Yeah, you are right - there is indeed only one additional complex integer solution. For the other two solutions to work (±1; 0) we need to choose different signs for the first and the second square root on the left (so that they add up to zero), which, strictly speaking, is theoretically possible (since complex roots are essentially multivalued), but normally is not regarded as a solution when we stick to the _principal_ complex square root definition.
@allozovsky5 ай бұрын
I asked Prime a question in a comment above about whether (and how) we solve irrational equations over the field of complex numbers ℂ (where complex nth roots are multivalued and have different sings/directions when n > 2, not just ±1), but he hasn't responded yet. Say, how do we solve equations like ³√z = −2 or ³√(z−6) = z (if we solve them at all over ℂ).
@sobolzeev5 ай бұрын
It still relies on the equality √(2-√3) + √(2+√3) = √6. This is not very difficult since (2±√3) = (3 ±2√3 + 1)/2 = (√3 ± 1)²/2.
@allozovsky5 ай бұрын
@TheMathManProfundities It gives 𝒊√6 for both sides (as a _principal_ value).
@EnochAkintayo5 ай бұрын
You're extremely smart, sir.
@courbe4535 ай бұрын
I like your way of explaining, a real mathematics teacher, thank you.
@beapaul44534 ай бұрын
11:46. Wonderful
@dawkinsfan66024 күн бұрын
03:37 😂😂😂😂😂😂 I watched it at 2x...it is even more hilarious than the original one! 😂
@AlirezaNabavian-eu6fz5 ай бұрын
Excellent
@mathpro9265 ай бұрын
Teacher always shows us a good way ❤
@chaosredefined38345 ай бұрын
From 2sqrt(x^2 - y) = xy - 2x. Note that x | RHS. Therefore, x | LHS. So, x | 2 sqrt(x^2 - y). Therefore, x^2 | 4x^2 - 4y. So, x^2 | 4y. Therefore, there exist some k such that 4y = k x^2. But we know that y >= 0, so 4y and x^2 are both positive, and therefore k has to be positive. Going back to the equation, we can move the 2 into the sqrt by squaring it, giving us sqrt(4x^2 - 4y) = xy - 2x. Then, replacing that 4y with k x^2, we get sqrt(4x^2 - kx^2) = xy - 2x. Extracting out the x^2 from the sqrt, we get x sqrt(4 - k) = xy - 2x. Option 1 is that x = 0, which leads us to 4y = k 0^2, and therefore 4y = 0, so y = 0. Option 2 is that sqrt(4 - k) = y - 2. Since k has to be positive, and sqrt(4 - k) has to be an integer, this means that sqrt(4-k) = 0, 1 or 2, or k = 4, 3 or 0. Case 1: k = 4 4y = 4 x^2, so y = x^2. Therefore, 2 sqrt(0) = x(4x^2) - 2x, so 4x^3 - 2x = 0, or 2x(2x^2 - 1) = 0. 2x^2 - 1 has no integer solutions, so the only solution is the (0, 0) solution we already covered. Case 2: k = 3 4y = 3x^2. So, y = 3x^2 / 4. 2 sqrt(x^2 - 3x^2 / 4) = 3x^3 / 4 - 2x => 2 sqrt(x^2 / 4) = 3 x^3 / 4 - 2x => 2 (x/2) = 3x^3 / 4 - 2x => 0 = 3x^3 / 4 - 3x => 3x( x^2 / 4 - 1) = 0. x = 0 is a solution again, which gives us y = 0. But x^2 / 4 - 1 has integer solutions, specifically x = 2 and x = -2. We reject x=-2 because we need x >= sqrt(y) >= 0. Therefore, x = 2 is the only new value we get from this case, and it gives us that y = 3 (2)^2 / 4 = 3. so, (2,3). Case 3: k = 0 This means that 4y = 0 x^2, so y = 0. Shoving that in, we get 2 sqrt(x^2) = -2x. But x >= 0, and now we need -2x
@iqtrainer5 ай бұрын
You did a fantastic job PN! You guys should do more collabs
@dirklutz28185 ай бұрын
An "illegal", but nice solution is: (x=sqrt(2), y=2)
@allozovsky5 ай бұрын
Square roots of integers are _algebraic integers,_ though (i.e. roots of some _monic_ polynomial with a leading coefficient of 1, like in x² = 2, for example), so one might call it an _algebraic_ integer solution to an _irrational_ Diophantine equation, and thus it makes sense from this point of view.
