Awww It looks like the pi creatures from 3 blue 1 brown
@karthiksriram4632
4 жыл бұрын
The new mic is awesome😍...3blue1brown will be happy😄...
@jonathangrey6354
4 жыл бұрын
“Let me put y iiiiin blue.” “No why would I do that, I’m blackpenredpen.”
@blackpenredpen
4 жыл бұрын
Nole Cuber Lol
@lazydoctor9207
4 жыл бұрын
Lol😂😂😂😂
@johannesgoeth3448
4 жыл бұрын
That line killed me!!! XD XD XD
@mariomestre7490
4 жыл бұрын
Ets genial . Des de Catalonia
@YoshBruh
4 жыл бұрын
"You guys should try and solve first" Me at 3 am: No, no I dont think I will
@xavierplatiau4635
4 жыл бұрын
Lol, it’s exactly what I did, just watch the thumbnail, was like « Ok (1;1) is a solution when x=y » and then... « If you multiply the second equation by 5 we get a polynomial expression that we can factorise by (x-y) and something of degree 2...damn I’ve got to try it » and so I did lol, it works fine.
@abs0lute-zer061
4 жыл бұрын
That's a golden answer for y.
@justaconcrete4789
4 жыл бұрын
Yeah thats omega golden ratio
@ameerunbegum7525
3 жыл бұрын
.
@CRT-n9k
2 ай бұрын
that's phi not Omega@@justaconcrete4789
@shahbazsheikh3545
4 жыл бұрын
The seamless switching of the pens was just as impressive as the math.
@blackpenredpen
4 жыл бұрын
8:46 golden ratio!!!!!!
@myrus5722
4 жыл бұрын
6:30 instead of doing polynomial division, if you add y - y and split the -2y^2 into -y^2 and -y^2 you get (y^3 - y^2) - (y^2 - y) - (y + 1). You can now see that you can take out a (y-1) from each of those. I got this, but admittedly I did do the polynomial division first then looked for a prettier solution once I knew the end result, but it’s still cool imo: A reminder that factoring non-obvious polynomials like this is in some ways about “creating or finding the symmetry” which I think is a helpful way to look at it for other factoring problems too.
@linguinelabs
4 жыл бұрын
This is my favourite content. These videos are keeping my math fresh while in quarantine. Thank you for making these!
@blackpenredpen
4 жыл бұрын
You’re welcome! I am happy to hear : )
@przemezio
4 жыл бұрын
First try: check if x=1, y=1 Result: it works 😳 Conclusion: I am a god. Greetings from Poland 🇵🇱 (now, it's 3.57 AM here, so I should go sleep 😴).
@pmj9925
4 жыл бұрын
siema byku
@incoherentproductions992
4 жыл бұрын
I also solved it that way
@anirvinvaddiyar7671
4 жыл бұрын
Wait I actually solved it using that
@GamingDragon07
4 жыл бұрын
Same way I did
@-kat
4 жыл бұрын
Guess and check ftw
@drpkmath12345
4 жыл бұрын
Shout out! "DO NOT Automatically substitute!" Haha especially differential equation. I like your pi. Would like to buy that one
@scoutskylar
4 жыл бұрын
The pi plushie is from 3blue1bown's store: store.dftba.com/collections/3blue1brown?_=pf&pf_pt_product_type=Plushies#bc-sf-filter-products
@mathadventuress
4 жыл бұрын
@@scoutskylar but how did he make it into a microphone lol
@scoutskylar
4 жыл бұрын
@@mathadventuress Probably just a lapel mic clipped onto the back
@equilibrium7756
4 жыл бұрын
This mic looks like a logo of one of the math youtube channel, but i don’t remember which exactly
@greatestlists7810
4 жыл бұрын
@@equilibrium7756 3blue1Brown
@MuPrimeMath
4 жыл бұрын
3blue1brown gang!!!
@blackpenredpen
4 жыл бұрын
Yea lol!!!
@F1U7R2Y9
4 жыл бұрын
Plz do a collab on a topic of math plzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzz
@southernkatrina8161
4 жыл бұрын
And Eddie Woo!
@blightedcrowmain8236
4 жыл бұрын
3brown1blue
@julianmejiac
4 жыл бұрын
6:32, I called this "Ruffini's rule", a particular version of "Horner's method" ( usually called synthetic division). I remembered I learned this in highschool, and I used to do it without thinking. Then I understood, how division of polynomials work, and never used it again until the point I completely forgot about the algorithm.
