Solving a Radical Parametric Equation

[@utkarshdubey8046]

Your channel is not famous like many others. But, I need to admit that it is the best.

[@SyberMath]

Wow, thank you!

[@MelomaniacEarth]

hi utkarsh remember me?

[@utkarshdubey8046]

No, can you please tell who you are? Sorry, I can't remember.

[@MelomaniacEarth]

@@utkarshdubey8046 no i just remind we watched a premeir together once......forget😅

[@dhruvladdha4789]

@@MelomaniacEarth bruh

[@sekarganesan]

The problems you solve are quite unique...less intimidating and so is the solution. Keep it coming.

[@SyberMath]

Thank you!

[@RamAurelius]

Your video ideas are absolutely radical, keep it up 😎

[@SyberMath]

Thanks, will do!

[@manojsurya1005]

The problems u choose are good👍 and pls do differential equation or integration soon

[@antoine5571]

You deserve more followers, thanks for all the problems you share :)

[@SyberMath]

You're welcome! I appreciate that!

[@shekhar9750]

Very well solution with nice explanation

[@SyberMath]

Thank you!

[@sahilsinghbhandari444]

We can solve this equation by transforming graphs , taking y = a2 , and y = 2|x|-x2

[@angelmendez-rivera351]

The three cases to consider are a < 0, a = 0, a > 0. If a < 0, then there are no solutions, due to the definition of the square root function. If a = 0, then sqrt(2·|x| - x^2) = 0, which is equivalent to 2·|x| - |x|^2 = a^2, because x^2 = |x|^2. This gives the solution set {-2, 0, 2}. If a > 0, then by the definition of the square root, 2·|x| - |x|^2 = a^2, with 2·|x| - |x|^2 > 0. This is equivalent to |x|^2 - 2·|x| = -a^2 & |x|^2 - 2·|x| < 0, which is equivalent to |x|^2 - 2·|x| + 1 = 1 - a^2 & |x|^2 - 2·|x| + 1 < 1. Notice that |x|^2 - 2·|x| + 1 = (|x| - 1)^2, so (|x| - 1)^2 = 1 - a^2 and (|x| - 1)^2 < 1. Notice that if a > 1, then 1 - a^2 < 0, and (|x| - 1)^2 = 1 - a^2 has no solutions. (|x| - 1)^2 < 1 is equivalent to -1 < |x| - 1 < 1, which is equivalent to 0 < |x| < 2. If a = 1, then (|x| - 1)^2 = 0, which is equivalent to |x| = 1, which gives solutions {-1, 1}. If 0 < a < 1, then (|x| - 1)^2 = 1 - a^2, which is equivalent to |x| - 1 = sqrt(1 - a^2) or |x| - 1 = -sqrt(1 - a^2), equivalent to |x| = 1 + sqrt(1 - a^2) or |x| = 1 - sqrt(1 - a^2), which always satisfy 0 < |x| < 2 for 0 < a < 1. This results in the solutions {-1 - sqrt(1 - a^2), 1 - sqrt(1 - a^2), -1 + sqrt(1 - a^2), 1 + sqrt(1 - a^2)}. As it happens, the same set is obtained when a = 0 or a = 1, only that the solutions collapse to have multiplicity greater than 1. Therefore, in summary, the solution set is {} if a < 0 or a > 1, {-1 - sqrt(1 - a^2), 1 - sqrt(1 - a^2), -1 + sqrt(1 - a^2), 1 + sqrt(1 - a^2)} otherwise.

[@farhatali2221]

Awesome explanation

[@SyberMath]

Glad you think so!

[@JefiKnight]

It is interesting that there are complex solutions when a

[@Ramesh-it4kk]

Thank u very nice please make videos of math like this only

[@Qermaq]

Only? That's like eating chocolate cake every day ;)

[@SyberMath]

I totally agree! Vegetables before dessert! 😁

[@mchowdhury7708]

Keep it. Awesome job. You are overpowered.

[@SyberMath]

Thank you! Will do!

[@Bareq_Raad]

Hi, I have questions from where did you learn math? how did you make your learning journey easier? how to deep understand math? Where to start solving and training to become advanced and where to start ? how to look at the problems? how to do everything thing you do in the question to simplify it and turn the form of it even if it looks nothing can be manipulated (how to transform the question using methods)?

