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.
@@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
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?"
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. :)
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.
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.
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.
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.
Alternative solution: Suppose x is a real number. Using 1:01 , you'll find +-1 as solutions. Now, suppose x is a complex number with non zero imaginary part. From the original equation: x²=6-5|x| For q=6-5|x|, q has to be a real number. But if q>0, x²=q makes x a real number, and we already now the solution; so, let q x = +-2i Then x must be in {-1,+1,+2i,-2i,+3i,-3i} ---- I judge the solution as alternative as it wasn't necessary to use x=a+bi, which reduces for only 2 quadratic equations to solve. However, it was needed a longer argumentation.
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
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
I found it even easier to do with a complex phase approach: x = re^(iy). The equation then becomes r^2 * e^(2iy) + 5r - 6 = 0. Since everything but the phase is real, e^(2iy) must also be real, so either 1 or -1. If it is 1, the solutions are +r and -r, while if it is -1 they are +ir and -ir. Now we just have to get solve both cases for r. For phase 1 we get r = 1 and r = -7. We accept only r = 1 because it must be positive by definition. For phase -1 we get r = 2 and r = 3, which are both positive and thus both acceptable. Therefore, we have 1, -1, 2i, -2i, 3i, -3i.
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.
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.
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.
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 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
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.)
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?
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.
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!
I foind the 2 real splution, however for the complex ones I just guessed they were the factkrs of 6 and it just happends to work out. After finding out there were complex solutions, I added 2 mkre cases for ±i5x which only gave me ±3i as new solutiond, and afterwards I was stumped
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
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)
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.
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
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!
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.
@@Ðęłẽțëď 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.
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.
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.
Another way: x^2 + 5|x| - 6 = 0 implies that x^2 is real because x^2 = 6 - 5|x|. If x^2 is real, then x must be either real or purely imaginary. If x is real, then x^2 = |x|^2 and the original equation can be written as |x|^2 + 5|x| - 6 = 0. Solving this quadratic equation in |x|, we get |x| = 1 (|x| = -1 must be rejected because |x| cannot be negative), therefore we get the solutions x = 1 or x = -1. On the other hand, if x is purely imaginary, then x = iy, with yu real. Substituting this in the original equation, -y^2 +5|iy| - 6 = 0. Again, as y is real, y^2 = |y|^2. Also, |iy| = |y| always, as |i| = 1. Therefore, -|y|^2 + 5|y| - 6 = 0. Solving this quadratic equation in |y|, we get |y| = 2 or |y| = 3, therefore y = 2, y = -2, y = 3 or y = -3, therefore we arrive at the other 4 solutions x = 2i, x = -2i, x = 3i, x = -3i.
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 ..
I agree, because if we just create more "imaginary numbers" for expressions that are undefined, there might be more solutions. So this is why I consider "solutions" as real solution and I call complex solutions, well complex solutions
@@thexoxob9448 Exactly, always when you introduce a variable in maths you ALWAYS have to tell the set in which it is included, otherwise you can always create new things that satisfy what you're talking about
@@thecrazzxz3383i thought complex numbers aren’t new, and he did explain in the video, which i feel like is why you’d click on it in the first place if you knew what solutions there are (also, aren’t there 4 complex and 2 real ones?) i have no idea what i’m going with this but just my opinion ig
Why is the absolute value of a complex number is the distance of that number from the origin? Like, the absolute value of a+bi is √(a²+b²). But abs(x) is also equal to √(x²). So abs(a+bi)=√((a+bi)²)=√(a²-b²+2abi). Are you saying that √(a²+b²)=√(a²-b²+2abi) ?
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
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.
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.
@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!
5:23 "consider cases ..." Nope. If b is real, then b² = |b|². So, the equation b² - 5|b| - 6 = 0 is just the equation |b|² - 5|b| - 6 = 0 so (|b| - 2)(|b| - 3) = 0 which gives b = ±2, ±3 immediately. The same thing applies to the equations involving a in the end.
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?
@@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.
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
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.
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?
Solving log-power equation x^ln(4)+x^ln(10)=x^ln(25)
ua-cam.com/video/xBpZRWCGw30/v-deo.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
Your definition of absolute value is undefined for x=0. It should return x if x >= 0.
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.
Thought it'd be way harder when applying complex numbers but that ab = 0 condition made it trivial, really cool
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.
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.
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.
x⁶-21x⁴+5x³+54x²+30x-24 = 0
has 6 solutions too.
Is it random? Because one root is φ⅔+φ⁻⅔ !
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 :)
Alternative solution:
Suppose x is a real number. Using 1:01 , you'll find +-1 as solutions.
Now, suppose x is a complex number with non zero imaginary part. From the original equation:
x²=6-5|x|
For q=6-5|x|, q has to be a real number. But if q>0, x²=q makes x a real number, and we already now the solution; so, let q x = +-2i
Then x must be in {-1,+1,+2i,-2i,+3i,-3i}
----
I judge the solution as alternative as it wasn't necessary to use x=a+bi, which reduces for only 2 quadratic equations to solve. However, it was needed a longer argumentation.