@Danicker
4 жыл бұрын
Yes it's a good way to crunch the numbers quickly in an exam but it's important to understand what's really going on behind the scenes
@friendtilldawn4380
4 жыл бұрын
just like almost every single math problem u start losing urself throughout the process and question if 1 + (-1) is 0 :))
@osmeridium
4 жыл бұрын
This video is very mathematically satisfying
@younesabid5481
4 жыл бұрын
Grant would be proud of your new microphone :)
@chillfrost3170
Жыл бұрын
about the division made in 8:00, I can affirm that in Portugal, my country, we learn it in high school (10th grade), but it is called as Ruffini's Rule
@rohitjacob803
4 жыл бұрын
What I did is that I first took x^3 common from the first equation and y^3 from the 2nd.Now I took (y/x) as some value 'a' and divided the two equations.When we simplify this we get a cubic equation having only the value 'a'. If u solve that equation, we get many more terms like the solutions of x^3 = 1 and y^3 = 1 and some more, which also includes the answers u found.But yours was an elegant method and much more easier. Great videos btw
@dmitryweinstein315
3 жыл бұрын
Man, you use colours so well! A very efficient lecturer, too...
@superg6787
4 жыл бұрын
I finally got one of these questions right!!! You guys don’t know how happy I am!!!
@amalwijenayaka410
3 жыл бұрын
kzitem.info/news/bejne/zKWtnp2gkauHgIY
@JohnDoe-lm6oe
3 жыл бұрын
Factor second equation: Y^2(X+Y)=2 1 is the only square factor of 2, therefore Y is equal to 1 Substitute known value of Y into the equation and get 1(X+1)=2 1(1+1)=2 X=1
@sergioh5515
4 жыл бұрын
"Hopefully we can get points for this question" 😅👍 yes you do Steve. And I loveeeeeeee your little pi figure from 3blue1brown? And excellent video of course.
@blackpenredpen
4 жыл бұрын
Yes. Thank you Sergio!
@henrique3194
4 жыл бұрын
Man I love the solutions you bring, they are always so fun to see!
@idrisShiningTimes
4 жыл бұрын
Great video. My expansions concept got cleared! Thanks!
@michaelvarney.
4 жыл бұрын
By inspection I noted that x = 1 and y = 1 are solutions. I then factored out the solution for x or y as (x-1) or (y-1) and then solved the system for the other solutions.
@a2te45
4 жыл бұрын
Ahhh that's very clever! Finding (1,1) by inspection was not very difficult, but I forgot about factoring out those solutions to reduce each equation to a problem in quadratics. Excellent!
@Skandalos
4 жыл бұрын
How do you factor these out?
@achyuththouta6957
4 жыл бұрын
@@Skandalos Using the division method probably. There are many methods but this is the method they teach in most schools.
@mhasan137
4 жыл бұрын
The system is symmetric with respect to the exchange of x and y. Then replacing y by x in either of the equations gives us x³=1. Then it is straightforward to find the three roots of x. The real solution is x=1, and the two complex roots are 120° symmetric with respect to the real root, on the complex plane.
@satyapalsingh4429
4 жыл бұрын
Your approach is praiseworthy .
@vari1535
4 жыл бұрын
Glad to see 3b1b spreading around just as well as *the thing*
@eleazaralmazan4089
4 жыл бұрын
Nice! The solution presented in this video is super elegant!
@calcul8er205
4 жыл бұрын
You can also divide equation 1 by equation 2 to reduce it to a cubic in terms of t=x/y however substituting back in to solve for x and y is much messier
@MAXYANKEE5
4 жыл бұрын
Are you sure you performed the division correctly. How did you manage to get a cubic of x/y?
@SuperThePaja
4 жыл бұрын
what a nice video!, youtube algorithm making my quarentine night better
@blackpenredpen
4 жыл бұрын
Hahaha thank you.
@JeffreyLByrd
4 жыл бұрын
(1,1) is really easy to find just by inspection, especially if you factor out an x^2 and y^2 from each equation. Then you can skip straight to finding the other two solutions.
@Math342010
11 ай бұрын
On 7:59, Bprp, you said "I'm not sure whether this is a popular method". Yes, in Indonesia, the method (If I am not mistaken: Horner Method) was popular just for a while (maybe until 2013). But since Indonesian math curriculum did not add this in the newest curriculum (which is from 2014), not many know this method.
