I solved it without the benefit of the formula, but effectively using the same method. I let E = √(5√3 + 6√2) and F = √(5√3 - 6√2). Then EF = √[(5√3)² - (6√2)²] = √(75 - 72) = √3. And E² + F² = 5√3 + 5√3 = 10√3. From these two equations, in turn we have (E+F)² = E² + F² + 2EF = 10√3 + 2√3 = 12√3 = 4√27 and (E-F)² = E² + F² - 2EF = 10√3 - 2√3 = 8√3 = 4√12 Taking square roots: (E+F) = 2√√27 and (E-F) =2√√12 Adding, and dividing by 2: E = √√27 + √√12
Mine was a very intuitive . Here it goes :- Sqrt{5•sqrt(3) + 6•sqrt(2)} = sqrt{sqrt(75) + sqrt(72)} ---------[because 5(sqrt3) = sqrt(75) and 6(sqrt2) = sqrt(72) ] We take sqrt(3) common inside the root and get Sqrt[sqrt(3)*{sqrt(25) + sqrt(24)}] It can be simplified to : Sqrt{sqrt(3)*(5+2(sqrt6))} 5+(2*sqrt(6)) = {sqrt(2) + sqrt(3)}² So , now we get : Sqrt{sqrt(3) * {sqrt(2) + sqrt(3)}² We can take the square out and get {Sqrt(2) + sqrt(3)}* {3^(1/4)} [Because : sqrt(sqrt3) = 3^(¼) Now it is very simple to solve : (2^½ + 3^½)3^¼ = ( 2^(2/4) + 3^(2/4)) * 3^¼ = (4^¼ + 9^¼)3¼ = {(4•3)^¼} + {(9•3)^¼} = (12^¼) + (27^¼) Pretty simple steps in my opinion.
Thanks for the interesting video. But I have a question : assuming sqrt(x+y) = sqrt(a)+sqrt(b) is understandable. But the second one sqrt(x-y) = sqrt(a)-sqrt(b) is harder to figure out (for me). It has to be independent of the first so we can solve for a and b. How does it come out of the binomial theorem without using the first equation? thanks in advance.
In Italy, secondary school teachers are used to teach the formula sqrt(sqrt(a) +- sqrt(b)) = sqrt( (sqrt(a) + sqrt(a-b)) /2)+- sqrt( (sqrt(a) - sqrt(a-b)) /2) rising the two sides to second power. Of course, you need a >=b>=0. I really appreciate your work. Thank you.
@@murkotron I always watch videos and comment as a way to interact with @SyberMath. but you, who have nothing to do with it, come to meddle and disrespect my comment: the typical boring!
Set the radical equal to a Divide both sides by 4th root of 3 You now have root(5 + 2root(6)) which can be simplified into root(3) + root(2), and then multiply both sides again to get a = 4th root(3)(root(3)+root(2))
5Root(3) -> Root(75), 6Root(2) -> Root(72) Both radical values are similar to each other, considering that they both sit between "64" and "81" in the graph of y=root(x).
Hello everyone! This was an idea that I had for a while but SILVIA TOTARO also commented about it...🥰 You can see the other video and the comment here: ua-cam.com/video/mnbuurVQARI/v-deo.html
Sir You remind me of my calculus teachers when I lived in England. He did what you do and made me hate math. He always thought everyone knew what he was talking about. You really need to slow down and explain what you are doing.
@Syber Maths sir i am a ten standard student preparing for ioqm of india so sir can u plz tell me which topics of geometry and number theory i need to master also sir plz make a vdo on how can we switch various ideas in brain while solving.high level tricky problems
Hi Imran! Thank you for the question. I'm not familiar with IOQM. I would just grab a book for prep and study through it. For both number theory and geometry, you need to master the basics. Find an easy book to get started
@@SyberMath How do you know x minus y always gives radical a minus radical b..you say it's easy to prove but if you didn't knkw that beforehand why would anyone think of that..do you know what I mean? Why not justbise the addition x plus y^1/2 formula by itself to solve?
speaking of formulas for radical expressions, any way to contact you via email? i'd like to share the formula i found years ago, for estimating value of square roots , i mean formula + proof
I denested it to 27^(1/4) + 12^(1/4)
I used the identity
sqrt(a + b) = (S + b / S) / sqrt(2)
where S = sqrt(sqrt(a^2 - b^2) + a)
Derivation of a standart formula was a nice piece of work, thanks.
