Amazing Math Tricks to Ace Radical Algebra Exam!

[@akhildhawan5728]

8sqrt(13)-16

[@ManojkantSamal]

Respected Sir, Good evening

[@Quest3669]

Crux.... X= (√13-1)/2 ;
x^6= 77-20√13
? = (X^6-1)/X-1= 8√13-(16) soln.

[@Usershima-c4v]

(11√3-3)/8

[@cjvs1604]

~578/45

[@offrampc]

(19 sqrt(13) - 61)/2

[@Fjfurufjdfjd]

Μετα απο πραξεις βρισκω χ=[(13)^(1/2)-1]/2. Προκειται για γεωμετρικη προοδο με λ=χ=[(13)^(1/2)-1]/2>1.α_1=1 ν=6.αρα Ε=S_ν=α_1×(λ^ν -1)/(λ-1).E=(χ^6-1)/χ-1). χ^6=[(χ)^2]^3....Ε=8[(13)^(1/2)-2]

[@ManojkantSamal]

^=read as to the power
*=read as square root
Let's explain RHS
N=7+(7.*7)+(7.*13)+(*91)
={7+(7.*7)}+{*13(7+*7)}
=({*7(7+*7)}+{*13(7+*7)}
=(7+*7)(*7+*13)
Let (7+*7)=R, (*7+*13)=A
So,
N=RA
Now explain D
D=42+(3.*7)+(3.*91)
=3{14+*7+*91}
=3[7+*7+7+*91]
=3[(7+*7)+*7(*7+*13)]
=3{R+(*7A)}
Inverse of N/D=D/N
D/N=3{R+(*7A)}/RA
SO,
D/3N={R+(*7A)}/RA
={R/RA}+(*7A)/RA}
=(1/A)+(*7/R)
Now explain (1//A)
1/A)=1/(*7+*13)
=(*13-*7)/{(*13+*7)(13-*7)}
=(*13-*7)/(13-7)
=(*13-*7)/6
Again explain (*7/R)
(*7/R)=*7/(7+*7)
={*7(7-*7)/{(7+*7)(7-*7)}
=(7.*7-7)/(49-7)
=7(*7-1)/42
=(*7-1)/6
So,
D/3N={(*13-*7)/6}+{(*7-1)/6}
={*13-*7+*7-1}/6
=(*13-1)/6
LHS=(1/X)
Inverse =X
So,
(X/3)=(*13-1)/6
X=(*13-1)/2
X^2=(13+1- 2.*13)/4
=(14-2.*13)/4
=2(7-*13)/4
=(7-*13)/2
X^4={49+13-(14.*13}/4
={62-14.*13)/4
=2(31- 7.*13)/4
=(31-7.*13)/2
1+x=1+{(*13-1)/2}
=(2+*13-1)/2
=(1+*13)/2
Now the question is
1+x+x^2+x^3+x^4+x^5
=(1+x)+x^2(1+x)+x^4(1+x)
=(1+x){1 x^2+x^4}}
Now,
(1+x^2+x^4)
=1+{(*7-13)/2}+{(31-7.*13)/2}
=(2+7-*13+31-7.*13)/2
=(40-8.*13)/2
=2(20-4.*13)/2
=20- 4.*13
Now
(1+x)(1+x^2+X^4)
={(*13+1)/2}×(20- 4.*13)
=(20.*13-52+20-4.*13)/2
=(16.*13-32)/2
=2(8.*13 - 16)/2
=8.*13 - 16