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.