@apotheos-i7qАй бұрын
@@allozovskywow thats brilliant
@geekandnerd51425 ай бұрын
Man, i love his videos so much. Is just so great to learn from someone who really likes what he is talking about. I got into maths because of this guy when I was like 15 and have absolutely no regrets. I don’t, know how often do you read your comments, but if you’re reading this, man, thank you. I’m currently battling to get a bachelor on maths, and I have no idea how my life would’ve turned out if I haven’t found that video of yours three years ago.
@PrimeNewtons5 ай бұрын
Wow! Hey I am so happy for you. You just inspired me with your story. I am honored. You are unstoppable.
@geekandnerd51425 ай бұрын
@@PrimeNewtons aaaah! I’m so glad you’ve read it. Is great to know I’ve inspired you as you inspired me with those easy explanations of precalc years ago! I’m sure there’s many people out there who, like me, felt completely greatful to have found that one teacher they can look up to, with the easy explanations which makes maths look as fun and elegant as it is. You help more people than you’ll ever know.
@Abby-hi4sf2 ай бұрын
@12:05 I love your explanation, it gives real depth how to see possible solutions, with brutal force x^2 (4 - y) = 4 = (4)(1) x^2 =4 , x= 2 and 4 - y = 1 , so y = 3 is
@ThePhotonMan1105 ай бұрын
Loved your energy in this video!
@surendrakverma5555 ай бұрын
Excellent explanation Sir. Thanks 👍
@Alephŋull3 ай бұрын
You can also do one thing Take 4 on rhs Then you will have X * X * (4-Y) = 4 One two options left 1*1*4 Or 2*2*1 1st case does not work Therefore x=2 and y=4-1=3
@futuregenerationinstitute96135 ай бұрын
Thanks for making maths easier. Would you please make a video about parallelogram formulas practically using DIAGONALS. Thank you sir.
@martinmolander5425Ай бұрын
I really like your videos, they are fantastic. However... I enjoyed this until the end; but when you have a radikal equation, you must always, ALWAYS, check your solutions to the original equation.
@chintu43985 ай бұрын
6:11 yea they're gone...they're goners💀
@treybell405015 ай бұрын
Don’t be afraid to express faith especially if it’s not in an offensive way. Peace to everyone
@jumpman82825 ай бұрын
It's always offensive to someone. I think the bible says somewhere not to do to others what you don't want them to do to you. And I don't think neither you nor Newton would want anyone to rub their faith in your faces. Peace.
@robertholder5 ай бұрын
@@jumpman8282 I agree. Much better to express faith through actions like providing free math videos. Flashing bible verses is just preaching.
@TheMathManProfundities5 ай бұрын
Why can't you take the root of negative numbers? √-6=i√6. x=-2 is therefore a valid candidate solution and turns out to be valid (y=3). Also in this vein, you should have added a ± when you multiplied the roots together (both can be imaginary). As it happens, the next thing you did was to square this so it wouldn't have made any difference. Also, later on you divided by y-4 without checking whether this can be zero. It can easily be seen that it can't be but this should have been included especially as this is designed to teach other people.
@allozovsky5 ай бұрын
@TheMathManProfundities > later on you divided by y-4 without checking whether this can be zero Yeah, adding one extra step at 10:50 (where Prime was actually going to factor out x²) to explicitly show that x²(4−y) = 4 has no solutions when y = 4, would have been much more rigorous and this checkup should have definitely been mentioned.