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
I found it even easier to do with a complex phase approach: x = re^(iy). The equation then becomes r^2 * e^(2iy) + 5r - 6 = 0. Since everything but the phase is real, e^(2iy) must also be real, so either 1 or -1. If it is 1, the solutions are +r and -r, while if it is -1 they are +ir and -ir. Now we just have to get solve both cases for r. For phase 1 we get r = 1 and r = -7. We accept only r = 1 because it must be positive by definition. For phase -1 we get r = 2 and r = 3, which are both positive and thus both acceptable. Therefore, we have 1, -1, 2i, -2i, 3i, -3i.
I love your channel bro, you’re the reason that I can understand calculus and algebra so easily as a freshman, thank you🙏
Honestly maths would not be the same for me without these creative problems on this channel
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.
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
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.
Now show us the graph, so we can visualise it. You've got a four-dimensional blackboard, haven't you...?
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.
great video and great solution!
You should try stuff like z^2 +az* + b = 0 where z* is the complex conjugate of z.
The marker switching in this video is almost as impressive as the math itself.
this new video really made my day
so satisfying... literally! I am disappointed that the video has already ended
Thank you sir ❤
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 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!
Please make a video about Bessel functions!
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 😭
The absolute value sign with obvious potential to add more solutions.
This guy is the only person who could pull an april fools joke on me.
Wow this is amazing ❤
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?
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.
Thank you,sir
Solution:
for real values:
|x| = √(x²)
for complex values (x = a + bi):
|x| = √(a² + b²)
so |i| = |0 + 1i| = √(0² + 1²) = √1 = 1
So for the real solutions, we have:
x² + 5√(x²) - 6 = 0
And for the complex solutions, we have:
(a + bi)² + 5√(a² + b²) - 6 = 0
a² + 2abi - b² + 5√(a² + b²) - 6 = 0
The "2abi" component *has* to be 0 for this equation. Therefore either a or b has to be 0.
If b = 0, x is not a complex value, therefore a = 0
(0 + bi)² + 5√(0² + b²) - 6 = 0
(bi)² + 5√(b²) - 6 = 0
-b² + 5√(b²) - 6 = 0
Let's solve for the real values:
x² + 5√(x²) - 6 = 0 |-x² +6
5√(x²) = 6 - x² |² → need to check for extraneous solutions!
25x² = 36 - 12x² + x⁴ |-25x²
x⁴ - 37x² + 36 = 0 |x² = u
u² - 37u + 36 = 0
u = -(-37)/2 ± √((-37/2)² - 36)
u = 37/2 ± √(1369/4 - 144/4)
u = 37/2 ± √(1225/4)
u = 37/2 ± 35/2
u₁ = 72/2 = 36
u₂ = 2/2 = 1
Since x² = u
x₁ = 36
x₂ = -36
x₃ = 1
x₄ = -1
check for extraneous solutions:
x² + 5√(x²) - 6 = 0
x₁ → 36² + 5√(36²) - 6 = 0 → left side is way to big, x₁ is an extraneous solution
x₂ → (-36)² + 5√((-36)²) - 6 = 0 → left side is way to big, x₂ is an extraneous solution
x₃ → 1² + 5√(1²) - 6 = 0 → 1 + 5 - 6 = 0 → 0 = 0
x₄ → (-1)² + 5√((-1)²) - 6 = 0 → 1 + 5 - 6 = 0 → 0 = 0
Therefore, the only real solutions are:
x₁ = 1
x₂ = -1
Now let's focus on the complex solutions:
-b² + 5√(b²) - 6 = 0 |+b² +6
5√(b²) = b² + 6 |² → need to check for extraneous solutions!
25b² = b⁴ + 12b² + 36 |-25b²
b⁴ - 13b² + 36 = 0 |b² = v
v² - 13v + 36 = 0
v = -(-13)/2 ± √((-13/2)² - 36)
v = 13/2 ± √(169/4 - 144/4)
v = 13/2 ± √(25/4)
v = 13/2 ± 5/2
v₁ = 18/2 = 9
v₂ = 8/2 = 4
Since b² = v
b₁ = 3
b₂ = -3
b₃ = 2
b₄ = -2
check for extraneous solutions:
-b² + 5√(b²) - 6 = 0
b₁ → -3² + 5√(3²) - 6 = 0 → -9 + 15 - 6 = 0 → 0 = 0
b₂ → same as b₁, as there is only b²
b₃ → -2² + 5√(2²) - 6 = 0 → -4 + 10 - 6 = 0 → 0 = 0
b₄ → same as b₃, as there is only b²
No extraneous solutions.
With x = a + bi and a = 0, we end up with
x₃ = 3i
x₄ = -3i
x₅ = 2i
x₆ = -2i
Finally:
x ∈ {-3i, -2i, -1, 1, 2i, 3i}
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
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!
