The answer is 1/2[sqrt(14)+sqrt(10)]. This is an example of a number theory adjacent radical simplification. One that requires knowing what a semiprime really is. I shall find similar and use that in order to *become* a mathematician!!!
Na década de 80, eu me deparava com esse tipo de questão em provas militares, não sabia resolver. Eu não tinha onde pesquisar , hoje, encontramos fácil na internet soluções de questões difíceis. A Internet é um prato cheio de conhecimento para quem quer aprender.
The point would be to remove the nested root. Much like the point of rationalising the denominator is to move a root from the bottom to the top of a fraction. It’s easier to deal with in further calculations.
Why are you using implication symbols (=>) between expressions? Each expression is exactly equal to the one before, so denote this with the equality symbol (=).
The answer is 1/2[sqrt(14)+sqrt(10)]. This is an example of a number theory adjacent radical simplification. One that requires knowing what a semiprime really is. I shall find similar and use that in order to *become* a mathematician!!!
Brilliant solution!👍👍👍
Есть более понятное и лёгкое решение.Могу научить блогера.
Observing that sqrt35 = sqrt7*sqrt5 = 2*sqrt(7/2)*sqrt(5/2). Then seeing that 6 = 12/2 = [sqrt(7/2)]^2 + [sqrt(5/2)]^2 = 7/2 + 5/2 = 12/2 = 6.
Thus sqrt(6 + sqrt35) = sqrt[7/2 + 5/2 + sqrt7*sqrt5] = sqrt{[sqrt(7/2)]^2 + [sqrt(5/2)]^2 + 2*sqrt(7/2)*sqrt(5/2)} = sqrt{[sqrt(7/2) + sqrt(5/2)]^2} = sqrt(7/2) + sqrt(5/2).
Mas o menos como cortarte un brazo para que no te duela un dedo.
Na década de 80, eu me deparava com esse tipo de questão em provas militares, não sabia resolver. Eu não tinha onde pesquisar , hoje, encontramos fácil na internet soluções de questões difíceis.
A Internet é um prato cheio de conhecimento para quem quer aprender.
Neat transformations, but they don't give anything because there are still two roots to calculate. So what's the point of all this?
The point would be to remove the nested root. Much like the point of rationalising the denominator is to move a root from the bottom to the top of a fraction. It’s easier to deal with in further calculations.
Why are you using implication symbols (=>) between expressions? Each expression is exactly equal to the one before, so denote this with the equality symbol (=).
let sqrt(6+sqrt 35) = x, sqrt(6-sqrt35) = y, then xy=1, x^2 + y^2 = 12, (x+y)^2 =x^2+y^2 +2xy = 12 + 2 =14 , so x+y=sqrt14 (x, y>0)
so, y=sqrt14-x ....xy=x(sqrt14-x)=1 -->x^2-(sqrt14)x +1 =0 --> x=(sqrt14 +sqrt10)/2 ( x>0)
@@문병인-r8u Perfeito!
You have a notation error at 6:48 - the lower line should be SQRT(2) x SQRT(2), not SQRT (2 x 2).
Negativo , o vídeo está certo.
Those are the same