A Radical Math Problem Solved with EASE!

[@RashmiRay-c1y]

Note that 1/2[47+21√5] = 1/4[94 + 2(7)(3√5)] = 1/4[7^2 +(3√5)^2 + 2(7)(3√5)] = [(7+3√5)/2]^2. Again, 1/2[7+3√5] = 1/4[14 +6√5] = 1/4[3^2+(√5)^2 + 2(3)(√5)]= [(3+√5)/2]^2. Further, 1/2(3+√5) = 1/4(6+2√5) = 1/4[(√5)^2 + 1^2 +2(√5)(1)] = [(√5+1)/2]^2. Thus, 1/4[94 + 2(7)(3√5)] = [ {((√5+1)/2)^2}^2]^2 = [(√5+1)/2]^8. So, E = 1/2(√5+1).

[@faladodososomeme6739]

How long this take u to write??

[@RyanLewis-Johnson-wq6xs]

Surd[((47+21Sqrt[5])/2),8]=0.5+0.5Sqrt[5]=(1+Sqrt[5])/2 It’s in my head.

[@RealQinnMalloryu4]

{x^8+x^8 ➖ }47x+47x ➖ +{21x+21x ➖ }+10={x^16+94x^2}+42x^2+10={94x^18+42x^2}+10={136x^18+10}=146x^18/2=73x^18 3^4^3x^2^9 1^2^2^3x^2^3^2 1^1^1^1x^2^3^1x2^3(x ➖ 3x+2).

[@ManojkantSamal]

*= read as square root
^=read as to the power
Ans =(1+*5)/2........May be
As per question
{47+(21.*5)/2}^(1/8)=?
Let R^8={47+(21.*5)}/2
Now explain
{47+(21.*5)}/2
Multiply 2, So
{94+(42.*5)}/4
={7^2+(3.*5)^2+(2×7×3.*5)}/4
=[{7+(3.*5)}/2]^2 eqn1
Again explain
{7 +(3.*5)}/2
Multiply 2
{14+(6.*5)}/4
=3^2+(*5)^2+(2×3×*5)}/4
={3+*5}^2/2^2
={(3+*5)/2}^2.......eqn2
Again explain (3+*5)/2
Multiply 2
{6+(2.*5)}/4
={1^2+(*5)^2+(2×1×*5)}/4
=(1+*5)^2/2^2
={(1 *5)/2}^2...... Eqn3
Compairing eqn1, eqn2 &eqn3 we shall get the following
{(1+*5)/2}^8={47 +(21×*5)}/2
So,
R^8={(1+*5)/2}^8
Hence R=(1+*5)/2

[@潘博宇-k4l]

E=[3+(5)^(1/2)]/2

[@Shobhamaths]

((47+21√5) /2)^1/8
((94+42√5) /4) ^1/2) 1/4
((7+√45) /4) ^1/4
((14+2√45) /8) ^1/2) ^1/2)
((3+√5) /8) ^1/2
((6+2√5) /16) 1/2
(√5+1) /2👍

[@Fjfurufjdfjd]

Εχω [47+21(5)^(1/2)]/2=2×[47+21(5)^(1/2)]/4=[94+42(5)^(1/2)]/4=[7^2+(3×5^(1/2) +2×21×(5)^(1/2)]/4=[7+3(5)^(1/2)]^2/4. Αρα Ε=[[7+3(5)^(1/2)]^2]/4=[[7+3(5)^(1/2)]/2]^(1/4) = [[14+6(5)^(1/2)]/4]^(1/4) = [9+(ριζα5)^2+6(ριζα5)]/4]^(1/4)=[[3+(5)^(1/2)]/2]^(1/2)=[[6+2(5)^(1/2)]2]^(1/2)=[[(5)^((1/2)+1]/2 δηλαδη πολλαπλασιαζω και διαιρω με το 2 καθε φορα. Ε=[(5)^(1/2)+1]/2