Glad you liked it!
I solved it without the benefit of the formula, but effectively using the same method.
I let E = √(5√3 + 6√2) and F = √(5√3 - 6√2).
Then EF = √[(5√3)² - (6√2)²] = √(75 - 72) = √3.
And E² + F² = 5√3 + 5√3 = 10√3.
From these two equations, in turn we have
(E+F)² = E² + F² + 2EF = 10√3 + 2√3 = 12√3 = 4√27
and (E-F)² = E² + F² - 2EF = 10√3 - 2√3 = 8√3 = 4√12
Taking square roots:
(E+F) = 2√√27
and (E-F) =2√√12
Adding, and dividing by 2:
E = √√27 + √√12
Nice!
Mine was a very intuitive . Here it goes :-
Sqrt{5•sqrt(3) + 6•sqrt(2)}
= sqrt{sqrt(75) + sqrt(72)} ---------[because 5(sqrt3) = sqrt(75) and 6(sqrt2) = sqrt(72) ]
We take sqrt(3) common inside the root and get
Sqrt[sqrt(3)*{sqrt(25) + sqrt(24)}]
It can be simplified to :
Sqrt{sqrt(3)*(5+2(sqrt6))}
5+(2*sqrt(6)) = {sqrt(2) + sqrt(3)}²
So , now we get :
Sqrt{sqrt(3) * {sqrt(2) + sqrt(3)}²
We can take the square out and get
{Sqrt(2) + sqrt(3)}* {3^(1/4)}
[Because : sqrt(sqrt3) = 3^(¼)
Now it is very simple to solve :
(2^½ + 3^½)3^¼
= ( 2^(2/4) + 3^(2/4)) * 3^¼
= (4^¼ + 9^¼)3¼
= {(4•3)^¼} + {(9•3)^¼}
= (12^¼) + (27^¼)
Pretty simple steps in my opinion.
Thanks for the interesting video. But I have a question : assuming sqrt(x+y) = sqrt(a)+sqrt(b) is understandable. But the second one sqrt(x-y) = sqrt(a)-sqrt(b) is harder to figure out (for me). It has to be independent of the first so we can solve for a and b. How does it come out of the binomial theorem without using the first equation? thanks in advance.
In Italy, secondary school teachers are used to teach the formula sqrt(sqrt(a) +- sqrt(b)) = sqrt( (sqrt(a) + sqrt(a-b)) /2)+- sqrt( (sqrt(a) - sqrt(a-b)) /2) rising the two sides to second power. Of course, you need a >=b>=0.
I really appreciate your work. Thank you.
Wow...
That's difficult!
I'll see again later, for my ultimate learning.
You can do it!
How's that difficult? It was literally brought on a plate...
@@murkotron
I always watch videos and comment as a way to interact with @SyberMath. but you, who have nothing to do with it, come to meddle and disrespect my comment: the typical boring!
@@notlin1976 wow sorry to bother you mr. respectable comments, please put away your sword
@@murkotron ok... you're forgiven! 😁😁😁
Set the radical equal to a
Divide both sides by 4th root of 3
You now have root(5 + 2root(6)) which can be simplified into root(3) + root(2), and then multiply both sides again to get a = 4th root(3)(root(3)+root(2))
√3 was common so I took it out. I was left with (5+2√6) which is equal to (√2+√3)²
So I think the answer should be (√2+√3)×3^(1/4)
yes i always use that method and it's always the quickest and easiest.
