Question Number 113913 by Aina Samuel Temidayo last updated on 16/Sep/20

Find the square root of (√(50))+(√(48))

Answered by 1549442205PVT last updated on 16/Sep/20

We have ((√(50))+(√(48)))^2 =98+2(√(50.48))  =98+2(√(25.16.6))=98+40(√6)  Hence (√((√(50))+(√(48))))= ^4 (√(((√(50))+(√(48)))^2 ))  = ^4 (√(98+40(√6)))= ^4 (√((a(√2)+b(√3))^4 )) (1)  We need find a,b so that  98+40(√6)=(a(√2)+b(√3))^4   =4a^4 +8a^3 (√2).b(√3)+6.2a^2 .3b^2 +4.3b^3 (√3).a(√2)  +9b^4 =(4a^4 +9b^4 +36a^2 b^2 +(8a^3 b+12ab^3 )(√6)  ⇔ { ((4a^4 +36a^2 b^2 +9b^4 =98)),((8a^3 b+12ab^3 =40)) :}  ⇔ { ((4a^4 +36a^2 b^2 +9b^4 =98)),((2a^3 b+3ab^3 =10(2))) :}  ⇔10(4a^4 +36a^2 b^2 +9b^4 )=98(2a^3 b+3ab^3 )  Put a=kb substituting into above  equality then divide two sides by k^4   we get 40k^4 +360k^2 +90=196k^3 +294k  ⇔20k^4 −98k^3 +180k^2 −147k+45=0  This equation has one root k=1  ⇒a=b.Replace into (2)we get  5a^4 =10⇔a= ^4 (√2).Replace into (1)  we have (√((√(50))+(√(48))))= ^4 (√(98+40(√6)))  = ^4 (√([ ^4 (√2)((√2)+(√3))]^4 )) = ^4 (√2) ((√2)+(√3))   Thus, (√((√(50))+(√(48))))= ^4 (√2) ((√2)+(√3))

Commented byAina Samuel Temidayo last updated on 16/Sep/20

Thanks.

Answered by mr W last updated on 16/Sep/20

(√(50))+(√(48))  =(√2)(5+2(√6))  =(√2)[((√2))^2 +2(√2)(√3)+((√3))^2 ]  =(√2)[(√2)+(√3)]^2   ⇒(√((√(50))+(√(48))))=(2)^(1/4) ((√2)+(√3))

Answered by Dwaipayan Shikari last updated on 16/Sep/20

Generally (√x)+(√y) =(√(a+(√b)))  x+y+2(√(xy))     =a+(√b)  (x+y)=a  4xy=b  (x−y)=(√(a^2 −b))  (√x)+(√y) =((√(a+(√(a^2 −b))))/( (√2)))+(√((a−(√(a^2 −b)))/2))                 =(√(((√(50))+(√2))/2))+(√((((√(50))−(√2))/2) ))       a=(√(50))    b=48

Commented byDwaipayan Shikari last updated on 16/Sep/20

(√((√(49))+(√(48)))) =(√(((√(49))+1)/2)) +(√(((√(49))−1)/2)) =2+(√3) =(√(8/2)) +(√(6/2))  (√((√(64))+(√(63)))) =(√((8+1)/2)) +(√(((8−1)/2) )) =(√(9/2))+(√(7/2))  (√((√(100))+(√(99)))) =(√(((11)/2) )) +(√(9/2))  (√((√(81))+(√(80))))  =(√((10)/2)) +(√(9/2))  Have you noticed any pattern?  (√1) =(√((2/2) ))+(√(0/2))  (√((√4)+(√3)))  =(√(3/2)) +(√(1/2))  (√((√9)+(√8))) =(√(4/2))+(√(2/2))  (√((√(16))+(√(15))))=(√(5/2)) +(√(3/2))  (√((√(25))+(√(24)))) =(√(6/2)) +(√(4/2))  (√((√(36))+(√(35))))=(√(7/2))+(√(5/2))  (√((√n^2 )+(√(n^2 −1))))   =(√((n+1)/2)) +(√((n−1)/2))  (√(n+(√(n^2 −1)))) =(√((n+1)/2)) +(√((n−1)/2))

Commented byRasheed.Sindhi last updated on 16/Sep/20

Actual learning from your  general method! Thank you.