Visualizing Extraneous Solutions
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- Опубліковано 21 жов 2024
- What are extraneous solutions, and why do they sometimes pop up, seemingly at random, when solving equations involving radicals? This video explains a way to visualize that algebra and understand exactly what's going on.
Correction: 2:05 This should read sqrt(1) = +/- 1, not sqrt(-1)
Correction: 2:42 A RHS of -x-1 gives no solutions at all, and for two extraneous solutions, we'd need something like (-1/2)x - 1
In germany, when doing "algebraic manipulations", we call them Äquivalenzumformungen, we explicitly write down the '⟺' sign. The "if and only if"-sign.
But when squaring both sides, the equations do not say the same thing anymore. They're not equivalent. That's why this operation of squaring both sides is only '⟹' and we have to check our solutions. But, hell yes, do I love this visual!
Germans are always ahead
Great job on your first youtube video! This video helps a lot in developing an intuition for a commonly confusing part of mathematics and helps the viewer aquire a curiosity for other parts of mathematics. The animation is smooth and nice to look at creating a good viewing experience. Achieving both of these qualities in the one video is impressive, especially for a first video. It is my pleasure to be the first person to comment on your channel, as i believe you have a great potential to become a major educational youtuber. :)
Looks like written from AI.
Thanks for demystifying the mystery of extraneous solutions.I hope high school teachers watch this video so they may explain it better to their students.
Now I understand why extraneous solutions pop up in equations with sqrt.
Bro this video, after watching the last one some time ago, is boggling.
Subscribed, and I hope you make many, MANY more like this.
There seems to be an error at 2:06 , because there shouldn't be a negative sign under the square root else the right hand side should say ±i
yeah it should say √1
Oof, you're right - silly typo on my part!
@@jaredbitz what you should of wrote is (±1)^2=1, or sqrt(1)=±1
@@erikhaag4250should have*
This is great, especially for a first video! I look forward to seeing more.
When manipulating equations, people are used to "iff and only" types of connections. a=b if and only if a-1=b-1. a=b if and only if 2a=2b. It is the case that any time we apply an operation to both sides of an equality, and we get another true equality. That is a primary property of equality (it comes from the substitution property). But all of the early examples in our math learning also create this two-way connection, so when we get to raising both sides to an even exponent -- an operation which does not do so, we are caught off guard. a=b only if a^2=b^2, but it is not necessarily the case that a=b if a^2=b^2.
Any manipulation we apply to an equation can be reframed as applying a function to both sides. If that function is not injective, then we do not get that two-way connection, and extraneous solutions can appear.
Exactly. The difference between ⇔ and ⇒.
a = b ⇔ 2a = 2b, meaning that for whatever values of A and B, if either equality is true then both are true and if either is false then both are false. This reads "a = b is equivalent to 2a = 2b" or "a = b if and only if 2a = 2b".
However, it is not true that a = b ⇔ a² = b². It is true that if a = b, then it has to be the case that a² = b², but that a² = b² doesn't necessarily mean that a = b. For example, it is true that (-2)² = 2², yet -2 ≠ 2.
In this case, we write a = b ⇒ a² = b² and we read "if a = b then a² = b²" or "a =b implies a² = b²".
Final (*funny, complicated) note, observe that is it true that a = b ⇒ 2a = 2b. But it is also true that 2a = 2b ⇒ a = b. Indeed, if statement p implies statement q AND statement q implies statement p. When that happens, when two statements imply each other, then both statements are equivalent, and the other way around, if two statements are equivalent, they imply each other, which means two statements imply each other is equivalent to both statements being equivalent, or (p ⇒ q ∧ p ⇒ q) ⇔ (p ⇔ q). Don't you love propositional logic?
