I don't know , I didn't solve it like that . it has other possible values more than 9 & 4 x=(7-y)^2 x=11-sqrt y (7-y)^2= 11-sqrt y sqrt y = 11-(7-y)^2 Let sqrt y = u u=11-(7-u^2)^2 you will get a quartic equation : u^4-14u^2+u+38=0 (u-2)(u^3+2u^2-10u-19)=0 u=2 , therefor y=4 , x= 9 (u^3+2u^2-10u-19)=0 (u+1.8)(u^2+0.2u-10.36)=0 u= -1.8 , can't be accepted as a solution u^2+0.2u-10.36=0 u=3.13 , therefor y= 9.79 , x = 7.8
Good to see you bro after a long time
y=4,x=9..
Right bro
Yes, we can do this by assuming the no.s from the very beginning as they are all natural numbers only these would satisfy given conditions
I don't know , I didn't solve it like that . it has other possible values more than 9 & 4
x=(7-y)^2
x=11-sqrt y
(7-y)^2= 11-sqrt y
sqrt y = 11-(7-y)^2
Let sqrt y = u
u=11-(7-u^2)^2
you will get a quartic equation :
u^4-14u^2+u+38=0
(u-2)(u^3+2u^2-10u-19)=0
u=2 , therefor y=4 , x= 9
(u^3+2u^2-10u-19)=0
(u+1.8)(u^2+0.2u-10.36)=0
u= -1.8 , can't be accepted as a solution
u^2+0.2u-10.36=0
u=3.13 , therefor y= 9.79 , x = 7.8
Last 2 values are not satisfying sqrt(x) + y = 7.
By the way well done.