The number of roots of the equation 1x+1(1−x2)=3512 is

Let 1x=u and 1(1−x2)=v then u+v=3512andu2+v2=u2v2 ⇒ (u+v)2=(3512)2 ⇒ u2+v2+2uv=(3512)2 ⇒ u2v2+2uv=(3512)2 [∵u2+v2=u2v2] ⇒ u2v2+2uv−(3512)2=0 ⇒ (uv+4912)(uv−2512)=0 uv=−4912,uv=2512Case I: If uv=−4912 then     1x⋅11−x2=−4912x4−x2+(12)2(49)2=0x=−(5+73)14Case II: If uv=2512 then         1x⋅1(1−x2)=2512 [herex>0]x4−x2+(12)2(25)2=0(x2−925)(x2−1625)=0⇒x=35,45On combining both cases,x=−(5+73)14,35,45Hence, number of roots = 3