@demogorgon2125
3 жыл бұрын
Salute to his highly developed brain and his courage to become so efficient in mathematics at such a young age
@demogorgon2125
3 жыл бұрын
Everytime I see his vids along with the solutions I wonder why such solutions don't come to my mind?😂
@lumina_
Жыл бұрын
"highly developed brain and his courage to become so efficient in mathematics at such a young age" bro what?
@sungwoojung9645
4 жыл бұрын
Are x and y real numbers?(I couldn't find the conditions). If complex numbers are also possible, x=y=(-1+sqrt(3)*i) / 2 and x=y=(-1-sqrt(3)*i) / 2 can be solutions as well.
@jomama3465
4 жыл бұрын
Golden ratio sure appears everywhere.... including in olympiad problems
@VibingMath
4 жыл бұрын
If there is a collab betwem bprp and 3b1b, little rabbit and pieeeeee will be very happy 😍
@killhean5042
4 жыл бұрын
Here's a nice trick : Multiply the bottom equation by 5 and subtract it from the first equation. The resulting equation is equal to zero. Now notice x = y is a root of the equation since the sum of coefficients is zero. Divide out the factor (x-y) using long division. Now solve for the other factors using the quadratic formula! Edit : Don't forget to substitute into either of the first two equations to find the 'initial condition' of the system
@lightyagami6647
4 жыл бұрын
Lol I just thought of same thing
@playcloudpluspc
4 жыл бұрын
Could you elaborate on this please? (x-y)(x^2 + 10xy + 5y^2)=0. Is that what you meant by dividing out the factor (x-y) from the equation? If so I don't see how I can apply the quadratic formula or how it helps me, I would be very grateful for your assistance. On further reflection using your method I get y= cuberoot(1/(-2 +- sqrt(5)) which gives the right answer but not in the same form.
@playcloudpluspc
Жыл бұрын
@Memes shorts Thank you.
@playcloudpluspc
Жыл бұрын
@Memes shorts Well KZitem informed me of your reply, so I wanted to thank you for taking the trouble. You're right it is amusing though 😂.
@zoromarak5649
4 жыл бұрын
Since both equations are homogenous. Let y = mx for some real number m and substitute that in both equations. This will eliminate the y variable. Now factor x^3 from both equations and divide the equations . Now youll get the cubic polynomial in m . Solve for m and back substitute to get values of x & y for corresponding values of m.
@TheManikarna
4 жыл бұрын
since the given equations are homogenous in nature put y=kx, later eliminate x^3 find values of k, we will get one solution as k=1 and another quadratic equation of k, let the roots be k1,k2,k3 we will get y1,y2,y3, and substitute in the given equations, we will get x1,x2,x3 now again put in y1=k1*x1, similarly y2=k2*x2, and y3 also
@vishalmishra3046
4 жыл бұрын
Adding Eq1 + 3 x Eq2 gives (x + 3y)^3 = 64 = 4^3. So, x + 3y = 4 x (1, w, w^2) [ w = cube-root of unity ]
@harshpanchal3131
4 жыл бұрын
You can also put y=m*x in the 2nd eq and then divide both eqs
@mathemagicallearning9593
4 жыл бұрын
BPRP’s videos are so good that it has 1 view and 7 likes
Here's an interesting problem: Find all integers "x" that satisfy x/a = x-a for a given integer "a"
@Ooopsi
4 жыл бұрын
