I Simplified A Radical Expression | 3 Methods 😮

🤩 Hello everyone, I'm very excited to bring you a new channel (aplusbi)
Enjoy...and thank you for your support!!! 🧡🥰🎉🥳🧡

When you purchase something from here, I will make a small percentage of commission that helps me continue making videos for you.
If you are preparing for Math Competitions and Math Olympiads, then this is the page for you!
You can find ARML books and many others here. CHECK IT OUT!!! ❤️ ❤️ ❤️
❤️ [sqrt(15)+sqrt(21)+)sqrt(35)+5]/[sqrt(3)+sqrt(7)+2sqrt(5)]

If you need to post a picture of your solution or idea:
#radicals #radicalequations #algebra #calculus #differentialequations #polynomials #prealgebra #polynomialequations #numbertheory #diophantineequations #comparingnumbers #trigonometry #trigonometricequations #complexnumbers #math #mathcompetition #olympiad #matholympiad #mathematics #sybermath #aplusbi #shortsofsyber #iit #iitjee #iitjeepreparation #iitjeemaths #exponentialequations #exponents #exponential #exponent #systemsofequations #systems
#functionalequations #functions #function #maths #counting #sequencesandseries

#algebra #numbertheory #geometry #counting #mathcontests #mathcompetitions
via @YouTube @Apple @Desmos @NotabilityApp @googledocs @canva

PLAYLISTS 🎵 :

[msmbpc]

Original expression is of the type ab/(a+b). Final answer is (√7 + √3)/2

[coreyyanofsky]

i solved it by thinking for a bit about (√3 + √7 + 2√5)², noticing where the "extra" terms come from, and then thinking about (√3 + √7 + 2√5)(√3 + √7)

[XJWill]

There is a general method for rationalizing any denominator that involves
the minimal polynomial of the denominator. If the denominator is D and the
minimal polynomial of D is m(x) then the rationalizing factor is found from
f(x) = c/x - m(x) / x
where c is chosen to eliminate the k / x term from m(x) / x and the desired
rationalizing factor is f(D)
The minimal polynomial of a number may be found by equating the number
to x and then manipulating both sides of the equation to eliminate all
irrational values.
I won't list out the math, but for this problem
x = sqrt(3) + sqrt(7) + 2*sqrt(5)
and after manipulation the minimal polynomial is
m(x) = x^8 - 120 * x^6 + 3632 * x^4 - 28800 * x^2 + 256
then choose c = 256 and
f(x) = - x^7 + 120 * x^5 - 3632 * x^3 + 28800 * x
and the desired rationalizing factor is
f( sqrt(3) + sqrt(7) + 2*sqrt(5) )
= 64*sqrt(105) - 256*sqrt(7) + 64*sqrt(125) - 128*sqrt(27)
Of course, the common factor of 64 may be removed without affecting
the utility: sqrt(105) - 4*sqrt(7) + sqrt(125) - 2*sqrt(27)
For this problem, the general method results in more arithmetic than other
methods. But for other denominators, the general method can be helpful.
For example, how to rationalize the denominator for
1 / ( 1 + sqrt(2) + cbrt(3) )
This is left as an exercise for the reader.

[mtaur]

I feel like we took it for granted that the numbers were conveniently rigged up to make a hidden shortcut possible. In general, such a quotient might require a linear combination of a rational, sqrt(3), sqrt(5), sqrt(7), and any product of these. Q adjoin square roots of three primes has dimension 2^3 over Q.

In solution 1, you saw that the numbers were super convenient because you had four equations with three variables. In solution 2, you were able to factor by grouping, which was very convenient and magical-looking. It's not clear to me how you would know in advance that the numbers are magical.

[broytingaravsol]

reckoning 5 as (√5)², that's everything

[scottleung]

Cool - although you only showed three methods.

[xwyl]

Your method is too complicated.
The numerator can be factored to (√3 + √5)(√5 + √7), and the denominator (√3 + √5)+(√5 + √7), then the whole thing becomes [(√3 + √5)^-1+(√5 +