Simplifying a Radical Expression in Three Ways

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PLAYLISTS 🎵 :

[MrLidless]

Multiplying numerator and denominator by √(2+√3) immediately gets the numerator to 1. Playing and simplifying the denominator thereafter is fairly trivial to get to √6.

[quantumobject]

I love how different the solutions are

[daddykhalil]

The first and third methods are long and complicated.
The second one is simple, easy and straightforward.
Thank you

[amoledzeppelin]

Ok, I found a fourth way: using the fact that (sqrt(3)-1)^2 = 4 - 2sqrt(3) = 2(2-sqrt(3)), we eliminate 1-sqrt(3) from both parts and are left with 1/(sqrt(3)*sqrt(2)), which is equal to sqrt(6)/6.

[subramaniankrishnaswami]

There is another method. Evaluate the numerator equating it to a-sqrt(b) and squaring both sides and solve for a and b. Take the positive value and it is (sqrt(3)-1)/sqrt(2).Use this value and compute the given expression. you get the answer as 1/sqrt(6)

[josephsilver]

All three methods are both unusual and unique. Very nice solutions.

[imonkalyanbarua]

What i really liked about your solution is that you chose the most obvious method for the last and started with the creative one first! Amazing! 👏👏👏

Another way that could have been done was by bringing the denominator inside the root and then solving it. 😇🙏

[ИльхамАбдуллаев-ьй]

Very nice and beautiful
solution .You are great .Bravo 👍👍👍👍👍

[佐藤広-cp]

Some people give the solution as 1/√6, but this should be √6/6. In circumstances where you have to calculate approximate numbers by hand, 2.44948.../6 is much easier than calculating 1/2.44948... . If it is a machine calculation, 1/√6 seems to be easier.

[Jha-s-kitchen]

Nice approaches!
Never saw using the white color though!

shorter method Simply square it and take 3 common from denominator and cancel 2 - root 3

[MrLeith]

Hello there and thanks for this nice expression. We can also solving it by noticing that 2-sqrt(3) = (sqrt(3/2) -sqrt(1/2))^2

[OrDoL]

What app are you using for the writing?

[walterwen]

Simplify a Radical Expression: sqrt(2 – sqrt3)/(3 – sqrt3) = ?
(sqrt2)[sqrt(2 – sqrt3)] = sqrt(4 – 2sqrt3) = sqrt[(sqrt3 – 1)^2] = ± (sqrt3 – 1)
3 – sqrt3 = (sqrt3)(sqrt3 – 1) > 0; sqrt(2 – sqrt3) = (sqrt3 – 1)/sqrt2
sqrt(2 – sqrt3)/(3 – sqrt3) = [(sqrt3 – 1)/sqrt2]/[(sqrt3)(sqrt3 – 1)] = (sqrt6)/6