Rearranging gives us a^2-ax+x=0 We will search for integer values of x where the solution a is an integer too. a=(x+-sqrt(x^2-4x))/2 For a to be integer we should have no square roots, thus x^2-4x must be a perfect square. x^2-4x can be written as (x-2)^2-4, which is some perfect square minus 4. Now this number is only a perfect square when it is equal to zero (can someone help me prove this?) x^2-4x=0 gives x=0 and x=4 Quickly checking gives us for x=0 a=0 and for x=4 a=2 I dont think I missed any solutions, I just need to prove that (x-2)^2-4 is only a perfect square when equal to zero, can someone tell me how to do it? Edit - nvm I figured it out, here you go: The question is: When is x(x-4)=n^2 for some integers x and n (n larger than zero) we quickly notice that for this to be the case, our 2 factors must be related to n,one of them must be divisible by n and the other must be a divisor of n , the larger should be divisible by n, so x is divisible by n and x-4 is a divisor of n by the same coefficient: meaning that x=bn for some integer b and n=b(x-4) for the same value of b. substituting n in the first equation: x=b[b(x-4)]=b^2(x-4) x=(b^2)x-4b^2 thus x(b^2-1)=4b^2 x=(4b^2)/(b^2-1) we can factor b^2-1 into (b-1)(b+1) and 4b^2 is (2b)^2 since x=((2b)^2)/(b-1)(b+1) and since x is an integer then (2b)^2 is divisible by both b-1 and b+1 using the logic we used earlier b+1=c(2b) and b-1=(2b)/c b+1=c^2(b-1) b(c^2-1)=c^2+1 b=(c^2+1)/(c^2-1) b=((c^2-1)+2)/(c^2-1) b=1+(2/(c^2-1)) since b is an integer and 1 is an integer, 2/(c^2-1) should also be an integer. c^2-1 has a minimum value of -1, which gives b=-1, and b=-1 gives no x solutions. For c=1 b is undefined (1+2/0) , for c greater than or equal to 2, c^2-1 is greater than 2, denominator greater than numerator aka 2/(c^2-1) wouldn't be an integer. Thus b can't be an integer and by contradiction there are no solutions (other than the case where n=0, which gives x=0 or x=4) One more thing, x=0 and a=0 is not a valid solution, since 0/0 is undefined. Thus the only solution we have is x=4 a=2 Edit 2 - I just noticed that c isn't necessarily an integer, but c^2 is. Solution isn't affected much though, no new x values appear. Edit 3 - Apparently b^2 and b do not have to be integers, but c^2 has to, so we can just substitute integer values of c^2 and check if we get a valid x solution. If anyone has a more efficient way of doing this please tell me.
@Ooopsi
4 жыл бұрын
Nevermind I found a much easier solution. Since both x and a are integers, x-a is an integer, thus x must be divisible by a. Let x=an for some integer n, we have: an/a=an-a n=an-a n=a(n-1) a=n/n-1 Now we need a to be an integer, so we can try some n values. For n=0 a=0 (rejected) For n=1 a=1/0 (rejected) For n=2 a=2, which gives x=4 (accepted solution) For n above 2, n-1 becomes greater than 1, n and n-1 would be relatively prime, and thus we wouldn't be able to divide n by n-1 and get an integer. For negative values of n, n=-1 a=1/2 n=-2 a=2/3 Etc Thus the only solution is the pair x=4 and a=2
@OJMorgan55
4 жыл бұрын
You can intuitively tell it's 4 and 2 just looking at it. What other pair of integers has the same difference and quotient
@thesixteenthstudent2497
3 жыл бұрын
imagine you saw 9 as "a" and tried to solve problem like 1 hours ...
@Azmidium
3 жыл бұрын
Wait it wasn't an A?!!!
@thesixteenthstudent2497
3 жыл бұрын
@@Azmidium You can't imagine how i tried to solve this problem i was about to cry lmao xD
@savitatawade2403
5 ай бұрын
thanks for telling me... wasted 20 mins😔😭@@thesixteenthstudent2497
@heungseoblee5980
4 жыл бұрын
What you got are full real solutions of the problem, but if we consider the complex number field, I think that there can be three cases in 4:21 As you know in complex field, (x+3y)^3 = 4^3 has three cases. x+3y = 4 or x+3y = 4e^{2pi i/3} or x+3y = 4e^{4pi i / 3} I don't know full statement of that problem in original olympiad contest. But if we want to find full 'complex' solutions, it would be troublesome!
