√x+y=11, √y+x=7

Question

Find the value of x and y

Answer 1

root(x)+y=11... x+root(y)=7... x=4 & y=9... root(x)+y=2+9=11... x+root(y)=4+3=7

Answer 2

Taking FROM SECOND eqn; root(x)=a,and root(y)=b. b=7-a^2 . Substitute value of b in first eqn We get x=a^2 and y=b^2 Now, a+{(7-a^2)}^2=11. {Vanishing Method . =>a^4-14a^2+a+49=11. factor} a+b^2 = 11_______(i) =>a^4-14a^2+a+38=0 b+a^2 = 7________(II). =>a^4-2a^3+2a^3-4a^2-10a^2+20a-19a+38=0 Solve these now its easy. =>a^3(a-2)+2a^2(a-2)-10a(a-2)-19(a- =>(a^3+2a^2-10a-19)(a-2)=0. * BY SAFAL DAS BISWAS (CLASS X)(S.A.V.M).(CHINSURAH)] *=>a=2, Thus a^2= x =2x2 = 4 and. b+4=7, or b=3 Thus b^2= y = 3x3 = 9. x=4 ; y=9 } Answer

Answer 3

root(x)+y=11                                                   root(y)+x=7

or,{root(x)-2}+y-9=0                                    or,{root(y)-3}+x-4=0

So,{root(x)-2}+y-9={root(y)-3}+x-4=0

Now,x-root(x)-2=y-root(y)-6=0         (cause they are factorable , either not possible)

so,factoring, x-root(x)-2 ={root(x)-2}{root(x)+1}=0 , Or, root(x)=2 as root(x)=-1 is absurd here

again,factoring, y-root(y)-6={root(y)-3}{root(y)+2}=0 , Or, root(y)=3 as root(y)=-2 is again absurd here

Thus,root(x)=2 Or,x=4 And root(y)=3 Or,y=9       [New Method Crack By Safal Das Biswas Savm Chinsurah]