You wouldn’t expect this "quadratic" equation to have 6 solutions! 

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

Then you need a defibrillator when watching his videos.

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

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

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.

Too many sol.

I think I’m going to solve a polynomial of degree 100,900,000

@@table5584 Go for it :)

Yes, but |x| also applies for complex values, |xi|=x, |-xi|=x. So your logic works.

incorrect assertions in the thumbnail is definitely an effective way to engage the math nerds.

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 ..

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.

if you only want the real solutions you can use the quadratic formula two times. ax² + b|x| + c x

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

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 😭

Absolute value doesn't tell you the value of x^2 when x is complex.

​@@Skyler827b is the imaginary part

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: ru-vid.com/video/%D0%B2%D0%B8%D0%B4%D0%B5%D0%BE-5-nyDHWTJ6c.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?

Is x allowed to be complex?

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)

@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

I think the coefficient on the linear term is also important

Looks like there were complex solutions. I never would have thought...

we have an equation where REAL + 2abi = 0 if 2ab is not 0, then the right hand side would have a non-zero complex part, but it does not. Maybe it's easier to understand if you see 0 as 0+0i. we are simply matching the complex parts on either side.

Three sixes, the cursed number.

And accidental factorial got in your comment too.

6! = 720, don't forget

pls

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!

Same question

I think he should have explained that the coefficient of 5|b| is still actually +/- 5|b| because of the +/- required by the quadratic equation. Thus, his case for b>0 has the +/- in it, but then four solutions for b>0 are b={+2, +3, "-2", "-3"}, but since the values of b={"-2", "-3"} violate the assumption of b>0, you discard these extraneous solutions for b>0, and thus for b>0 you are left with only b={+2, +3}. Proceeding then for b

No, you can't get 8 solutions. Please see my comment for a proof.

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

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

True if x is real. If x is not real, then it doesn’t hold. Try x=2i. The left hand side is 4 but the right hand side is -4.

​​@@blackpenredpen i thought that you were solving in IR

@@spacetimeslasher Why solve this in infrared radiation??

@kimtak805 Oh, my bad haha How silly of me. Of course he meant the real numbers. And I was about to tell everyone to put their sunscreen on! Thanks for clearing up this misunderstanding, kimtak805!

Yes possible by Lambert w function.

A complex number is 0 if both the real and imaginary part are 0. The imaginary part is 2ab.

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

No, you can't get 8 solutions. Please see my comment for a proof.

0*x^2+|x|-1=0 has infinity solutions.

Then the left part of the equation is complex, and therefore is not equal zero, despite the right part is zero.

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.