@ffggddss
4 жыл бұрын
x³ + 9x²y = 10 y³ + xy² = 2 First blush of my thought process: Hmmm..the LHS of the 1st equation looks like the start of the expansion of (x + 3y)³ - what's the rest? Let's see, x³ + 3·3x²y + 3·9xy² + 1·27y³ = x³ + 9x²y + 27xy² + 27y³ = x³ + 9x²y + 27(xy² + y³) And son-of-a-gun! There's the LHS of the 2nd equation! So (x + 3y)³ = 10 + 27·2 = 64 x + 3y = 4 [or 4 times either of the two complex roots of unity] Hey! How about x = y = 1? YES!! That works! That could have been guessed without going through any of that rigamarole above. What about other possible solutions, even without using the complex roots we set aside earlier? y = ⅓(4-x) 10 = x³ + 3x²(4-x) = -2x³ + 12x² x³ - 6x² + 5 = 0
@midorimashintaro2613
4 жыл бұрын
ffggddss how can you do exponents and fractions in this comment box
@soumyaranjantripathy7611
4 жыл бұрын
2x³+15x²y=25 5y³+6xy²=1 Then x=? y=? Challenge for you
@ffggddss
4 жыл бұрын
@@midorimashintaro2613 Some are available on my Mac keyboard using Option and Option-Shift (in Windows, "alt" & "alt-Shift"); some I've collected from other posts on Yahoo Answers & YT, by Copy-Paste into a text file, then from there to YT comments. In Windows, I understand there are ways to get these Unicode characters using the keyboard... Fred
@ayyubalam1771
4 жыл бұрын
I solved it within a 10 sec x=1 and y= 1
@JohnRandomness105
3 жыл бұрын
Solving the second equation for x, and substituting into the first equation is not very difficult. You get a cubic in y^3, and y^3=1 is one of the solutions. Divide it out, and you get a quadratic in y^3. If I didn't make a mistake, y^3 = 2 +- sqrt(5) is the other pair of solutions. x = (2 - y^3)/y^2 -- Now getting all pairs (x,y) might be a bit more grungy.
@hasanpezuk8757
4 жыл бұрын
There are 9 solutions x'=cos(2π/3), y'=isin(2π/3) and x"=cos(4π/3), y"=isin(4π/3) are olso other solutions for (1,1). for every solutions (including imaginary numbers) just rotate each solution by 2π/3 and 4π/3 {it means multiply e^(2πi/3) and e^(4πi/3)}
@spirosbaos575
4 жыл бұрын
Well your equation has a golden solution
@SACHINKUMAR-il2ez
4 жыл бұрын
In 1st equation just add & subtract X^2Y & XY take common (X-1) X-1=0 X=1 put in equation Y= 1
@stomoxe1
4 жыл бұрын
Question could you explain in other video the grobner methode for systematique resolution without math tricks ? {x*y^2+y^3 = 2, x^3+9*x^2*y = 10} With complexes solutions: is: {{x = 1, y = 1}, {x = -1/2-(1/2*I)*sqrt(3), y = -1/2-(1/2*I)*sqrt(3)}, {x = -1/2+(1/2*I)*sqrt(3), y = -1/2+(1/2*I)*sqrt(3)}, {x = 5/2-(3/2)*sqrt(5), y = 1/2+(1/2)*sqrt(5)}, {x = 5/2+(3/2)*sqrt(5), y = 1/2-(1/2)*sqrt(5)}, {x = -2*(-1/4-(1/4*I)*sqrt(3)-(1/4)*sqrt(-10+(10*I)*sqrt(3)))^3-4*(-1/4-(1/4*I)*sqrt(3)-(1/4)*sqrt(-10+(10*I)*sqrt(3)))^2-1/4+(7/4*I)*sqrt(3)+(7/4)*sqrt(-10+(10*I)*sqrt(3)), y = -1/4-(1/4*I)*sqrt(3)-(1/4)*sqrt(-10+(10*I)*sqrt(3))}, {x = -2*(-1/4-(1/4*I)*sqrt(3)+(1/4)*sqrt(-10+(10*I)*sqrt(3)))^3-4*(-1/4-(1/4*I)*sqrt(3)+(1/4)*sqrt(-10+(10*I)*sqrt(3)))^2-1/4+(7/4*I)*sqrt(3)-(7/4)*sqrt(-10+(10*I)*sqrt(3)), y = -1/4-(1/4*I)*sqrt(3)+(1/4)*sqrt(-10+(10*I)*sqrt(3))}, {x = -2*(-1/4+(1/4*I)*sqrt(3)-(1/4)*sqrt(-10-(10*I)*sqrt(3)))^3-4*(-1/4+(1/4*I)*sqrt(3)-(1/4)*sqrt(-10-(10*I)*sqrt(3)))^2-1/4-(7/4*I)*sqrt(3)+(7/4)*sqrt(-10-(10*I)*sqrt(3)), y = -1/4+(1/4*I)*sqrt(3)-(1/4)*sqrt(-10-(10*I)*sqrt(3))}, {x = -2*(-1/4+(1/4*I)*sqrt(3)+(1/4)*sqrt(-10-(10*I)*sqrt(3)))^3-4*(-1/4+(1/4*I)*sqrt(3)+(1/4)*sqrt(-10-(10*I)*sqrt(3)))^2-1/4-(7/4*I)*sqrt(3)-(7/4)*sqrt(-10-(10*I)*sqrt(3)), y = -1/4+(1/4*I)*sqrt(3)+(1/4)*sqrt(-10-(10*I)*sqrt(3))}} numeric : {x = 1., y = 1.} {x = -.8542, y = 1.6180}, {x = 5.8542, y = -.61800}, {x = -2.9271-5.0698*I, y = .30902+.53523*I}, {x = -2.9271+5.0698*I, y = .30902-.53523*I}, {x = -.50000-.86605*I, y = -.50000-.86605*I}, {x = -.50000+.86605*I, y = -.50000+.86605*I}, {x = .4268-.7391*I, y = -.80902+1.4013*I}, {x = .4268+.7391*I, y = -.80902-1.4013*I}
@mathwithjanine
4 жыл бұрын
Loving the new microphone!