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.
I foind the 2 real splution, however for the complex ones I just guessed they were the factkrs of 6 and it just happends to work out. After finding out there were complex solutions, I added 2 mkre cases for ±i5x which only gave me ±3i as new solutiond, and afterwards I was stumped
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
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)
I got scared when you pulled out the blue pen 😂
WOW!
for a real number a
*|a| = √a²*
for a complex number z = a + ib (a, b are real)
*|z| = √(a² + b²)*
wow!!
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?
The solutions come easily if you set y = |x|. Since x^2 = |x|^2. Then apply the quadratic formula
¡Felicidades! ¡Qué buen profesor!...
How come if b
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
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:
ua-cam.com/video/5-nyDHWTJ6c/v-deo.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?
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...
XX - 5X + 6 = 0
* 6 = 1x6 = 2x3
2+3 = 5 = (b)
** 2: 2x2 - 5x2 + 6
4 - 10 + 6 = 0
**. 3 : 3x3 - 5x3 + 6
9 - 15 + 6 = 0
***. X' = 3 , X" = 2./.
This is the beauty of math , it does not leave possible cases even if they are thousands in numbers ❤
Fun equation!
Michael Jordan might have 6 rings but Ariana Grande has 7
Please make video about Lambert Functions
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
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?
@@Ðęłẽțëď 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.
@@Ðęłẽțëď I think the limit is infinity, since W(x)^(1/W(x)) goes to 1 (i think)
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.
The Michael Jordan refrence is awesome 👏
I was disappointed when I found out the 4 other solutions are 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.
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
Hey Blackpenredpen, can you find the limit x->0((x!)^(((1)/(x))))?
pls
Another way: x^2 + 5|x| - 6 = 0 implies that x^2 is real because x^2 = 6 - 5|x|. If x^2 is real, then x must be either real or purely imaginary. If x is real, then x^2 = |x|^2 and the original equation can be written as |x|^2 + 5|x| - 6 = 0. Solving this quadratic equation in |x|, we get |x| = 1 (|x| = -1 must be rejected because |x| cannot be negative), therefore we get the solutions x = 1 or x = -1. On the other hand, if x is purely imaginary, then x = iy, with yu real. Substituting this in the original equation, -y^2 +5|iy| - 6 = 0. Again, as y is real, y^2 = |y|^2. Also, |iy| = |y| always, as |i| = 1. Therefore, -|y|^2 + 5|y| - 6 = 0. Solving this quadratic equation in |y|, we get |y| = 2 or |y| = 3, therefore y = 2, y = -2, y = 3 or y = -3, therefore we arrive at the other 4 solutions x = 2i, x = -2i, x = 3i, x = -3i.
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 ..
I agree, because if we just create more "imaginary numbers" for expressions that are undefined, there might be more solutions. So this is why I consider "solutions" as real solution and I call complex solutions, well complex solutions
@@thexoxob9448 Exactly, always when you introduce a variable in maths you ALWAYS have to tell the set in which it is included, otherwise you can always create new things that satisfy what you're talking about
@@thecrazzxz3383i thought complex numbers aren’t new, and he did explain in the video, which i feel like is why you’d click on it in the first place if you knew what solutions there are (also, aren’t there 4 complex and 2 real ones?) i have no idea what i’m going with this but just my opinion ig
Learned something new yay
Why is the absolute value of a complex number is the distance of that number from the origin? Like, the absolute value of a+bi is √(a²+b²). But abs(x) is also equal to √(x²). So abs(a+bi)=√((a+bi)²)=√(a²-b²+2abi). Are you saying that √(a²+b²)=√(a²-b²+2abi) ?
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
very cool, never solved an equation like this before
Very good. I figured out the +/-2i and +/-3i but didn't find the easy ones!!
Hi Dr.!
So cool!
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.
One of those "tap to erase" whiteboards. Yeah, i got one of those
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.
Stunning😮
property |x|²=x²
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!
On 2:50, the imaginary length is bi not just b
It’s on the i axis. Length (magnitude) is just b.
x = 1 or -1 ?
5:23 "consider cases ..."
Nope. If b is real, then b² = |b|². So, the equation
b² - 5|b| - 6 = 0
is just the equation
|b|² - 5|b| - 6 = 0
so
(|b| - 2)(|b| - 3) = 0
which gives
b = ±2, ±3
immediately. The same thing applies to the equations involving a in the end.
wow that is very cool
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.
What about (floor(x^2)) + 5x - 6 = 0
😂
This question came in my college Enterance exam!
I actually expected this to have eight 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😅
Awesome!
Look at the boxes and boxes of markers behind you!
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 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.
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
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
He is best mathematical scientist in today’s Era.❤
Hey what are your thoughts about the 'annals of mathematics studies' series?
Why 2ab = 0?
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.
Two solutions the real numbers, six solutions in the complex numbers, infinitely many solutions in the quaternions.
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