |bi|?
|b| = -b
|-5| = -(-5)
|-5|=5
@@BryanLu0 That's not what I meant. That's obvious.
And it's not like I can even doubt the solution. Obviously I can plug it in and demonstrate that it's correct.
What I didn't get is why it makes a difference. 5|b| = 5|-b|, by definition, so why does it allow us to get a different result at all? It should be the same.
But that was the insight I was missing easier: It's an equation. The whole point is that it's the same.
why does he define sqrt(a^2) = | a | ? isn't sqrt(a^2) = a ?
Sqrt is generally defined to only output the positive answer. If you also want the negative answer, you put +/- in front like we see in the quadratic formula.
So if sqrt only gives the positive root, sqrt(a^2) only gives a positive value, regardless of whether a was positive or negative- I.e. we get the absolute value.
|x| forces x^2 to be real, which forces x to be purely real or purely imaginary. very swag, cool eqn.
I didn't undersand why 2abi must be equal to zero, someone could explain?
lets supose you have A+Bi = C+Di.
since you can't add or substract the real part from the imaginary part (and vice versa) you can write this as A=C and B=D (in a complex number the Re and Im are linearly idependent)
In the case of the video, C = 2ab and D=0, so 2ab = 0
Two solutions the real numbers, six solutions in the complex numbers, infinitely many solutions in the quaternions.
Since x^2 is real, x can only be real or imaginary. You can divide it into real and imaginary cases and you don't have to use 2 real variables.
Wait so what are the x intercepts??
pls answer my question why i²=(-1)
since sqroot -1 times sqroot of -1 is sqroot of -1 whole square which is actually mod of |-1| bcz -1 is real
Sqrt(-1) * sqrt(-1) = -1, not sqrt(-1)
Maybe it's easier to see with examples
Sqrt(x) * sqrt(x) = x
Sqrt(2) * sqrt(2) = 2
Is it possible to solve:
x^(5x)=x+4
?
Please help 🙏
Yes possible by Lambert w function.
Hmmm, i'm confused. If we plugin X=2i in the first equation we get sqr(2i) + 5|2i| - 6 = 0. This becomes -10+10i = 0 and this doesn't seem right.
It confused me for a bit, but I think |2i| is actually 2 as the absolute value is just the (real) distance from the origin on the complex plane
@@robertchambers8725 right. It's easy to get lost in complex plane :)