I puzzled over this for a while and didn't see your neat solution at first. But I finally found a variation on it. The expression with root(2) and root(3) suggested trig to me, so I drew a right triangle with root(3) on the hypotenuse, 1 on the horizontal and root(2) on the vertical. Call the angle t. Then x = root(2sec(t)+2) - root(sec(t) - tan(t)). Then square x: x^2 = 2sec(t) + 2 + sec(t) - tan(t) - 2 root( 2 sec^2(t) - 2 sec(t) tan(t) + 2 sec(t) - 2 tan(t) ) The stuff inside the root is a perfect square of (sec(t) - tan(t) + 1), which is easy to prove using tan^2(t) + 1 = sec^2(t) so x^2 simplifies to sec(t) + tan(t) = root(3) + root(2) //.
Conjugate? More like conjuGREAT! I just had your channel recommended to me by a friend and I am so glad that he did so. I’m planning to become a regular, and I am sure I will share your channel in the future as well. Thanks so much for making and posting all of your videos!
When i solved this problem i used the first method, but the second method was honestly very instructive, i didn't know you that technique, thank you for teaching me
For this kind of problems there is the complex radical formula: sqrt(a+-sqrt(b))=sqrt((a+sqrt(a^2-b))/2)+-sqrt((a-sqrt(a^2-b))/2) So sqrt(sqrt(3)-sqrt(2))=sqrt((sqrt(3)+1)/2)-sqrt((sqrt(3)-1)/2) And sqrt(2*sqrt(3)+2)=2*sqrt((sqrt(3)+1)/2). So the difference is simply with another sign: sqrt((sqrt(3)+1)/2)+sqrt((sqrt(3)-1)/2)=sqrt(sqrt(3)+sqrt(2)).
I think both do the methods are elegant ! I mean obviously second method was simpler but the interesting thing in the first method was noticing that the expression under the radio was in fact a perfect square ! I think it’s better to learn all methods cause you can have more creativity and solve bigger problem that can only be solved in one way Thank you sybermath math :D
Excellent problem. Since we can make it in the form of sqrt(3)/2 and 1/sqrt(2) after taking some terms common( which can be considered in the form of sin and cos), I tried to approach the problem using Trigonometry; however, I couldn't go far. Should these types of problems be tried using trigonometry?
I call the first root x, second y, x-y=z x=y+z (y+z)^2=2root3+2 y^2+2yz+z^2=2root3+2 y^2=root3-root2 I also get yz=1 by using 5-2root6=(root3-root2)^2 finally z^2=root3+root2 the key to solve this problem is to know what is 5-2root6 as you mentioned in the beginning, but it was quite obvious for me.
Finding what's under that long square root is just impossible if you're not a computer. Also how would you know to do what you did in the second method?
As I said in the video, I know the 2nd method will work because that's how I came up with this problem. Then thinking about an alternative approach brings me to the 1st method which I think is better but I agree, it's not easy to see!
I have reached the result ( ( 3^0.5+ 1)^0.5 + (3^0.5 - 1)^0.5) /(2^0.5) Squaring and simplifying and then take a square root , giving (3^0.5 + 2^0.5 )^0.5
Wow. I gave up this problem. Finding that sqrt(3)-sqrt(2)+1 under the radical is pure genius... Second method is a little cheating though. I mean, you just did something so far from the original problem that it might be considered guessing and checking. Of course unless you saw this working in your mind. Ramanujan style. : )
Hmm what a rooty expression. Also adway kumar if your reading this,I'm terribly sorry for making u feel bad. But you shouldn't have made me look dumb,so let's both just apologize and forgive each other. Also keep up the content Syber.
Haha, Could hear your smile and satisfaction near the end of the second method. You wrinkled a lot of brains with this one!
😁
Write 2√3 +2 as (√3+√2)+(√3-√2)+2 which is equal to the square of [✓{√3+√2} +✓{√3-√2}]
Hi, can I ask a question? At 8:37 you said that -√2 +2√2 equal to -√2. Isn’t that supposed to be +√2. Am I missing something here?