5Root(3) -> Root(75), 6Root(2) -> Root(72)
Both radical values are similar to each other, considering that they both sit between "64" and "81" in the graph of y=root(x).
√√3 ( √3 + √2)
Hello everyone! This was an idea that I had for a while but SILVIA TOTARO also commented about it...🥰
You can see the other video and the comment here:
ua-cam.com/video/mnbuurVQARI/v-deo.html
after the video, I realize 5 sqrt(3) + 6 sqrt (2)= sqrt(3)* ( 5 + 2 sqrt(6)), then the problem is pretty straightforward
Thank you for this lesson
You are welcome!
If √3=a and √2=b, the given expression E = [a(a^2+b^2) + 2 a^2b]^1/2 = a^1/2 [a^2+b^2+2ab]^1/2 = a^1/2 (a+b) = 3^1/4(√3 + √2).
Excellent!, I'd change only sqrt(a + b)=sqrt(x) + sqrt(y) and sqrt(a - b)= sqrt(x) - sqrt(y).
Hello syber, I was wonder what program are you using. like write and stuff like that
Hi!
I use an iPad, an Apple Pencil and the Notability App.
I record the videos using screen recording.
Very cool formula, thanks for that!
Glad you liked it!
Thanks!
That's the way. I like it
There is a mistake at 6:11! (5-1)/2=4 and not 2
Sir You remind me of my calculus teachers when I lived in England. He did what you do and made me hate math. He always thought everyone knew what he was talking about. You really need to slow down and explain what you are doing.
I prefer the defenestrated form !
😲😂
sir i took ✓3 common inside the radical and then got ( ✓2 +✓3)^2 which i took out of the radical and finally my ans is (✓2+✓3)×✓(✓3)
really cool, man
Thanks!
It is wrong since 6√2 not equal to 2 √6
Denesting a Radical Using a Formula: sqrt(5sqrt3 + 6sqrt2) = ?
5sqrt3 + 6sqrt2 = (sqrt3)(5 + 2sqrt6) = (sqrt3)[3 + 2(sqrt3)(sqrt2) + 2]
= (sqrt3)[(sqrt3)^2 + 2(sqrt3)(sqrt2) + (sqrt2)^2] = (sqrt3)[(sqrt3 + sqrt2)^2]
sqrt(5sqrt3 + 6sqrt2) = sqrt{(sqrt3)[(sqrt3 + sqrt2)^2]} = ± [3^(1/4)](sqrt3 + sqrt2)
= ± [3^(1/4)][3^(1/2)+ 2^(1/2)] = ± {3^(3/4) + [3^(1/4)][4^(1/4)]}
= ± [27^(1/4) + 12^(1/4)]
@Syber Maths sir i am a ten standard student preparing for ioqm of india so sir can u plz tell me which topics of geometry and number theory i need to master also sir plz make a vdo on how can we switch various ideas in brain while solving.high level tricky problems
Hi Imran! Thank you for the question. I'm not familiar with IOQM. I would just grab a book for prep and study through it. For both number theory and geometry, you need to master the basics. Find an easy book to get started
We learned this formula at school, it's from Lagrange.
👍
I can solve it in a different way
nice^^😀
Bro ru turkish? cuz u pronounce turkish names pretty accurate
You are right
Please write in English.
@@robertveith6383 I did.?
Hii 😍
Hello!
@@SyberMath How do you know x minus y always gives radical a minus radical b..you say it's easy to prove but if you didn't knkw that beforehand why would anyone think of that..do you know what I mean? Why not justbise the addition x plus y^1/2 formula by itself to solve?
😢😢
αα
Hello teacher
Hi MAKARA!
@@SyberMath ☺☺
speaking of formulas for radical expressions, any way to contact you via email? i'd like to share the formula i found years ago, for estimating value of square roots , i mean formula + proof
SyberMathBiz at gmail dot com
Thanks!
So many new members recently!
No problem! Thank you, Roh4n! 🥰🥰