[@SyberMath]

Great questions! I majored in math and got my teaching credential. Since my focus was more on teaching not on the subject itself, I haven't taken any courses in Number Theory or Linear Algebra while I was in college. Most of my courses were based on pedagogy, teaching strategies, techniques, so on and so forth. As a student, I haven't participated in any math competitions because my grades were not high enough for the selection criteria. When I graduated, I got a job teaching high school math and started working very hard to keep up with the students. Problem solving was my daily routine. The more I did, the more I enjoyed it. At some point, I started teaching competition math and trained students for various math competitions. Some of my students have been very successful and were admitted to ivy league colleges. That was my first encounter with Number Theory, Divisibility, Fermat's Little Theorem, Euler's Theorem, CRT etc. I have been fascinated by the beauty of Number Theory which is expressed in the words of Gauss as "Mathematics is the queen of the sciences-and number theory is the queen of mathematics." Years after I graduated, I took "Advanced Linear Algebra" without taking the prerequisite course "Linear Algebra" and hated it. During these years, I continued to work with gifted students at different levels and enjoyed the process of teaching and at the same time learning from my students. My goal has always been making the math understandable and enjoyable because I 💗 math, needless to say, and I would like everyone that I work with to feel the same way! I fell in love with solving and writing challenging math problems, not to say that I'm a great problem solver. I just love doing it and learning from the process! Sorry for the long answer but this is my journey in a nutshell! (seriously! 😂) Math is hard and requires lots of lots of patience and perseverance. The journey is painful but definitely worth the effort! Start somewhere and build your skills. I hope this answers your question. Feel free to write more if it doesn't or if you have any other questions and again thanks for the great questions!

[@Bareq_Raad]

@@SyberMath Thank you

[@SyberMath]

@@Bareq_Raad You're welcome!

[@berzerksharma]

nice video. loved it

[@SyberMath]

Glad you liked it

[@Qermaq]

This is really childish, but if you graph a = root(2|x| - x^2) it kinda looks like a body part. :D In "reality", the real answers seem to form perfect semicircles on the a+ side of the x axis, having radii 1 and centers at {-1,0} and {1,0}.

[@SyberMath]

😁

[@mohamedomrane5481]

For this equation : think to check that solution must be in (-2,2) interval.

[@SyberMath]

Why?

[@mcbeaulieu]

@@SyberMath before even seeing the graph or watching the video, I can tell that 2|x| - x^2 >= 0 for the (-2, 2) interval. A quick non-integer example is x = -2.1, as it would make the radicant equal to 4.2 - 4.41 = -0.21. So that limits the domain to (-2,2) 😁

[@farhatali2221]

Can you please share which software you are using to write in yellow Font while explaining.. it s beautiful

[@SyberMath]

Thanks! Sure I use notability

[@WolfgangKais2]

Why use the strict inequality 0 < a < 1 ?

[@SyberMath]

I don't know. Why not? 😜

[@WolfgangKais2]

@@SyberMath …because 0

[@sahilsinghbhandari444]

a can only take value from 0 to 1

[@SyberMath]

Very good!

[@namunamu3424]

شرح رائع جداً واكرر طلبي ياريت لو تضع ترجمة الى اللغة العربية

[@SyberMath]

Thank you!

[@thfchris]

If this is in the exam, generally the condition of a would be specified on the question otherwise it should be scored at high marks, means that the candidate should be aware of specifying all the conditions of a to give all the answers.

[@SyberMath]

That's right! This is an equation with parameters so one should consider all possibilities

[@comingshoon2717]

why 0

[@SyberMath]

Sure. We can do that

[@blahblah6725]

I did not understand why a has to be greater than or equal to zero.

[@SimchaWaldman]

02:46 Why must you work so hard?! Just complete the square.

[@SyberMath]

How come I never thought about that? 😁

[@akvmaths]

👍🏻

[@comingshoon2717]

porque mis profesores no lo explicaban así de simple ? :(

[@SyberMath]

No idea!

[@mariomestre7490]

Genial

[@SyberMath]

Thanks!

[@tonyhaddad1394]

Niceee !!!! I solved wiiii !!!!

[@SyberMath]

You're good, Tony!!! 😊

[@tonyhaddad1394]

@@SyberMath thank u im good when i learn from you and other great channels ❤❤

[@Qermaq]

@@tonyhaddad1394 I've gotten so much better with math just by watching this channel. And I've watched all the famous ones for years. This guy can explain this stuff, not watered-down, in a language that I can understand. Seems there are a lot of people like us.

[@tonyhaddad1394]

@@Qermaq i completly agree with you 100% ❤❤

[@marienbad2]

Absolute cool! (> 0 lol)

[@SyberMath]

Thanks!

[@agnibeshbasu3089]

Well, this was an easy one i guess

[@SyberMath]

Was it?

[@agnibeshbasu3089]

@@SyberMath yeah i think so

[@agnibeshbasu3089]

@@SyberMath by the way, you can make a telegram channel for mathematical discussions. How is this idea?

[@chhomck5668]

Hello guy where are you from? I like video

[@SyberMath]

Thanks! I'm from Planet Earth. 😁

[@chhomck5668]

@@SyberMath what country are you from?

[@ManjulaMathew-wb3zn]

Actually a can be in between -1 and 0 and have real solutions. Take another look at your very first statement”no real solution if a is negative”. I really admire your math knowledge. I only wanted to point out the honest mistake.