@allozovsky5 ай бұрын
And going back to the issue with Wolfram Alpha - it gives no negative solutions for √x = √(x³) either, when restricted to integers. And when not restricted, gives x = −1 as well with a remark: (assuming a complex-valued square root).
@allozovsky5 ай бұрын
In the documentation for the Solve function it is stated: • If _dom_ is *Reals,* or a subset such as *Integers* or *Rationals,* then _all constants and function values are also restricted to be real._ Whether this is a wise choice is hard to tell - it depends on the constraints of the problem where an equation was formed. But now things are starting to get more clear. And in any case, it is up to Wolfram to decide how their tools are implemented. Say, for some strange reason Wolfram Mathematica gives *Indeterminate* to *0^0,* but at the same time gives *1* to both *a^0/.a->0* and *a^0/.a->Infinity,* assuming that *a^0* is *1* for _any_ base *a* (just like most other math tools do), which doesn't look very consistent.
@TheMathManProfundities5 ай бұрын
@allozovsky I'm not terribly familiar with Alpha, the main issue I had with it is that it treats 1/2x as (1/2)x whereas it treats 1/ax as 1/(ax). 0⁰ is a whole other conversation as a⁰/.a>0=1 but 0ᵃ/.a>0=0. As such 0⁰ had to be considered undefined but as I understand it, it can take a value consistent with it's surroundings. As such y=x⁰ and y=0ˣ are both continuous at x=0.
@allozovsky5 ай бұрын
@TheMathManProfundities > it treats 1/2x as (1/2)x whereas it treats 1/ax as 1/(ax) Oh, yeah! 🙂 That (in)famous "issue" increases the extremely popular 6/2(1+3) confusion even more. And the same goes with 2x^2x vs ax^ax. Not very consistent indeed.
@janimed92665 ай бұрын
Bravo. Mais Pour la dernière réponse il fallait mieux expliquer D'après la dernière égalité on a 4-y doit être supérieur a zéro:4-y>0--------->y
@MrJasbur15 ай бұрын
If it’s a Diophantine equation, isn’t the only requirement that x and y be integers? Like with the more complicated expressions involving square roots being whatever they want whether irrational or even complex? If so, wouldn’t (-2,3) also be a solution?
@marekpiatek015 ай бұрын
Correct since inputting that solution would result in i√6 on both sides of the original equation, making it a true statement. Good observation
@allozovsky5 ай бұрын
Yeah, you are right - that's the only requirement. Though Wolfram Alpha gives only the two non-negative solutions we obtained in the video when asked to solve this equation over the integers, while it certainly knows how to evaluate complex roots.
@iqtrainer5 ай бұрын
@@allozovskyStop talking about complex roots.
@JamesWanders4 ай бұрын
For 4x^2-4-x^2•y=0, (x,y)=(1,0) looks to be a valid solution--but that doesn't work in the original equation. I'm struggling to see why the (2,3) solution works and (1,0) does not as both satisfy that side of the zero product step 🤨
@PrimeNewtons4 ай бұрын
If x =1 and y= 0, the right hand side would be 0. The left would be 2.