@toddtrimble2555
4 жыл бұрын
Funny thing is, I "substituted right away" x = 1, y = 1. Another attack is to divide the first equation by the second to get (t^3 + 9t^2)/(1+t) = 5 where t = x/y. So t^3 + 9t^2 - t - 5 = 0 where we know t-1 divides the LHS since the solution x=1, y=1 gives t = 1 as a root. The other two solutions (after synthetic division, quadratic formula) are t = -5 + 2\sqrt{5}, t = -5 - 2\sqrt{5}. It's not too hard from there. Also one shouldn't worry about real versus complex: by homogeneity of the left-hand sides, if (x, y) is a solution, then so is (cx, cy) where c is a cube root of unity.
@er.devvrat181
3 жыл бұрын
x^3+9yx^2=10 x^2(x+9y) =10 We can take x^2=1, 2,5 and x+9y=10, 5,2 On solving 2nd equation y^3+xy^2=2 y^2(y+x) =2 Here we can take y^2=1, 2 And (x+y) =2, 1 after putting these values we have to solve the equation and we will get x=1and y=1
@IamRajaneesh
4 жыл бұрын
X,y is 1. I Solved by multiplying 27 to second equation and maing it (X+3y)^3 = 4^3. X+3y is 4 Y is 4-x/3 Substituting in first equation.
@Grundini91
3 жыл бұрын
At y^3-2y^2+1=0, first thing I notice is that the coefficients add up to 0, if they add to 0 then (y-1) is a factor.
@jamboree1953
4 жыл бұрын
y3 - 2y2 + 1 =0 y3 - y2 -y2 + 1 =0 (y3 - y2) - (y2 - 1) = 0 y2(y - 1) - (y-1)(y+1)=0 (y-1)(y2 - (y+1)) =0 (y-1)(y2 - y -1) =0 Yes, y=1 is a solution. The other two solutions are irrational.
@oloyt6844
2 жыл бұрын
Idk about you guys but I just instantly saw 1 as a solution….
@VaradMahashabde
4 жыл бұрын
0:25 Man talking to his dolls during isolation, 2020, colorized
@tanishislam4953
4 жыл бұрын
I have a better method at the stage we get the cubic equation => y^3 - 2y^2 +1 =0 Plugging in y=1 in the equation, we know that it is a solution, so (y-1) should be a factor. So simply factorize using that Information. => y^3 - 2y^2 + 1 = 0 => y^3 - y^2 = y^2 -1 => y^2(y-1) = (y+1)(y-1) => (y-1)[ y^2 - (y+1)] = 0 Thus (y-1)(y^2-y-1)=0 , solve for y.
@Gerserh
3 жыл бұрын
Красиво. Особенно разложение на множители. Разделил оба на xy^2, замена x/y=u, доумножил на -5, сложил, получил кубическое уравнение u^3+4u-5=0. Дальше, только разглядев u=1, смог разложить на множители, без подсказки видео не додумался. Здоровское решение, спасибо.