This is the more general idea yes
@@adb012 I always interpreted the two-way arrow as saying you can go both ways.
x-1 = 0
You can go from this to x = 1
But because it's two way, you can also go back to x-1 = 0
You can't do this with squaring
x = 2
You can go to x² = 4
But because it's one-way, you can't go back. Notice that when we try to go back we don't apply a function, we say √(x²) = ±√(4). The plus-minus part makes it not a function
@@xXJ4FARGAMERXx Exactly!
x-1=0 ⇔ x=1 because x-1=0 ⇒ x=1 AND x=1 ⇒ x-1.
But while x=2 ⇒ x²=4, it is NOT ture that x²=4 ⇒ x=2, hence it is NOT true that x=2 ⇔ x²=4
i was here when you had 202 subs. just commenting for when you get famous 💙
Nice video! I stumbled across your channel due to the algorithm and I'm glad I clicked on it. I hope your channel will grow, and I'm sure it will if you continue uploading such quality original content!
Personally, I skipped past the 2min of the video, when you narrated the solving of the original equation. Given my high school math background, I wasn't going to watch 3min of it. You could consider just flashing the algebra for a while if you think your audience can grasp it quickly enough. Depends on your target audience.
That said, I really enjoyed the graphical visualisations and I really learnt something new! Thank you
Very lucid explanation, on par with 3blue1brown's. You just earned a subscriber, keep it up!!
Same!
Another way to think about it is when you square both sides you're effectively expanding the domain of the funtion, which would explain the extraneous roots.
This is a great video! I'm rewatching it in the morning because this was my "fall asleep to math" video and I passed out a minute in and I'm still curious about extraneous solutions
Actually, there is a way to solve those types of equations without plugging back the roots to the original equations for checking which one is extraneous. For example, consider the equation: sqrt(2x-1)=x-5. Since when we are using only the sqrt symbol, we are talking about the positive square root, hence sqrt( of any function f(x)) is always greater than or equal to 0. So sqrt(2x-1)>=0. The equation states that sqrt(2x-1)=x-5, because the left side of the equation>=0, so the right side of it must also be >=0. So x-5>=0 which means x>=5. As long as the roots fulfill the condition x>=5, they will be the actual roots of the original equation. Furthermore, since sqrt(2x-1)=x-5 => 2x-1=(x-5)^2. (x-5)^2 always >=0, which means 2x-1>=0, so you don't have to worry about the condition 2x-1>=0 anymore. To sum up, similar to when solving equations containing x in the denominator you have to ensure that the denominator is different than 0, or when solving any equations containing f(x) inside square roots you have to ensure f(x)>=0, when solving equations like sqrt(f(x))=g(x), you have to ensure that g(x)>=0
Hi, i just subscribed to your channel. Your explanations are very clear! May i know where are you animating/creating your graphs? This is really helpful for me (i am currently a stem teacher undergraduate). Thanks!
I use a tool called manim (www.manim.community/) to create the animations using Python code, then kdenlive (a free video editor) to stitch them together.
I also studied math education in university - good luck with your degree!
For more complex shapes, my graphical intuition says that the maximum number of intersections between the graph of a function (that has a nonzero second derivative) and a straight line should be two, plus the number of inflection points that the function has. For each new intersection point beyond the two that you can always get between any curved line and a straight line, the curved line has to bend back towards the straight line, and that requires an inflection point where the curvature changes sign.
Very beautifull and awesome.this lesson has tought me some thing that i never understood before.YOU ARE GREAT TEACHER.THANKS ALOT.
If you have the bottom line -(3/4)x - 3/4, you will have only 1 point touching the bottom curve, so basically it's the same but you'll have 1 solution one solving algebraically, and that solution will be extraneous. (When solving you'll only get one solution because the ±√(b²-4ac) part of the quadratic formula will become 0 so it just goes away)
Continue making videos about maths and computer science, they're very good :)
This video is wonderful, but that -1 typo will confuse students. Is there any way you can fix it and reupload? I really want to share this. Very well done.
