Question:
(x2 +1)2 – x2 = 0 has
(a) four real roots
(b) two real roots
(c) no real roots
(d) one real root
Solution:
(c) Given equation is $\left(x^{2}+1\right)^{2}-x^{2}=0$
$\Rightarrow \quad x^{4}+1+2 x^{2}-x^{2}=0 \quad\left[\because(a+b)^{2}=a^{2}+b^{2}+2 a b\right]$
$\Rightarrow \quad x^{4}+x^{2}+1=0$
Let $x^{2}=y$
$\therefore \quad\left(x^{2}\right)^{2}+x^{2}+1=0$
$y^{2}+y+1=0$
On comparing with $a y^{2}+b y+c=0$, we get
$a=1, b=1$ and $c=1$
Discriminant, $D=b^{2}-4 a c$
$=(1)^{2}-4(1)(1)$
$=1-4=-3$
Since, $D<0$
$\therefore y^{2}+y+1=0$ i.e., $x^{4}+x^{2}+1=0$ or $\left(x^{2}+1\right)^{2}-x^{2}=0$ has no real roots.