@xamtplays1933
4 жыл бұрын
bet nobody was expecting the 1,1 until they saw it and substituted it into the equation
@think_logically_
4 жыл бұрын
Less tricky, but more intuitive. This is a homogeneous equation (all terms apart from free terms have same degree 3). So get rid of free terms. Multiply second equation by 5: x³ + 9x²y=10 5y³ + 5xy² = 10 Subtract second equation from first: x³ + 9x²y - 5xy² -5y³ = 0 Divide by y³ (which is different from 0 because of second given equation): (x/y)³ + 9(x/y)² - 5(x/y) - 5 = 0 Substitution z=x/y gives a cubic equation z³+9z²-5z-5=0, that obviously has root z=1, so it can by factorised: (z-1)(z²+10z+5) = 0, z=1, z=-5±2√5, => x = y, or x=(-5±2√5)y Substitution into the second given equation gives the expression for y³, from which y and x can be found.
@rasheedmohammed2227
4 жыл бұрын
Blackpenredpen did u can solve the question in by like 2 seconds by using cross method multiplication. (10*1-9*2)/1*1-9*1= - 8/-8 = 1 thus the first row =1 therefore x and y equal 1(the row only works in this case because of x and y having no coefficient usually it is only x and u sub in to get y)
@jkstudyroom
4 жыл бұрын
Nicely done sir! Really like your videos!!!
@wajahatrasheed2193
3 жыл бұрын
these are homogeneous equation .can be solved by taking commom x^3 and y^3v respectively from both equations and then divide.
@jarikosonen4079
3 жыл бұрын
It looks give more questions than answers. How to calculate the number of roots/solutions for this equation (it looks like having total 5 roots in WFA)? Why can you divide the equations (1) and (2) with (x-1)*(y-1) and then solve rest of the roots?
@antoniosouza1213
4 жыл бұрын
Thanks teacher from Brazil!
@noahtaul
4 жыл бұрын
Don’t forget the complex solutions by multiplying x and y by a third root of unity
@ianmoseley9910
4 жыл бұрын
noahtaul He said specifically that they are only looking for real numbers.
@danielbrstak5730
7 ай бұрын
I was trying to solve this for 3 months with no success....and it was so simply after I've watched you...
@somanathdash8143
4 жыл бұрын
Press f for those who tried substitution before watching the video 😂
@integralboi2900
4 жыл бұрын
F
@asdfghjkl6506
4 жыл бұрын
F(1)
@Hexanitrobenzene
4 жыл бұрын
I tried 3 different subtractions of these equations, only to find that the solution was a sum... (facepalm).
@Hexanitrobenzene
4 жыл бұрын
By the way, what does this "press f" cliche mean ?
@asdfghjkl6506
4 жыл бұрын
Press f to pay repesct
@CHAS1422
4 жыл бұрын
Love your show. Now you need a bigger board so you won't have to erase the first half to present the second.
@milanmitreski7657
3 жыл бұрын
I would rather multiply the second equation with 5 and then eliminate the constants by subtracting equation. Then divide the formula by y³ and you get a cubic equation in q=x/y. Then you solve for q, and plugin x=qy in any of the given equation and find x and y. I think this is better because if you get uglier constants it will work
@entropiccroissant4619
4 жыл бұрын
This night I weirdly dreamt about watching a long bprp video and forty minutes in we start hearing violin and he's kinda mad since it's his children and they are not practicing the piece they're supposed to. Given that Steve seemed the same as he is now, his children were likely very young while playing at professional level. Weird dream, but fun dream.
@m.matoori343
4 жыл бұрын
easier way: divide the first sentence by the secone one so you get x+9y/x+y=5 so after making this one right you get 4x=4y which means x=y and use it with the x+3y=4 and solve the problem. im not very good at use english words for math but i hope ive sum it up!
@MarioFanGamer659
4 жыл бұрын
The division is incorrect as you can't just multiply or divide multiple equations with each other. I also don't see how the second equation can cancle anything from the first solution anyway which is the second reason why your approach is wrong.
@scarbotheblacksheep9520
4 жыл бұрын
Divide first equation by x^2. Divide second equation by y^2. Add them together, get 10y + 2x = 10/x^2 + 2/y^2. Hm...could y = 1/x^2 and x = 1/y^2? I look at that and think x=1 and y=1? Well, it works! But I guess that isn't the only solution pair.
@theofanomarvelrevayawantan3296
4 жыл бұрын
I substitute x=(2-y^3)/y^2 and the answer i get is: Real solution. x=1,y=1 , x=(20-9sqrt5)(cbrt(161+72sqrt5)),y=cbrt(2+sqrt5) , x=(20+9sqrt5)(cbrt(161-72sqrt5)),y=cbrt(2-sqrt5). Am i wrong?
