How do I simplify radical equations with nested radicals?

Agent_J
Messages
13
Reaction score
0
Simplify
sqrt(11+sqrt72)) + sqrt(11-sqrt(72))

better picture here
http://members.rogers.com/agentj/images/math.jpg

I don't know where to begin for this one, but apparently my calculator says the answer is 6 :redface:
 
Last edited by a moderator:
Mathematics news on Phys.org
let x be the surd, square x, what do you get?
 
let x be the "surd"?
 
ok, set sqrt(11+sqrt72)) + sqrt(11-sqrt(72)) = x
(by the way, sqrt(72) = 6*sqrt(2), which i'll just call 6r2 for simplicity)
square both sides, we get
(sqrt(11+sqrt72)) + sqrt(11-sqrt(72)))^2 = x^2
simplify and you get
11+6r2+2*sqrt((11+6r2)(11-6r2))+11-6r2=x^2
22+2sqrt(121+11*6r2-11*6r2-36*2)=x^2
22+2*sqrt(121-72)=x^2
22+2*sqrt(49)=x^2
22+2*7=x^2
36=x^2
x=6 (well, plus or minus, but we know it must be positive since the addition of two non-complex roots must be >= 0)
 
Yes, surd: an expression involving radicals. I just didn't want to have to type it out. Just square the expression, simplify and take the square root, et voila, we have 6, with the appropriate choice of sign.
 
Insights auto threads is broken atm, so I'm manually creating these for new Insight articles. In Dirac’s Principles of Quantum Mechanics published in 1930 he introduced a “convenient notation” he referred to as a “delta function” which he treated as a continuum analog to the discrete Kronecker delta. The Kronecker delta is simply the indexed components of the identity operator in matrix algebra Source: https://www.physicsforums.com/insights/what-exactly-is-diracs-delta-function/ by...
Fermat's Last Theorem has long been one of the most famous mathematical problems, and is now one of the most famous theorems. It simply states that the equation $$ a^n+b^n=c^n $$ has no solutions with positive integers if ##n>2.## It was named after Pierre de Fermat (1607-1665). The problem itself stems from the book Arithmetica by Diophantus of Alexandria. It gained popularity because Fermat noted in his copy "Cubum autem in duos cubos, aut quadratoquadratum in duos quadratoquadratos, et...
I'm interested to know whether the equation $$1 = 2 - \frac{1}{2 - \frac{1}{2 - \cdots}}$$ is true or not. It can be shown easily that if the continued fraction converges, it cannot converge to anything else than 1. It seems that if the continued fraction converges, the convergence is very slow. The apparent slowness of the convergence makes it difficult to estimate the presence of true convergence numerically. At the moment I don't know whether this converges or not.
Back
Top