Prove the following

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.

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