@nothingbutmathproofs71505 ай бұрын
Nice job, as always. When you had (xy-2x) you should have factored out the x. Life would have been easier from there
@PrimeNewtons5 ай бұрын
I really didn't see it. That's smart 🤓
@m.h.64705 ай бұрын
Solution: √(x - √y) + √(x + √y) = √(xy) |² x - √y + 2√(x - √y)√(x + √y) + x + √y = xy 2x + 2√((x - √y)(x + √y)) = xy 2x + 2√(x² - y) = xy |-2x 2√(x² - y) = xy - 2x 2√(x² - y) = x * (y - 2) |² 4(x² - y) = x² * (y - 2)² 4x² - 4y = x² * (y² - 4y + 4) 4x² - 4y = x²y² - 4x²y + 4x² |-4x² +4y 0 = x²y² - 4x²y + 4y 0 = x²y² - (4x² - 4)y This is a quadratic equation in terms of y with x² as a parameter y = (-(-(4x² - 4)) ± √((-(4x² - 4))² - 4(x²)(0)))/(2x²) y = ((4x² - 4) ± √((-(4x² - 4))²))/(2x²) y = ((4x² - 4) ± |-(4x² - 4)|)/(2x²) The absolute value creates 2 cases: -(4x² - 4) ≥ 0 → -4x² + 4 ≥ 0 → -4x² ≥ -4 → 4x² ≤ 4 → x² ≤ 1 → as x ∈ Z, only x = 0 and x = 1 are valid -(4x² - 4) < 0 → -4x² + 4 < 0 → -4x² < -4 → 4x² > 4 → x² > 1 → any x ∈ Z > 1 are valid Since x = 1 leads to 0 inside the absolute value term, it can be used in either case Case x = 0: y = ((4x² - 4) ± |-(4x² - 4)|)/(2x²) y = ((4(0)² - 4) ± |-(4(0)² - 4)|)/(2(0)²) y = (-4 ± 4)/0 y = -8/0 OR 0/0 Since this doesn't give a valid answer, let's plug it into the original equation: √(x - √y) + √(x + √y) = √(xy) √(0 - √y) + √(0 + √y) = √(0 * y) √(-√y) + √(√y) = √0 √(-√y) + √(√y) = 0 Since √ only returns positive values, y HAS to be 0 √(-√0) + √(√0) = 0 √(-0) + √0 = 0 0 + 0 = 0 0 = 0 Therefore x = y = 0 is a valid solution Case x ≥ 1: y = ((4x² - 4) ± (4x² - 4))/(2x²) Since y = ((4x² - 4) - (4x² - 4))/(2x²) = 0/(2x²) = 0, we already have that solution. y = ((4x² - 4) + (4x² - 4))/(2x²) y = (2(4x² - 4))/(2x²) y = (4x² - 4)/x² y = 4(x² - 1)/x² Given, that x ∈ Z, the right side is alway positive. Given that y ∈ Z, the right side has to be an integer. But with x² in the denominator, the right side can only be an integer in two cases: if x = ±1, because the denominator becomes 1 if x = ±2, because the denominator becomes 4 and cancels out with the factor 4 Since we are in the case x ≥ 1, only x = 1 and x = 2 are relevant. Case x = 1: y = 4(x² - 1)/x² y = 4(1 - 1)/1 y = 4(0)/1 y = 0 This is an extraneous solution, as y = 0 leads to x = 0 in the original equation: √(x - √y) + √(x + √y) = √(xy) √(x - √0) + √(x + √0) = √(x * 0) √(x) + √(x) = √0 2√(x) = 0 |:2 √x = 0 |² x = 0 Case x = 2: y = 4(x² - 1)/x² y = 4(2² - 1)/2² y = 4(4 - 1)/4 y = 3 So there are 2 integer solutions: x = 0 and y = 0 x = 2 and y = 3
@Rishab_Sharma_Python_Teacher5 ай бұрын
Sir please make a video on how to find intersection coordinates of two circles
Can you use this idea says theres 2 cases to solve Case 1 when x=y Case 2 when x≠y or x=ay Where a is constant
@flowingafterglow6295 ай бұрын
Something is bothering me about your x^2 = 4/(4 - y) equation Yes, it works to get y = 3 and x = 2 But what if you put in y = 0? In that case, you get x^2 = 4/4 = 1. However, y = 0, x = 1 is not a solution. Is that just an extraneous solution introduced by squaring? I wrote it as y = 4 - 4/x^2 but the result is the same
@chaosredefined38345 ай бұрын
It's an extraneous solution. Extraneous solutions are also why you should test the (2,3) solution.
@iqtrainer5 ай бұрын
this is a good point. i watched both videos and thats what dr pk did in the solution
@iqtrainer5 ай бұрын
this is a good point. i watched both videos and thats what dr pk did in the solution
@rabotaakk-nw9nm5 ай бұрын
8:55 "Testing y=0" !!! 😁
@flowingafterglow6295 ай бұрын
@@rabotaakk-nw9nm But that doesn't address my comment. He derived the expression for x^2, and if you use that you get x = 1 as a solution. As noted, it is extraneous
@foisalmahdi5 ай бұрын
Can you please factor this polynomial: -2x²+6y²+xy+8x-2y-8 ? I can't factor it.
