Amazing Math Tricks to Ace Radical Algebra Exam!

[@akhildhawan5728]

8sqrt(13)-16

[@ManojkantSamal]

Respected Sir, Good evening

[@Usershima-c4v]

(11√3-3)/8

[@Quest3669]

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

[@offrampc]

(19 sqrt(13) - 61)/2

[@cjvs1604]

~578/45

[@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

[@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]