The roots of the equation `(x^2+1)^2-x^2=0` are

Options

A. four real roots
B. two real roots
C. no real roots
D. one real root

Answer

C

Explanation

Given equation is `(x^(2)+1)^(2)-x^(2)=0`
`implies x^(4)+1+2x^(2)-x^(2)=0 [ :' (a+b)^(2)=a^(2)+b^(2)+2ab]`
`impliesx^(4)+x^(2)+1=0`
Let `x^(2)=y`
`:. (x^(2))^(2)+x^(2)+1=0`
`y^(2)+y+1=0`
on comparing with `ay^(2)+by+c=0` we get
`a=1,b=1` and `c=1`
Discriminant `D=b^(2)-4ac`
`=(1)^(2)-4(1)(1)`
`=1-4=-3`
Since `Dlt0`
`:. y^(2)+y+1=0` i.e. `x^(4)+x^(2)+1=0` or `(x^(2)+1)^(2)-x^(2)=0` has no real roots.