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.
I tried the second method, because I noticed the product over sum pattern. The galaxy brain move of inverting and looking for 1/x instead did not occur to me. Facepalm! I was so close. :)
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.
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 + √7)^-1]^-1=[(√5-√3+√7-√5)/2]^-1=(√7+√3)/2
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)
Original expression is of the type ab/(a+b). Final answer is (√7 + √3)/2
reckoning 5 as (√5)², that's everything
conjugate or carnagugate(a terror).
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.
Competition type problems always have magical numbers. Some people call it ‘contrived’
@@SyberMath That's fair enough. It is what it is.
I tried the second method, because I noticed the product over sum pattern. The galaxy brain move of inverting and looking for 1/x instead did not occur to me. Facepalm! I was so close. :)
Thanks for sharing!
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.
Wow! Pretty good
Nice problem
Thanks
2/(√7-√3)=2(√7+√3)/4=(√7+√3)/2
👍
You said 4methods
I also got the answer with the third method
Oops. Sorry. I fixed it
Cool - although you only showed three methods.
Uh-oh! Did I forget the fourth one? 🤪
@@SyberMath yea u did the first, then the third, and the second. Maybe u could post the last one in a separate video?
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 + √7)^-1]^-1=[(√5-√3+√7-√5)/2]^-1=(√7+√3)/2
That’s one of the methods 🧐
@@SyberMath um I just skipped most part and didn't see it.