@williamdragon10235 ай бұрын
(-x+2y+2)(2x+3y-4) according to wolfram alpha.
@danobro5 ай бұрын
8 second ago no way!!!
@sobolzeev5 ай бұрын
From the very form of the equation we have (1) y≥0 (since having √y); (2) x≥√y≥0 (since having √(x-√y)). After the first squaring we get 2√(x² - y) = xy - 2x = x(y-2) ≥ 0, so x=0 (and hence y=0) or x>0 and y>2 (y=2 implies x²=2, which is impossible). Finally, x,y>0 imply √(x² - y) x(y-2) and y
@allozovsky5 ай бұрын
But the (−2; 3) solution also works, if we consider complex principal square roots. It's a Diophantine equation, after all, and x, y, ∈ ℤ is the only requirement.
@sobolzeev5 ай бұрын
@@allozovsky Can you produce your calculations here? You see, I am not sure what is the principle square root of a negative number. For a positive A, a principle square root is the bigger of the two roots of the equation x²=A. For a negative, they are incomparable.
@allozovsky5 ай бұрын
@@sobolzeev You are right, a principal complex root is a bit vague concept and can be defined in different ways, depending upon how we define the principal argument of a complex number. If −π < φ = arg(z) ≤ π, then the principal square root of z = r·exp(𝒊φ) is normally taken to be √z = √r·exp(𝒊φ/2) and for a negative z we have an argument of π/2, that is the upward looking square root. But surely we can choose some other half-open interval of length 2π to define our principal argument and then the principal square root may switch direction (for example, if we choose the interval [−π; π) instead, including the left end and excluding the right one).
@allozovsky5 ай бұрын
But if we do not define a principal root but instead treat our square root expression as a multivalued complex root, then (−2; 3) is a solution anyway, along with the two more additional solutions of the form (±1; 0). But multivaluedness may eventually lead us to undesired properties of mixed irrational expressions with complex roots.
@allozovsky5 ай бұрын
I asked Prime a question in a comment above about whether (and how) we solve irrational equations over the field of complex numbers ℂ, but he hasn't responded yet. Say, how do we solve equations like ³√z = −2 or ³√(z−6) = z (if we solve them at all over ℂ).
@doctorno16265 ай бұрын
x=y=0
@IammybrothersBro5 ай бұрын
What is this ²3⁴ 2 superpower 3 times 4
@allozovsky5 ай бұрын
Funny notation indeed 😊
@allozovsky5 ай бұрын
Also !3! - is it a factorial of a subfactorial or a subfactorial of a factorial? 🤔
@allozovsky5 ай бұрын
But I guess ²3⁴ should be evaluated as ²3⁴ = (3³)⁴ = 27⁴ = 3¹² = 531441, because superpower should have a higher priority (and also we evaluate left to right in this case anyway).
@IammybrothersBro4 ай бұрын
And also the order is 2,3,4
@cyberagua4 ай бұрын
An irrational Diophantine equation???!!! 😱 Sounds like some sort of a math oxymoron. Is it "find integer solutions of an irrational equation" what is really meant here?
@cyberagua4 ай бұрын
Searched the web for "irrational Diophantine equation" and found noting. Only "irrational Diophantine quadruples" and "irrational Diophantine numbers". And of course there are rational/fractional/algebraic/exponential Diophantine equations, but an irrational Diophantine equation sounds really weird.
@cyberagua4 ай бұрын
If there are roots involved, does it mean that we may also look for integer solutions that make the root expressions complex-valued? Sounds pretty reasonable, since we are already breaking the rules and mixing up different conceptions.