Correct
You're not missing anything he just wrote it wrong.
He corrected himself later tho
Thank you so much for all the reply, I used this video to learn mathematics so I thought I missed a step or something.
Watch a little more
I puzzled over this for a while and didn't see your neat solution at first. But I finally found a variation on it. The expression with root(2) and root(3) suggested trig to me, so I drew a right triangle with root(3) on the hypotenuse, 1 on the horizontal and root(2) on the vertical. Call the angle t. Then x = root(2sec(t)+2) - root(sec(t) - tan(t)). Then square x:
x^2 = 2sec(t) + 2 + sec(t) - tan(t) - 2 root( 2 sec^2(t) - 2 sec(t) tan(t) + 2 sec(t) - 2 tan(t) )
The stuff inside the root is a perfect square of (sec(t) - tan(t) + 1), which is easy to prove using tan^2(t) + 1 = sec^2(t) so x^2 simplifies to sec(t) + tan(t) = root(3) + root(2) //.
Very nice!
Nice video!
Just for supplement, the discussion would be more detailed if you discuss the radical expression is either positive or negative.
Good point! Thanks
Conjugate? More like conjuGREAT! I just had your channel recommended to me by a friend and I am so glad that he did so. I’m planning to become a regular, and I am sure I will share your channel in the future as well. Thanks so much for making and posting all of your videos!
Welcome aboard! 💖😊
@@SyberMath I'm glad to be here, and thank you for having me!
When i solved this problem i used the first method, but the second method was honestly very instructive, i didn't know you that technique, thank you for teaching me
For this kind of problems there is the complex radical formula:
sqrt(a+-sqrt(b))=sqrt((a+sqrt(a^2-b))/2)+-sqrt((a-sqrt(a^2-b))/2)
So sqrt(sqrt(3)-sqrt(2))=sqrt((sqrt(3)+1)/2)-sqrt((sqrt(3)-1)/2)
And sqrt(2*sqrt(3)+2)=2*sqrt((sqrt(3)+1)/2).
So the difference is simply with another sign: sqrt((sqrt(3)+1)/2)+sqrt((sqrt(3)-1)/2)=sqrt(sqrt(3)+sqrt(2)).
The expression 6 -2√6 + 2√3 -2√2 contains 2*√2, 2*√3, and 2*√2*√3,
whereas (√2)^2 + (√3)^2 + 1*1 = 6
Hereby, 6 -2√6 + 2√3 -2√2
= (1+√3 -√2) ^2
this is helpful to simplify this solution
Good point!
Why 8:50 X= √(√3-√2) ?
I think both do the methods are elegant ! I mean obviously second method was simpler but the interesting thing in the first method was noticing that the expression under the radio was in fact a perfect square ! I think it’s better to learn all methods cause you can have more creativity and solve bigger problem that can only be solved in one way
Thank you sybermath math :D
Absolutely!
U make it very simple to understand.we enjoy watching & learning your methods.Thank you.
Thank you! I'm glad to hear that! 💖
Excellent problem. Since we can make it in the form of sqrt(3)/2 and 1/sqrt(2) after taking some terms common( which can be considered in the form of sin and cos), I tried to approach the problem using Trigonometry; however, I couldn't go far. Should these types of problems be tried using trigonometry?
That's quite radical!! (i saw your comment on my channel hehe)
This is a pretty nice question since the end result should look nice
replying to my comment maybe im not spam to youtube anymore
@@MathElite haha lol don't worry now
I also take formula of a+b square and I done it is good radical simplification questions
Loved both methods.
The subject of Mathematics is simply the most beautiful subject ever. 😍😍😍😍
It definitely is!!! 🤩💖
@@SyberMath 🥰🥰🥰
I call the first root x, second y, x-y=z
x=y+z
(y+z)^2=2root3+2
y^2+2yz+z^2=2root3+2
y^2=root3-root2
I also get yz=1 by using 5-2root6=(root3-root2)^2
finally z^2=root3+root2
the key to solve this problem is to know what is 5-2root6 as you mentioned in the beginning, but it was quite obvious for me.