Just wanted to know if anyone of you has noticed tiny dots converging towards the blue dot and the point 2 on the x-axis of the straight line in the thumbnail everytime you slide the thumbnail upwards or downwards
Muy interesante el vídeo. Siempre me había preguntado qué rayos con las soluciones extrañas. Supongo que es algo similar para el resto de ella. Me pregunto cómo se vería el caso de las ecuaciones racionales con soluciones extrañas
Nice job!. Great video! Keep it up .
Great video! Keep it up 👏🏻
So the solutions exist in the (terribly named) imaginary world, which, through... Euler conversion*?... creates the space (complex plane {stylized C}) where you can see these solutions. You've also displayed them simply in the real plane by handjamming in the completed equation (terrible phrasing), it's just the equation that exists if we don't want y to be a function of x. I had a question here and I think it was: is this "view" just a different way of looking at the complex plane? And is the relationship that polar Euler conversion thing that I mentioned before? Awesome content, by the way.
* - let me know if that's not a thing
These extraneous solutions are just when the values meet at the negative square root rather than the positive one. No need for ℂomplex numbers. ℂomplex numbers only get involved when looking at the equations with seemingly no solutions, extraneous or otherwise.
@@angeldude101 appreciate you!!
At 2:23. The square root of 1 negative 1 is never ±1. It's i the imaginary number. I had to pause there and say, "What?". The square root of +1 is +/- 1. Small error but it is enough to confuse a student and distract from the longer lesson.
Great video
Is there some way to avoid these extraneous solutions without exiting the world or algebra and many using modulus or something and not squaring the both sides?
Any time you do an operation that could introduce extraneous solutions, simply write down the domain restrictions that it ignores.
For instance, if you are squaring both sides to get rid of the square root on sqrt(x+3), write down x+3 >=0, to remind yourself that if you get a value inconsistent with this constraint, that it is extraneous. Or, if you multiply by a (x-1) to clear a denominator, write x not equals 1. Then when you get your solutions at the end, reject all of them that don't fit these constraints.
Squaring both sides of an equation is an illegal step. It's necessary for solving the equation but is illegal. This illicit step converts the original equation to a non-equivalent one. Fortunately, the solution's set of the original equation is included in the solution's set of the further equation.
Nice job!😊
Beautiful
I think these appear because branches of the square root like yea squaring both sides does a weird sometimes
-x=x
x²=x²
1:06
2:18 uh wait what
I, for one, ALWAYS mean the positive and negative roots, even if complex.
What? Are you saying you interpret the principal square root sign to mean both square roots? Because that’s very unconventional if so.
@@tamarleigh That's what I was taught. I always thought that's what everyone did. Unless the directions specifically say to give only one root, all roots are to be given. Not just ±sqrt, but also all 3 roots of cube root, all 4 of forth root, etc.
If you ignore the neg branch, you may miss a valid positive real solution and you've missed the chance to learn something deeper about math.
Obviously, if I was working on a real problem, like woodworking or such, I would not use negative lengths, complex areas, or 4D shapes, but for "paper" math, yes, I would find all valid answers.
@@sdspivey But there's a big difference between encountering the principal square root symbol and taking a square root. That symbol is *not* merely the square root symbol, it's the *principal* square root symbol, which means, the positive square root symbol. If you look up "square root" on Wikipedia it's clearly explained in the first paragraph that this symbol is the PRINCIPAL square root symbol.
On the other hand, if you come along and TAKE the sqare root of something, you DO need to account for the positive and negative roots. That's not the same thing as the principal square root.
This is why if you graph y=(principal square root symbol) x, you will get values only in the first quadrant, whereas if you graph y^2=x, you will get a horizontal parabola.
new veritasium
Unusueful and too complex ! Just write 2 conditions ( in Real and not Complex world ....:) 1/ Expression under radical must be positive and 2/ The radical itself positive too ! Just exclude the solution(s....) which do not respect these 2 conditions ! ....