@cyberagua4 ай бұрын
May I have a question? Where does this equation come from? Is it some tournament or olympiad? Doesn't look like a legitimate test question, At least, being posed as a "Diophantine equation".
@ahmetd.yazgan7185 ай бұрын
x² = 4 / (4-y) 》{1,0} but not correct. Why?
@niloneto16085 ай бұрын
Because you must substitute these values in the original equation and see whether it holds true or not.
@niloneto16085 ай бұрын
Everytime you square both sides of the equation, extraneous roots appear. For instance, take x=y, when x²=y², when for this second equation, x=-y is a extraneous root which doesn't belong in the original equation.
If y = 2 or 1 then x squared would be 4/2 or 4/3 and x would therefore not be an integer but irrational
@sklolss16385 ай бұрын
For y = 2, you get x^2 = 4/2 = 2. so x = sqrt2. sqrt2 is not in Z. For y = 1 you get x^2 = 4/3. so x = 2/sqrt3. sqrt 3 is not in Z.
@skyking98355 ай бұрын
Dr PK Math didn't seem to have any tricks up his sleeve. He did what you did but faster (and sloppier)
@iqtrainer5 ай бұрын
Did almost the same but Dr PK method was more analytic. PN method was more algebraic. Better cool with that mouth
@johnka54075 ай бұрын
This at the end is a joke I don’t get or a fragment from a Bible without any context?
@luminator9115 ай бұрын
idk why he's preaching, it's not like easter or anything
@sadeqirfan55825 ай бұрын
Dude…. You’re religious?
@erenshaw5 ай бұрын
huh?
@icetruckthrilla5 ай бұрын
@@erenshawcheck out the last second of the video
@niloneto16085 ай бұрын
Yeah, and what? Let him profess some parts of the Bible, it's not the main topic of his videos.
@icetruckthrilla5 ай бұрын
Doesn’t bother me. I was just responding to someone who was confused.
@niloneto16085 ай бұрын
@@icetruckthrilla And I was responding the op who asked if Prime Newtons was religious
@harshplayz318825 ай бұрын
I didnt liked dr pk math video😅
@iqtrainer5 ай бұрын
What a born hater with hater gene. I liked both actually as an ardent viewer of both channels
@Aiellosfetano5 ай бұрын
But if i take y=0 i have x^2= 4/(4-0)-> 4/4->1 X^2=1->x=1. Is this possible?
@allozovsky5 ай бұрын
If we check this solution by substituting it back into the original equation, we will get 1 + 1 = 0, so if we consider the principal real-valued square root, this is an extraneous solution.
@Montegasppa5 ай бұрын
I got a questions: doesn’t √(x - √y)² equal to |x - √y|?
@treybell405015 ай бұрын
He corrected himself I think
@niloneto16085 ай бұрын
No need for the absolute values, as the only restrictions are x,y>=0 and x²>=y. With these in mind, x-√y can never be negative.
@vitotozzi19725 ай бұрын
Very very good!!!
@badralshammari80045 ай бұрын
Only God we can call , only god
@Rizzlers_Edits5 ай бұрын
Sir we can also put 0 as a value of y instead of putting three Is it a solution to the equation?
@allozovsky5 ай бұрын
That would give us two extraneous solutions (±1; 0) for which the LHS is non-zero but the RHS is zero.
@allozovsky5 ай бұрын
For these solutions to work, we need to choose different signs for the first and the second square root on the left (so that they add up to zero), which, strictly speaking, is theoretically possible (since complex roots are essentially multivalued), but normally is not regarded as a solution when we stick to the _principal_ complex square root definition.
@allozovsky5 ай бұрын
Thank you for another great video and a suggestion for a new math channel (it appeared to be very interesting indeed). I've got a question/suggestion to you (in a comment below) 👇
@allozovsky5 ай бұрын
The question goes like this: do we solve irrational equations over the field of complex numbers, where complex roots are essentially multivalued? Say, an equation with a cube root like ³√z = −2 or ³√(z−6) = z - do they have solutions over ℂ and how do we solve them?