Rất thú vị bằng việc tính toán căn thức bằng hằng thức, và đặt ẩn phụ. Cảm ơn.
wow syber!
thanks for these joyful moments !
My pleasure!
@@SyberMath ❤️❤️❤️
Second solution - Brilliant!!
Loved the constructive beauty of the 2nd method.
Glad you liked it!!
2nd method is briliant.thanks alot
Most welcome 😊
Darn, I couldn't have a single clue to this!
very nice syber!!!
i will keep supporting you!!!!!!!
Thank you so much!
Finding what's under that long square root is just impossible if you're not a computer.
Also how would you know to do what you did in the second method?
As I said in the video, I know the 2nd method will work because that's how I came up with this problem. Then thinking about an alternative approach brings me to the 1st method which I think is better but I agree, it's not easy to see!
The first method is more intuitive, but the second method is more elegant. I'd say I like the 2nd one more.
nice solutions sir thanks
Most welcome
3:26 That explains a lot..
Oooh so interesting .Both the methods are praiseworthy . I salute you from the bottom of my heart , dear professor . God bless you .
It's my pleasure! Thank you! 💖
I liked both method , mann you re awesome !!!!
Glad to hear that! Thanks! 💖
Great problem and solution. I like how you de-nested the radical by turning it into a perfect square, twice!
Thank you! 😊
was hard but you mastered it bro
Oh, thank you, brother! 💖
The first method was really amazing!
Glad to hear that!
Holy smokes! That factorisation was insane!
😂
I have reached the result
( ( 3^0.5+ 1)^0.5 + (3^0.5 - 1)^0.5) /(2^0.5)
Squaring and simplifying and then take a square root , giving
(3^0.5 + 2^0.5 )^0.5
Is this a perfectly radical problem or what? Good one 👍
Thank you!
Love your videos 😍😍
Thank you! 💖
😃 wonderful solution!
Thank you, Roh4n! 😊
Perfect
Wow. I gave up this problem. Finding that sqrt(3)-sqrt(2)+1 under the radical is pure genius... Second method is a little cheating though. I mean, you just did something so far from the original problem that it might be considered guessing and checking. Of course unless you saw this working in your mind. Ramanujan style. : )
I know! The second one is reverse engineering sort of 😁
@@SyberMath Exactly 😂😂
This is magic
Hmm what a rooty expression.
Also adway kumar if your reading this,I'm terribly sorry for making u feel bad.
But you shouldn't have made me look dumb,so let's both just apologize and forgive each other.
Also keep up the content Syber.
I hope you guys can get along from now on
@@MathElite No worries we will try getting along.
Thank you!
Very clever
Thanks!
Good content again thank you :)
No problem. Thanks for watching! 😊
I liked the 2nd method better.
shouldn't the answer be sqrt(sqrt(3) + sqrt(2))
cause (-sqrt(2) + 2sqrt(2) = sqrt(2))
yes
My brain is buzzing
Sorry this was a little intense
@@SyberMath But the knowledge I gained is priceless, thank you so much
Error + not - :(, mechanical errors
Nice
Warning, on some of your videos, you've forgotten the subtitle (remember, I'm French...)
Hi
I don't know answer this time 😅😅😅
Hi 😲😁
Such a great radical change(Revolution) lol
can you tell the answer to the question you wrote in the live chat?
@@resilientcerebrum the answer is cosecθ-cotθ,you can check this out on math elite's channel also
👍
Lovely 🥰🥰🥰
😊😊😊
errou ai o sinal e raiz de 3 mais raiz de 2
eu consertei isso
who uses flag division here ?
What's flag division?
@@SyberMath It's a method in Vedic math. ua-cam.com/video/ykNSwSBe8Lw/v-deo.html
Hello I’m early :D
Sorry I'm late 😁
:3