@isura.m
2 жыл бұрын
There are 6 more non-real solutions
@jongyon7192p
4 жыл бұрын
x+3y=64^(1/3) What if we set x+3y=4e^(+-2/3pi*i), then solved the rest again like that?
@paulerhard1252
4 жыл бұрын
Cool solution! Alas... I would have never figured it out. So congratulations are in order ... 👍
@GourangaPL
4 жыл бұрын
About that dividing polynomials method you used in 8:00 what i can confirm is in Poland where i live about 10 years ago while i was in high school we had it but it was class with advanced math profile, classes with basic math didn't have it and last time i checked like 2-3 years ago advanced classes didn't have that method anymore
@김무명-c4w
4 жыл бұрын
can you comment about my approach? my approach: set y as constant variable 5 times under equation and subsititue then we will get x^3+9yx^2-5y^2x-5y^3=0 and move y in real number. then we will get a lot of real number solution x which correspond each y then there will be uncountable x and y am i wrong?
@littyfam5136
4 жыл бұрын
Glad to see a 3Blue1Brown Pi mic
@Azoz2013azoz
4 жыл бұрын
you could have factored the y^3 -2y^2 + 1 = y^2 ( y -1) - (y^2 -1) = y^2(y-1) -(y-1)(y+1) = (y-1)(y^2 - y - 1)
@patryslawfrackowiak6690
4 жыл бұрын
Nice way for dividing polynomials... I need to memorize it.
@michaelempeigne3519
4 жыл бұрын
easy. if the coefficients add to the RHS, then the solution is unity in all variables. Therefore the answer is x = 1 and y = 1 since 1 + 9 = 10 ad 1 + 1 = 2
@rmela4501
4 жыл бұрын
Man, did I ever do this the long hard way. I rearranged first equation to solve for y and plugged into 2nd equation. After some algebra I got a degree 9 equation. Made a sub u=x^3 to reduce it to a cubic ( all the powers were multiples of 3). After using x=1 for initial guess, used synthetic division to get a quadratic and used formula to get u and took cube root to get x. So my solution is the same, just looks diff (1,1), ((100+45(5)^1/2)^1/3),(-10-5(5)^1/2)/x^2)) last one inverts signs in the middle. The cubic i got was u^3 - 201u^2 + 175u + 125, the quadratic was u^2 - 200u - 125.
@Hexanitrobenzene
4 жыл бұрын
...and that is exactly why one shouldn't do it this way. 9th degree equation...
@intrawachira
4 жыл бұрын
(x+3y)^3=10+27*2=64 x+3y=4
@SlidellRobotics
3 жыл бұрын
Rephrasing the solutions: (1,1),(φ⁴-1,φ),(φ⁻⁴-1,φ⁻¹) Added: or even sillier, (φ²-φ¹,φ²-φ¹),(φ⁴-φ⁰,φ²-φ⁰),(φ⁻⁴--φ⁰,φ⁰-φ¹)
@nosnibor800
4 жыл бұрын
You make maths fun.
@ealejandrochavez
4 жыл бұрын
I solved this question the same way you did. But the solution is incomplete: there are three cubic roots of 64, two of which are complex. So x + 3y = 4e^{2 pi * k * i / 3), where k = 0, 1, 2. Thus, this produces 6 additional nontrivial solutions.
@rabindranathghosh31
4 жыл бұрын
Isn't x=y=omega (complex cube root of unity) and x=y=omega^2 also solutions?
@knighttt
4 жыл бұрын
Nice 3b1b pi plushie!
@chillanagaraju879
4 жыл бұрын
Just addicted to your channel
@limitXbreaker
4 жыл бұрын
Hey but i got x=2/y^2 - y and when i put it in the first equation i got 9 values of y out of which 6 were imaginary and real were 1, 2+√5 and 2-√5 Can uh explain this?
@sailingteam1minecraft124
3 жыл бұрын
1:52 LITTERAL CHILLS
@BurningPandama
2 жыл бұрын
Do i get any points for just guessing the x=1m y=1 solution at the start?
@conceptispower9268
4 жыл бұрын
x=1 and y=1 just by observation
@symnick
4 жыл бұрын
I have another solution if i put y=kx and then divide two equatios by part. Then i calculate k etc. We have the same results.
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