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.
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).
I watched the PMath method of solution as you mentioned. His method was the same as your approach, but your explanation was by fan superior! Thank you for your videos.
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
@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
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.
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
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?
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.
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.
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.
@@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.
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.
@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.
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).
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.
@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.
@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.
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 🤨
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.
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
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
@@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
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.
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.
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.
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).
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?
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.
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.
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".
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
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.
@@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.
@@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).
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.
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).
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) 👇
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?
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When you said "brute force", I expected you to plug integers into the given equation to see what happens.
That would be explosive
BONUS: I allowed sqrt(-N) = i * sqrt(N), and that allowed for another solution (and only one more): (-2, 3)
Let's check:
√(x-√y) + √(x+√y) = √(xy)
√(-2-√3) + √(-2+√3) = √(-2.3)
(√(-2-√3) + √(-2+√3))² = (√(-2.3))²
(-2-√3) + 2 √(-2-√3).(-2+√3)+(-2+√3) = (i√6)²
-4 + 2 √(-1)(2+√3).(-1)(2-√3) = -6
-4 + 2 i²√(4-3) = -6
-4 - 2.1 = -6
-6 = -6
(-2, 3) looks ok.
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.
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).
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.
@TheMathManProfundities It gives 𝒊√6 for both sides (as a _principal_ value).
Your hand writing is really sensational!
You're extremely smart, sir.
11:46. Wonderful
I like your way of explaining, a real mathematics teacher, thank you.
Excellent
I watched the PMath method of solution as you mentioned. His method was the same as your approach, but your explanation was by fan superior! Thank you for your videos.
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
both did great sir
I'm sure someone has noted to you that if x^2 = 4/(4-y), then you can also have (1,0), but that solution doesn't work in the original equation.
@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
Don’t be afraid to express faith especially if it’s not in an offensive way. Peace to everyone
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.
@@jumpman8282 I agree. Much better to express faith through actions like providing free math videos. Flashing bible verses is just preaching.
Teacher always shows us a good way ❤
Very very good!!!
Loved your energy in this video!
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
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?
Correct since inputting that solution would result in i√6 on both sides of the original equation, making it a true statement. Good observation
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.
@@allozovskyStop talking about complex roots.
An "illegal", but nice solution is: (x=sqrt(2), y=2)
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.
@@allozovskywow thats brilliant
Excellent explanation Sir. Thanks 👍
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.
Wow! Hey I am so happy for you. You just inspired me with your story. I am honored. You are unstoppable.
@@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.
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.
@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.
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).
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.
@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.
@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.
Thanks for making maths easier. Would you please make a video about parallelogram formulas practically using DIAGONALS. Thank you sir.
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
03:37 😂😂😂😂😂😂 I watched it at 2x...it is even more hilarious than the original one! 😂
6:11 yea they're gone...they're goners💀
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 🤨
If x =1 and y= 0, the right hand side would be 0. The left would be 2.
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.
You did a fantastic job PN! You guys should do more collabs
Sir please make a video on how to find intersection coordinates of two circles
شكرا أستاذ
في الدقيقة 12 من بين الحلول أرى (1،0) لماذا لم تأخذها بعين الاعتبار وشكرا
Nice job, as always. When you had (xy-2x) you should have factored out the x. Life would have been easier from there
I really didn't see it. That's smart 🤓
But if y=0, x can also equal 1
How so?
@@assassin01620 Nevermind. I was wrong.
vʼ(1-vʼ0)+vʼ(1+vʼ0)≠vʼ(1•0) 2≠0 !!!
8:55 "Testing y=0" !!!
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
I got a questions: doesn’t √(x - √y)² equal to |x - √y|?
He corrected himself I think
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.
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
It's an extraneous solution. Extraneous solutions are also why you should test the (2,3) solution.
this is a good point. i watched both videos and thats what dr pk did in the solution
this is a good point. i watched both videos and thats what dr pk did in the solution
8:55 "Testing y=0" !!! 😁
@@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
Sir we can also put 0 as a value of y instead of putting three
Is it a solution to the equation?
That would give us two extraneous solutions (±1; 0) for which the LHS is non-zero but the RHS is zero.
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.
Can you please factor this polynomial: -2x²+6y²+xy+8x-2y-8 ? I can't factor it.
(-x+2y+2)(2x+3y-4) according to wolfram alpha.
Only God we can call , only god
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
But if i take y=0 i have x^2= 4/(4-0)-> 4/4->1
X^2=1->x=1. Is this possible?
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.
✓(x - ✓y) + ✓(x + ✓y) = ✓(xy)
[✓(x - ✓y) + ✓(x + ✓y)]^2 = [✓(xy)]^2
2x + 2✓(x^2 - y) = xy
xy - 2x = 2✓(x^2 - y)
(xy - 2x)^2 = [2✓(x^2 - y)]^2
(xy)^2 + 4x^2 - 4yx^2 = 4x^2 - 4y
(xy)^2 - 4yx^2 + 4y = 0
if y = 0, x = 0
if y ≠ 0
yx^2 - 4x^2 + 4 = 0
(y - 4)x^2 + 4 = 0
x^2 = 4/(4 - y)
y = 3, x = 2 (x cannot be -2)
(x, y) = (0, 0), (2, 3)
x² = 4 / (4-y) 》{1,0} but not correct. Why?
Because you must substitute these values in the original equation and see whether it holds true or not.
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.
@@niloneto1608 thank you.
What is this
²3⁴
2 superpower 3 times 4
Funny notation indeed 😊
Also !3! - is it a factorial of a subfactorial or a subfactorial of a factorial? 🤔
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).
And also the order is 2,3,4
8 second ago no way!!!
Why y=3 not y=2 or y=1?
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
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.
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?
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.
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.
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".
x=y=0
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
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.
@@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.
@@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).
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.
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).
This at the end is a joke I don’t get or a fragment from a Bible without any context?
idk why he's preaching, it's not like easter or anything
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) 👇
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?
Dr PK Math didn't seem to have any tricks up his sleeve. He did what you did but faster (and sloppier)
Did almost the same but Dr PK method was more analytic. PN method was more algebraic. Better cool with that mouth
Dude…. You’re religious?
huh?
@@erenshawcheck out the last second of the video
Yeah, and what? Let him profess some parts of the Bible, it's not the main topic of his videos.
Doesn’t bother me. I was just responding to someone who was confused.
@@icetruckthrilla And I was responding the op who asked if Prime Newtons was religious
I didnt liked dr pk math video😅
What a born hater with hater gene. I liked both actually as an ardent viewer of both channels